Co-ordinate Geometry

In the equation 2x − 5 = 0, we need to find the ordinate, which means the y-coordinate of a point on the graph of this equation.

2x - 5 = 0

⇒ 2x = 5

⇒ x = $\frac{5}{2}$

⇒ x = 2.5

Notice that this equation only involves x and has no y-term. This tells us that the equation represents a vertical line parallel to y-axis at x = 2.5. Hence, ordinate = 0 (i.e., y = 0).

Hence, option 1 is the correct option.

Given:

$y = 3 + \frac{4x}{5}$

In the above equation, the value of y depends on the value of x, so y is said to be dependent variable and x is said to be independent variable.

Hence, option 2 is the correct option.

Since two ordered pairs are equal, their first components are equal and their second components are separately equal.

(2x - 1, y + 5) = (3, 10)

⇒ 2x - 1 = 3 and y + 5 = 10

⇒ 2x = 3 + 1 and y = 10 - 5

⇒ 2x = 4 and y = 5

⇒ x = $\frac{4}{2}$ and y = 5

⇒ x = 2 and y = 5

Hence, option 3 is the correct option.

The point (5, -7) has a positive x-coordinate and a negative y-coordinate. This places it in the 4th quadrant.

Hence, option 4 is the correct option.

Given:

3x - 2 = 7

⇒ 3x = 7 + 2

⇒ 3x = 9

⇒ x = $\frac{9}{3}$

⇒ x = 3

And, 8 - 3y = 2

⇒ 3y = 8 - 2

⇒ 3y = 6

⇒ y = $\frac{6}{3}$

⇒ y = 2

Therefore, the co-ordinates of the point = (3, 2).

Hence, option 1 is the correct option.

(i) $y = \frac{4}{3}x - 7$

In the above equation, the value of y depends on the value of x, so y is said to be dependent variable and x is said to be independent variable.

Hence, x = independent variable and y = dependent variable.

(ii) x = 9y + 4

In the above equation, the value of x depends on the value of y, so x is said to be dependent variable and y is said to be independent variable.

Hence, x = dependent and y = independent.

(iii) $x = \frac{5y + 3}{2}$

In the above equation, the value of x depends on the value of y, so x is said to be dependent variable and y is said to be independent variable.

Hence, x = dependent and y = independent.

(iv) $y = \frac{1}{7}(6x + 5)$

In the above equation, the value of y depends on the value of x, so y is said to be dependent variable and x is said to be independent variable.

Hence, x = independent and y = dependent.

Since two ordered pairs are equal, their first components are equal and their second components are separately equal.

(i) (x - 1, y + 3) = (4, 4)

⇒ x - 1 = 4 and y + 3 = 4

⇒ x = 4 + 1 and y = 4 - 3

⇒ x = 5 and y = 1

Hence, x = 5 and y = 1.

(ii) (3x + 1, 2y - 7) = (9, - 9)

⇒ 3x + 1 = 9 and 2y - 7 = -9

⇒ 3x = 9 - 1 and 2y = -9 + 7

⇒ 3x = 8 and 2y = -2

⇒ x = $\frac{8}{3}$ and y = -$\frac{2}{2}$

⇒ x = $2\frac{2}{3}$ and y = -1

Hence, x = $2\frac{2}{3}$ and y = -1.

(iii) (5x - 3y, y - 3x) = (4, - 4)

⇒ 5x - 3y = 4 ................(1)

And, y - 3x = - 4

⇒ y = -4 + 3x ................(2)

Putting the value of y in (1), we get

⇒ 5x - 3(-4 + 3x) = 4

⇒ 5x + 12 - 9x = 4

⇒ 5x - 9x = 4 - 12

⇒ - 4x = - 8

⇒ x = $\frac{8}{4}$

⇒ x = 2

Putting the value of x in (2),

y = - 4 + 3x

⇒ y = - 4 + 3 x 2

⇒ y = - 4 + 6

⇒ y = 2

Hence, x = 2 and y = 2.

(i) The abscissa is 2.

The abscissa represents the x-coordinate, so x = 2.

Observing the graph, the y-coordinate for this point is also 2.

Hence, the point is A = (2, 2).

(ii) The ordinate is 0.

The ordinate represents the y-coordinate, so y = 0.

Observing the graph, the x-coordinate for this point is 5.

Hence, the point is B = (5, 0).

(iii) The ordinate is 3.

The ordinate represents the y-coordinate, so y = 3.

Observing the graph, the x-coordinates for this point are -4 and 6.

Hence, the points are C = (-4, 3) and E = (6, 3).

(iv) The ordinate is - 4.

The ordinate represents the y-coordinate, so y = - 4.

Observing the graph, the x-coordinate for this point is 2.

Hence, the point is D = (2, -4).

(v) The abscissa is 5.

The abscissa represents the x-coordinate, so x = 5.

Observing the graph, the y-coordinates for this point are 0, 5 and -3.

Hence, the points are B = (5, 0), H = (5, 5) and G = (5, -3).

(vi) The abscissa is equal to the ordinate.

The abscissa represents the x-coordinate, and ordinate represents the y-coordinate so x = y.

Observing the graph, the points are (2,2), (5, 5) and (4, 4).

Hence, the points are A = (2, 2), H = (5, 5) and I = (4, 4).

(vii) The ordinate is half of the abscissa.

The abscissa represents the x-coordinate, and ordinate represents the y-coordinate so y = $\frac{\text{x}}{2}$.

Observing the graph, the point is (6, 3).

Hence, the point is E = (6, 3).

(i) False

Reason :

The ordinate is the y-co-ordinate of a point.

(ii) False

Reason :

The origin (0, 0) is the point where the x-axis and y-axis intersect.

(iii) True

Reason :

The y-axis is the vertical number line and the x-axis is the horizontal number line in the coordinate plane.

(iv) False

Reason :

Points that lie on the x-axis or y-axis are not located in any of the four quadrants.

(v) False

Reason :

If the ordinate (y-coordinate) equals the abscissa (x-coordinate), (x = y), the point lies either in the first quadrant (x > 0, y > 0) or the third quadrant (x < 0, y < 0).

(vi) True

Reason :

The origin (0, 0) lies on both the x-axis and y-axis, as it is the intersection point of the two axes.

(vii) False

Reason :

The point (a, b) lies on the y-axis if the x-coordinate a = 0, not b = 0.

(i) 3 - 2x = 7

⇒ 2x = 3 - 7

⇒ 2x = - 4

⇒ x = - $\frac{4}{2}$

⇒ x = - 2

2y + 1 = 10 - $2 \frac{1}{2}y$

⇒ 2y + $\frac{5}{2}y$ = 10 - 1

⇒ $\frac{4 + 5}{2}y$ = 9

⇒ $\frac{9}{2}y$ = 9

⇒ y = $\frac{2 \times 9}{9}$

⇒ y = 2

Hence, the co-ordinates of the point = (-2, 2).

(ii)

$$\begin{aligned}\frac{2a}{3} - 1 = \frac{a}{2}\\[1em] ⇒ \frac{2a}{3} - \frac{a}{2} = 1\\[1em] ⇒ \frac{4a - 3a}{6} = 1\\[1em] ⇒ \frac{a}{6} = 1\\[1em] ⇒ a = 6\\[1em]\end{aligned}$$

$$\begin{aligned}\frac{15 - 4b}{7} = \frac{2b - 1}{3}\\[1em] ⇒ 3(15 - 4b) = 7(2b - 1)\\[1em] ⇒ 45 - 12b = 14b - 7\\[1em] ⇒ 45 + 7 = 14b + 12b\\[1em] ⇒ 52 = 26b\\[1em] ⇒ b = \frac{52}{26}\\[1em] ⇒ b = 2\\[1em]\end{aligned}$$

Hence, the co-ordinates of the point = (6, 2).

(iii)

$$\begin{aligned}5x -(5 - x) = \frac{1}{2}(3 - x)\\[1em] ⇒ 5x - 5 + x = \frac{1}{2}(3 - x) \\[1em] ⇒ 6x - 5 = \frac{1}{2}(3 - x) \\[1em] ⇒ 2(6x - 5) = 3 - x\\[1em] ⇒ 12x + x = 3 + 10 \\[1em] ⇒ 13x = 13 \\[1em] ⇒ x = \frac{13}{13}\\[1em] ⇒ x = 1 \\[1em]\end{aligned}$$

$$\begin{aligned}4 - 3y = \frac{4 + y}{3}\\[1em] ⇒ 3(4 - 3y) = 4 + y\\[1em] ⇒ 12 - 9y = 4 + y\\[1em] ⇒ 12 - 4 = 9y + y\\[1em] ⇒ 8 = 10y\\[1em] ⇒ y = \frac{8}{10}\\[1em] ⇒ y = \frac{4}{5}\\[1em]\end{aligned}$$

Hence, the co-ordinates of the point = $\Big(1, \frac{4}{5}\Big)$.

(i) After plotting the given points A(2, 0), B(8, 0) and C(8, 4) on a graph paper, join A with B and B with C. From the graph, it is clear that the vertical distance between the points B(8, 0) and C(8, 4) is 4 units. Therefore, the vertical distance between points A(2, 0) and D must be 4 units.

Now, complete the rectangle ABCD and read the coordinates of point D. As shown on the graph, D = (2, 4).

Hence, D = (2, 4).

(ii) After plotting the given points A(4, 2), B(-2, 2) and D(4, -2) on a graph paper, join A with B and A with D.

From the graph it is clear that the vertical distance between the points A(4, 2) and D(4, -2) is 4 units and the horizontal distance between the points A(4, 2) and B(-2, 2) is 6 units. Therefore, the vertical distance between points B(-2, 2) and C must be 4 units and the horizontal distance between points B(-2, 2) and C must be 6 units.

Now, complete the rectangle ABCD and read the coordinates of point C. As shown on the graph, C = (-2, -2).

Hence, C = (-2, -2).

(iii) After plotting the given points A(-4, -6), C(6, 0) and D(-4, 0) on a graph paper, join A with D and C with D.

From the graph it is clear that the vertical distance between the points A(-4, -6) and D(-4, 0) is 6 units and the horizontal distance between the points C(6, 0) and D(-4, 0) is 10 units. Therefore, the vertical distance between the points A(-4, -6) and B must be 10 units and the horizontal distance between the points C(6, 0) and B must be 10 units.

Now complete the rectangle ABCD and read the coordinates of point B. As shown on the graph, B = (6, -6).

Hence, B = (6, -6).

(iv) After plotting the given points B (10, 4), C (0, 4) and D (0, - 2) on a graph paper, join B with C and C with D.

From the graph it is clear that the vertical distance between the points C(0, 4) and D(0, -2) is 6 units and the horizontal distance between the points C(0, 4) and D(0, -2) is 10 units. Therefore, the vertical distance between the points B(10, 4) and A must be 6 units and the horizontal distance between the points D(0, -2) and B must be 10 units.

Now complete the rectangle ABCD and read the coordinates of point A. As shown on the graph, A = (10, -2).

Hence, A = (10, -2).

Plot the points A (-2, 2), B (8, 2) and C (4, -4) on the graph paper. Join point A with B and B with C.

From the graph, it is clear that the horizontal distance between the points A (-2, 2) and B (8, 2) is 10 units and the vertical distance between the points B (8, 2) and C (4, -4) is 6 units. Therefore, the vertical distance between the points A (-2, 2) and D must be 6 units and the horizontal distance between the points C (4, -4) and D must be 10 units.

Now, complete the parallelogram ABCD and read the coordinates of point D. As shown on the graph, D = (-6, -4).

The midpoint of AB lies exactly halfway between A(-2, 2) and B(8, 2). On the graph, this midpoint is at (3, 2), as it is 5 units from both A and B.

The midpoint of CD lies exactly halfway between C (4, -4) and B (-6, -4). On the graph, this midpoint is at (-1, -4), as it is 5 units from both C and D.

Hence, D = (-6, -4) and mid point of AB = (3, 2) and CD = (-1, -4).

Plot the points A (- 2, 4), C (4, 10) and D (- 2, 10) on the graph paper. Join point A with D and D with C.

From the graph, it is clear that the horizontal distance between the points C (4, 10) and D (-2, 10) is 6 units and the vertical distance between the points A (-2, 4) and D (-2, 10) is 6 units. Therefore, the vertical distance between the points C (4, 10) and B must be 6 units and the horizontal distance between the points A (-2, 4) and B must be 6 units.

Now, complete the square ABCD and read the coordinates of point B, as shown on the graph, B = (4, 4).

The midpoint of BC lies exactly halfway between B(4, 4) and C(4, 10). On the graph, this midpoint is at E(4, 7), as it is 3 units from both B and C.

The midpoint of CD lies exactly halfway between C(4, 10) and D(-2, 10). On the graph, this midpoint is at F(1, 10), as it is 3 units from both C and D.

The coordinates of the midpoint of diagonals of the square is G = (1, 7).

Hence, the co-ordinates of the mid-point of BC = (4, 7), the co-ordinates of the mid-point of CD = (1, 10) and the co-ordinates of the point of intersection of the diagonals of the square ABCD = (1, 7).

Given:

y + 7 = 0

⇒ y = -7

The graph of y = -7 is a straight line that is parallel to the x-axis and at a distance of -7 units from it.

Hence, option 1 is the correct option.

A line is parallel to y-axis means the equation of the line is in the form x = + a or -a units, where a is the distance from the y-axis.

If the line is at a distance of 5 units from the y-axis on the positive side, the equation will be x = 5.

Hence, option 3 is the correct option.

Given:

6x - 5y = 7

If x = 2,

⇒ 6 $\times$ 2 - 5y = 7

⇒ 12 - 5y = 7

⇒ 12 - 7 = 5y

⇒ 5 = 5y

⇒ y = $\frac{5}{5}$

⇒ y = 1

Hence, option 1 is the correct option.

Given:

$\frac{x}{3} - \frac{y}{2} = 1$

When x = 9,

$$\begin{aligned}⇒ \frac{9}{3} - \frac{y}{2} = 1\\[1em] ⇒ 3 - \frac{y}{2} = 1\\[1em] ⇒ 3 - 1 = \frac{y}{2}\\[1em] ⇒ 2 = \frac{y}{2}\\[1em] ⇒ y = 2 \times 2\\[1em] ⇒ y = 4\end{aligned}$$

Hence, option 3 is the correct option.

x - 4 = 0 ⇒ x = 4

This line represents a vertical line parallel to y-axis.

3y = 1 ⇒ y = $\frac{1}{3}$

This represents a horizontal line parallel to x-axis.

Since the first line has x = 4 and the second line has y = $\frac{1}{3}$, the point od intersection P is (4, $\frac{1}{3}$).

Hence, option 2 is the correct option.

x = 3

The graph of x = 3 is the straight line which is parallel to the y-axis at a distance of 3 units from it.

x + 3 = 0

⇒ x = -3

The graph of x + 3 = 0 is the straight line which is parallel to the y-axis at a distance of -3 units from it.

x - 5 = 0

⇒ x = 5

The graph of x - 5 = 0 is the straight line which is parallel to the y-axis at a distance of 5 units from it.

2x - 7 = 0

⇒ 2x = 7

⇒ x = $\frac{7}{2}$

⇒ x = 3.5

The graph of 2x - 7 = 0 is the straight line which is parallel to the y-axis at a distance of 3.5 units from it.

y = 4

The graph of y = 4 is the straight line which is parallel to the x-axis at a distance of 4 units from it.

y + 6 = 0

⇒ y = - 6

The graph of y + 6 = 0 is the straight line which is parallel to the x-axis at a distance of -6 units from it.

y - 2 = 0

⇒ y = 2

The graph of y - 2 = 0 is the straight line which is parallel to the x-axis at a distance of 2 units from it.

3y + 5 = 0

⇒ 3y = - 5

⇒ y = - $\frac{5}{3}$

⇒ y = - 1.6

The graph of 3y + 5 = 0 is the straight line which is parallel to the x-axis at a distance of - 1.6 units from it.

2y - 5 = 0

⇒ 2y = 5

⇒ y = $\frac{5}{2}$

⇒ y = 2.5

The graph of 2y - 5 = 0 is the straight line which is parallel to the x-axis at a distance of 2.5 units from it.

y = 0

The graph of y = 0 is the straight line which is on x-axis.

x = 0

The graph of x = 0 is the straight line which is on y-axis.

y = 3x

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then y = 3 $\times$ (-1) = -3

Let x = 0, then y = 3 $\times$ 0 = 0

Let x = 1, then y = 3 $\times$ 1 = 3

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	-3	0	3

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

y = - x

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then y = - (-1) = 1

Let x = 0, then y = 0 = 0

Let x = 1, then y = - 1 = - 1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	1	0	-1

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

y = - 2x

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then y = - 2 $\times$ (-1) = 2

Let x = 0, then y = - 2 $\times$ 0 = 0

Let x = 1, then y = - 2 $\times$ 1 = - 2

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	2	0	-2

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

y = x

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then y = -1

Let x = 0, then y = 0

Let x = 1, then y = 1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	-1	0	1

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

5x + y = 0

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then 5 $\times$ (-1) + y = 0 ⇒ y = 5

Let x = 0, then 5 $\times$ 0 + y = 0 ⇒ y = 0

Let x = 1, then 5 $\times$ 1 + y = 0 ⇒ y = -5

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	5	0	-5

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

x + 2y = 0

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then (-1) + 2y = 0 ⇒ y = 0.5

Let x = 0, then 0 + 2y = 0 ⇒ y = 0

Let x = 1, then 1 + 2y = 0 ⇒ y = -0.5

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	0.5	0	-0.5

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

4x - y = 0

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then 4 $\times$ (-1) - y = 0 ⇒ y = -4

Let x = 0, then 4 $\times$ 0 - y = 0 ⇒ y = 0

Let x = 1, then 4 $\times$ 1 - y = 0 ⇒ y = 4

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	-4	0	4

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

3x + 2y = 0

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then 3 $\times$ (-1) + 2y = 0 ⇒ y = 1.5

Let x = 0, then 3 $\times$ 0 + 2y = 0 ⇒ y = 0

Let x = 1, then 3 $\times$ 1 + 2y = 0 ⇒ y = -1.5

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	1.5	0	-1.5

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

x = - 2y

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then (-1) = - 2y ⇒ y = 0.5

Let x = 0, then 0 = - 2y ⇒ y = 0

Let x = 1, then 1 = - 2y ⇒ y = - 0.5

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	0.5	0	-0.5

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

y = 2x + 3

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then y = 2 $\times$ (-1) + 3 ⇒ y = 1

Let x = 0, then y = 2 $\times$ 0 + 3 ⇒ y = 3

Let x = 1, then y = 2 $\times$ 1 + 3 ⇒ y = 5

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	1	3	5

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

$y = \frac{2}{3} x - 1$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -3, then $y = \frac{2}{3} \times (-3) - 1 ⇒ y = -3$

Let x = 0, then $y = \frac{2}{3} \times 0 - 1 ⇒ y = -1$

Let x = 3, then $y = \frac{2}{3} \times 3 - 1 ⇒ y = 1$

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-3	0	3
y	-3	-1	1

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

y = -x + 4

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then y = -(-1) + 4 ⇒ y = 5

Let x = 0, then y = 0 + 4 ⇒ y = 4

Let x = 1, then y = -1 + 4 ⇒ y = 3

Step 2:

Make a table i(as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	5	4	3

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

$y = 4x - \frac{5}{2}$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -$\frac{3}{2} = -1.5$, then $y = 4 \times \Big(\frac{-3}{2}\Big) - \frac{5}{2} ⇒ y = -\frac{17}{2} = -8.5$

Let x = $\frac{1}{2} = 0.5$, then $y = 4 \times \frac{1}{2} - \frac{5}{2} ⇒ y = -\frac{1}{2} = -0.5$

Let x = $\frac{3}{2} = 1.5$, then $y = 4 \times \frac{3}{2} - \frac{5}{2} ⇒ y = \frac{7}{2} = 3.5$

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1.5	0.5	1.5
y	-8.5	-0.5	3.5

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

$y =\frac{3}{2}x + \frac{2}{3}$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -2, then $y =\frac{3}{2} \times (-2) + \frac{2}{3} ⇒ y = -2.3$

Let x = 0, then $y =\frac{3}{2} \times 0 + \frac{2}{3} ⇒ y = 0.6$

Let x = 2, then $y =\frac{3}{2} \times 2 + \frac{2}{3} ⇒ y = 3.6$

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-2	0	2
y	-2.3	0.6	3.6

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

2x - 3y = 4

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -4, then 2 $\times$ (-4) - 3y = 4 ⇒ y = -4

Let x = 2, then 2 $\times$ 2 - 3y = 4 ⇒ y = 0

Let x = 5, then 2 $\times$ 5 - 3y = 4 ⇒ y = 2

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-4	2	5
y	-4	0	2

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

$\frac{x-1}{3} - \frac{y+2}{2} = 0$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -5, then $\frac{-5-1}{3} - \frac{y+2}{2} = 0 ⇒ y = -6$

Let x = 1, then $\frac{1-1}{3} - \frac{y+2}{2} = 0 ⇒ y = -2$

Let x = 4, then $\frac{4-1}{3} - \frac{y+2}{2} = 0 ⇒ y = 0$

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-5	1	4
y	-6	-2	0

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

$x - 3 = \frac{2}{5}(y + 1)$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then $0 - 3 = \frac{2}{5}(y + 1) ⇒ y = -8.5$

Let x = 1, then $1 - 3 = \frac{2}{5}(y + 1) ⇒ y = -6$

Let x = 3, then $3 - 3 = \frac{2}{5}(y + 1) ⇒ y = -1$

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	0	1	3
y	-8.5	-6	-1

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

x + 5y + 2 = 0

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -2, then -2 + 5y + 2 = 0 ⇒ y = 0

Let x = 0, then 0 + 5y + 2 = 0 ⇒ y = - 0.4

Let x = 2, then 2 + 5y + 2 = 0 ⇒ y = - 0.8

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-2	0	2
y	0	-0.4	-0.8

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

3x + 2y = 6

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then 3 $\times$ (-1) + 2y = 6 ⇒ y = $\frac{9}{2} = 4.5$

Let x = 0, then 3 $\times$ 0 + 2y = 6 ⇒ y = $\frac{6}{2} = 3$

Let x = 1, then 3 $\times$ 1 + 2y = 6 ⇒ y = $\frac{3}{2} = 1.5$

Step 2: Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	4.5	3	1.5

Step 3: Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

From the graph we get that the line meets x-axis at (2, 0) and y-axis at (0, 3)

2x - 5y = 10

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then 2 $\times$ (-1) - 5y = 10 ⇒ y = $- \frac{12}{5} = -2.4$

Let x = 0, then 2 $\times$ 0 - 5y = 10 ⇒ y = $-\frac{10}{5} = -2$

Let x = 1, then 2 $\times$ 1 - 5y = 10 ⇒ y = $-\frac{8}{5} = -1.6$

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	-2.4	-2	-1.6

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

From the graph we get that the line meets x-axis at (5, 0) and y-axis at (0, -2)

$\frac{1}{2}x + \frac{2}{3}y = 5$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -2, then $\frac{1}{2} \times (-2) + \frac{2}{3}y = 5 ⇒ y = \frac{6 \times 3}{2} = 9$

Let x = 0, then $\frac{1}{2} \times 0 + \frac{2}{3}y = 5 ⇒ y = \frac{15}{2} = 7.5$

Let x = 2, then $\frac{1}{2} \times 2 + \frac{2}{3}y = 5 ⇒ y = \frac{4 \times 3}{2} = 6$

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-2	0	2
y	9	7.5	6

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

From the graph we get that the line meets x-axis at (10, 0) and y-axis at (0, 7.5)

$\frac{2x - 1}{3} - \frac{y - 2}{5} = 0$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then $\frac{2 \times (-1) - 1}{3} - \frac{y - 2}{5} = 0 ⇒ y = -3$

Let x = 0, then $\frac{2 \times 0 - 1}{3} - \frac{y - 2}{5} = 0 ⇒ y = 0.3$

Let x = 1, then $\frac{2 \times 1 - 1}{3} - \frac{y - 2}{5} = 0 ⇒ y = 3.6$

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	-3	0.3	3.6

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

From the graph we get that the line meets x-axis at (-0.1, 0) and y-axis at (0, 0.3)

3x - (5 - y) = 7

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then 3 $\times$ (-1) - (5 - y) = 7 ⇒ y = 15

Let x = 0, then 3 $\times$ 0 - (5 - y) = 7 ⇒ y = 12

Let x = 1, then 3 $\times$ 1 - (5 - y) = 7 ⇒ y = 9

Let x = 4, then 3 $\times$ 4 - (5 - y) = 7 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1	4
y	15	12	9	0

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

The area of the triangle ABO will be = $\frac{1}{2} \times base \times$ altitude

= $\frac{1}{2} \times OA \times$ OB

= $\frac{1}{2} \times 4 \times$ 12 square unit

= 24 square unit

Hence, the area of triangle = 24 square unit.

7 - 3 (1 - y) = - 5 + 2x

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 7 - 3 (1 - y) = - 5 + 2 $\times$ 0 ⇒ y = -3

Let x = 4.5, then 7 - 3 (1 - y) = - 5 + 2 $\times$ 4.5 ⇒ y = 0

Let x = 6, then 7 - 3 (1 - y) = - 5 + 2 $\times$ 6 ⇒ y = 1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	0	4.5	6
y	-3	0	1

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

The area of the triangle ABO will be = $\frac{1}{2} \times base \times$ altitude

= $\frac{1}{2}$ $\times$ OA $\times$ OB

= $\frac{1}{2}$ $\times$ 3 $\times$ 4.5

= 6.75 square unit

Hence, the area of triangle = 6.75 square unit.

First equation = y = 3x - 1

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then y = 3 $\times$ (-1) - 1 ⇒ y = -4

Let x = 0, then y = 3 $\times$ 0 - 1 ⇒ y = -1

Let x = 1, then y = 3 $\times$ 1 - 1 ⇒ y = 2

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	-4	-1	2

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation = y = 3x + 2

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -2, then y = 3 $\times$ (-2) + 2 ⇒ y = -4

Let x = -1, then y = 3 $\times$ (-1) + 2 ⇒ y = -1

Let x = 0, then y = 3 $\times$ 0 + 2 ⇒ y = 2

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-2	-1	0
y	-4	-1	2

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Hence, the two lines are parallel to each other.

First equation = y = x - 3

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then y = (-1) - 3 ⇒ y = - 4

Let x = 0, then y = 0 - 3 ⇒ y = - 3

Let x = 1, then y = 1 - 3 ⇒ y = - 2

Let x = 4, then y = 4 - 3 ⇒ y = 1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1	4
y	-4	-3	-2	1

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation = y = - x + 5

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then y = - (-1) + 5 ⇒ y = 6

Let x = 0, then y = - 0 + 5 ⇒ y = 5

Let x = 1, then y = - 1 + 5 ⇒ y = 4

Let x = 4, then y = - 4 + 5 ⇒ y = 1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1	4
y	6	5	4	1

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Hence, the two lines are perpendicular to each other.

First equation = 2x - 3y = 6

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then 2 $\times$ (-1) - 3y = 6 ⇒ y = - 2.6

Let x = 0, then 2 $\times$ (0) - 3y = 6 ⇒ y = - 2

Let x = 1, then 2 $\times$ 1 - 3y = 6 ⇒ y = - 1.3

Let x = 3, then 2 $\times$ 3 - 3y = 6 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1	3
y	-2.6	-2	-1.3	0

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation = $\frac{x}{2} + \frac{y}{3} = 1$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then $\frac{-1}{2} + \frac{y}{3} = 1$ ⇒ y = 4.5

Let x = 0, then $\frac{0}{2} + \frac{y}{3} = 1$ ⇒ y = 3

Let x = 1, then $\frac{1}{2} + \frac{y}{3} = 1$ ⇒ y = 1.5

Let x = 2, then $\frac{2}{2} + \frac{y}{3} = 1$ ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1	2
y	4.5	3	1.5	0

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Hence, the two lines are perpendicular to each other.

First equation = 3x + 4y = 24

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -2, then 3 $\times$ (-2) + 4y = 24 ⇒ y = 7.5

Let x = 0, then 3 $\times$ 0 + 4y = 24 ⇒ y = 6

Let x = 2, then 3 $\times$ 2 + 4y = 24 ⇒ y = 4.5

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-2	0	2
y	7.5	6	4.5

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation = $\frac{x}{4} + \frac{y}{3} = 1$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then $\frac{-1}{4} + \frac{y}{3} = 1$ ⇒ y = 3.7

Let x = 0, then $\frac{0}{4} + \frac{y}{3} = 1$ ⇒ y = 3

Let x = 1, then $\frac{1}{4} + \frac{y}{3} = 1$ ⇒ y = 2.2

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	3.7	3	2.2

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Hence, the two lines are parallel to each other.

The given equations are y = x - 2, y = 2x + 1 and y = 4.

Make a table (as given below) for the different pairs of the values of x and y:

x	-4	0	3
y = x - 2	-6	-2	1
y = 2x + 1	-7	1	7
y = 4	4	4	4

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

The given equations are y = 2x - 1, y = 2x and y = 2x + 1.

Make a table (as given below) for the different pairs of the values of x and y:

x	-2	0	4
y = 2x - 1	-5	-1	7
y = 2x	-4	0	8
y = 2x + 1	-3	1	9

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Yes, the lines drawn are parallel to each other.

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -2, then 3 $\times$ (-2) + 2y = 6 ⇒ y = 6

Let x = -1, then 3 $\times$ (-1) + 2y = 6 ⇒ y = 4.5

Let x = 0, then 3 $\times$ 0 + 2y = 6 ⇒ y = 3

Let x = 1, then 3 $\times$ 1 + 2y = 6 ⇒ y = 1.5

Let x = 2, then 3 $\times$ 2 + 2y = 6 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-2	-1	0	1	2
y	6	4.5	3	1.5	0

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

The graph intersects the x-axis at P(2,0).

The graph intersects the y-axis at Q(0,3).

Hence, the coordinates of P = (2, 0) and Q = (0, 3).

The inclination of a line is 60°, then θ = 60°.

The slope of the line = m = tan 60° = $\sqrt3$

Hence, option 3 is the correct option.

⇒ 2x - 5y = 8

⇒ - 5y = 8 - 2x

⇒ 5y = 2x - 8

⇒ y = $\frac{2}{5}\text{x} - \frac{8}{5}$

∴ Slope = coefficient of x = $\frac{2}{5}$

Hence, option 4 is the correct option.

⇒ 5x - 6y = 9

⇒ -6y = 9 - 5x

⇒ 6y = 5x - 9

⇒ y = $\frac{5}{6}\text{x} - \frac{9}{6}$

∴ y-intercept = constant term = -$\frac{9}{6} = - \frac{3}{2}$

Hence, option 4 is the correct option.

slope = -2 ⇒ m = -2

y-intercept = -7 ⇒ c = -7

∴ Equation is : y = mx + c

⇒ y = (-2)x + (-7)

⇒ y = -2x - 7

⇒ 2x + y + 7 = 0

Hence, option 1 is the correct option.

The given equation : x - y + 1 = 0

⇒ y = x + 1

General form of equation: y = mx + c

m = 1, c = 1

Hence, option 1 is the correct option.

MO = NO

∠ NMO = ∠ MNO

∠ MON = 90°

Let ∠ MNO = ∠ NMO = x.

As we know that sum of all angles of triangle = 180°

⇒ ∠ NMO + ∠ MNO + ∠ MON = 180°

⇒ x + x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ x = $\frac{90°}{2}$

⇒ x = 45°

Hence, the inclination = 45°.

MO = NO

∠ NMO = ∠ MNO

∠ MON = 90°

Let ∠ MNO = ∠ NMO = x.

As we know that sum of all angles of triangle = 180°

⇒ ∠ NMO + ∠ MNO + ∠ MON = 180°

⇒ x + x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ x = $\frac{90°}{2}$

⇒ x = 45°

Inclination = 180° - 45° = 135°

Hence, the inclination = 135°.

∠ NMO = 2x

∠ MNO = x

∠ MON = 90°

As we know that sum of all angles of triangle = 180°

⇒ ∠ NMO + ∠ MNO + ∠ MON = 180°

⇒ 2x + x + 90° = 180°

⇒ 3x = 180° - 90°

⇒ x = $\frac{90°}{3}$

⇒ x = 30°

Hence, the inclination = 30°.

(i) A line parallel to the x-axis has an inclination of 0°, as it does not make an angle with the x-axis.

Hence, the inclination of the line = 0°.

(ii) A line perpendicular to the x-axis is parallel to the y-axis and makes a 90° angle with the x-axis.

Hence, the inclination of the line = 90°.

(iii) A line parallel to the y-axis is perpendicular to the x-axis, so its inclination is 90°.

Hence, the inclination of the line = 90°.

(iv) A line perpendicular to the y-axis is parallel to the x-axis, so its inclination is 0°.

Hence, the inclination of the line = 0°.

(i) 0°

The inclination of a line is 0°, then θ = 0°.

The slope of the line = m = tan 0° = 0

Hence, the slope of the line whose inclination is 0° is 0.

(ii) 30°

The inclination of a line is 30°, then θ = 30°.

The slope of the line = m = tan 30° = $\frac{1}{\sqrt3}$

Hence, the slope of the line whose inclination is 30° is $\frac{1}{\sqrt3}$.

(iii) 45°

The inclination of a line is 45°, then θ = 45°.

The slope of the line = m = tan 45° = 1

Hence, the slope of the line whose inclination is 45° is 1.

(iv) 60°

The inclination of a line is 60°, then θ = 60°.

The slope of the line = m = tan 60° = $\sqrt3$

Hence, the slope of the line whose inclination is 60° is $\sqrt3$.

(i) The slope of the line = m = 0 ⇒ tan θ = 0

⇒ tan θ = tan 0°

⇒ θ = 0°

Hence, the inclination is 0°.

(ii) The slope of the line = m = 1 ⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°

Hence, the inclination is 45°.

(iii) The slope of the line = m = ${\sqrt3}$ ⇒ tan θ = ${\sqrt3}$.

⇒ tan θ = tan 60°

⇒ θ = 60°

Hence, the inclination is 60°.

(iv) The slope of the line = m = $\frac{1}{\sqrt3}$ ⇒ tan θ = $\frac{1}{\sqrt3}$

⇒ tan θ = tan 30°

⇒ θ = 30°

Hence, the inclination is 30°.

(i) A line parallel to the x-axis has an inclination of 0°, as it does not make an angle with the x-axis.

Therefore, the inclination of the line is 0°.

The slope of the line = m = tan θ = tan 0° = 0

Hence, the slope of the line is 0.

(ii) A line perpendicular to the x-axis is parallel to the y-axis and makes a 90° angle with the x-axis.

Therefore, the inclination of the line is 90°.

The slope of the line = m = tan θ = tan 90° = not defined

Hence, the slope of the line is not defined.

(iii) A line parallel to the y-axis is perpendicular to the x-axis, so its inclination is 90°.

Therefore, the inclination of the line is 90°.

The slope of the line = m = tan θ = tan 90° = not defined

Hence, the slope of the line is not defined.

(iv) A line perpendicular to the y-axis is parallel to the x-axis, so its inclination is 0°.

Therefore, the inclination of the line is 0°.

The slope of the line = m = tan θ = tan 0° = 0

Hence, the slope of the line is 0.

(i) x + 3y + 5 = 0

⇒ 3y = -x - 5

⇒ y = -$\frac{1}{3}x - \frac{5}{3}$

∴ Slope = coefficient of x = $\frac{-1}{3}$

And, y-intercept = constant term = $\frac{-5}{3}$

Hence, the slope = $\frac{-1}{3}$ and y-intercept = $\frac{-5}{3}$.

(ii) 3x - y - 8 = 0

⇒ y = 3x - 8

∴ Slope = coefficient of x = 3

And, y-intercept = constant term = -8

Hence, the slope = 3 and y-intercept = -8.

(iii) 5x = 4y + 7

⇒ 4y = 5x - 7

⇒ y = $\frac{5}{4}$ x - $\frac{7}{4}$

∴ Slope = coefficient of x = $\frac{5}{4}$

And, y-intercept = constant term = $\frac{-7}{4}$

Hence, the slope = $\frac{5}{4}$ and y-intercept = $\frac{-7}{4}$.

(iv) x = 5y - 4

⇒ 5y = x + 4

⇒ y = $\frac{1}{5}$ x + $\frac{4}{5}$

∴ Slope = coefficient of x = $\frac{1}{5}$

And, y-intercept = constant term = $\frac{4}{5}$

Hence, the slope = $\frac{1}{5}$ and y-intercept = $\frac{4}{5}$.

(v) y = 7x - 2

∴ Slope = coefficient of x = 7

And, y-intercept = constant term = -2

Hence, the slope = 7 and y-intercept = -2.

(vi) 3y = 7

⇒ 3y = 0 $\times$ x + 7

⇒ y = 0 $\times$ x + $\frac{7}{3}$

∴ Slope = coefficient of x = 0

And, y-intercept = constant term = $\frac{7}{3}$

Hence, the slope = 0 and y-intercept = $\frac{7}{3}$.

(vii) 4y + 9 = 0

⇒ 4y = 0 $\times$ x - 9

⇒ y = 0 $\times$ x - $\frac{9}{4}$

∴ Slope = coefficient of x = 0

And, y-intercept = constant term = $\frac{-9}{4}$

Hence, the slope = 0 and y-intercept = $\frac{-9}{4}$.

(i) slope = 2 ⇒ m = 2

y-intercept = 3 ⇒ c = 3

∴ Equation is : y = mx + c

⇒ y = 2x + 3

Hence, the equation of the line is y = 2x + 3.

(ii) slope = 5 ⇒ m = 5

y-intercept = -8 ⇒ c = -8

∴ Equation is : y = mx + c

⇒ y = 5x - 8

Hence, the equation of the line is y = 5x - 8.

(iii) slope = -4 ⇒ m = -4

y-intercept = 2 ⇒ c = 2

∴ Equation is : y = mx + c

⇒ y = -4x + 2

⇒ 4x + y = 2

Hence, the equation of the line is 4x + y = 2.

(iv) slope = -3 ⇒ m = -3

y-intercept = -1 ⇒ c = -1

∴ Equation is : y = mx + c

⇒ y = -3x - 1

⇒ 3x + y + 1 = 0

Hence, the equation of the line is 3x + y + 1 = 0.

(v) slope = 0 ⇒ m = 0

y-intercept = - 5 ⇒ c = - 5

∴ Equation is : y = mx + c

⇒ y = 0 $\times$ x - 5

⇒ y + 5 = 0

Hence, the equation of the line is y + 5 = 0.

(vi) slope = 0 ⇒ m = 0

y-intercept = 0 ⇒ c = 0

∴ Equation is : y = mx + c

⇒ y = 0 $\times$ x + 0

⇒ y = 0

Hence, the equation of the line is y = 0.

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 3 $\times$ 0 + 4y = 12 ⇒ y = 3

Let x = 1, then 3 $\times$ 1 + 4y = 12 ⇒ y = 2.2

Let x = 4, then 3 $\times$ 4 + 4y = 12 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	0	1	4
y	3	2.2	0

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

From the graph, y-intercept of the line = OB = 3.

Hence, y-intercept of the line = 3.

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 2 $\times$ 0 - 3y - 18 = 0 ⇒ y = -6

Let x = 3, then 2 $\times$ 3 - 3y - 18 = 0 ⇒ y = -4

Let x = 6, then 2 $\times$ 6 - 3y - 18 = 0 ⇒ y = -2

Let x = 9, then 2 $\times$ 9 - 3y - 18 = 0 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	0	3	6	9
y	-6	-4	-2	0

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

From the graph, y-intercept of the line = OB = -6.

Hence, y-intercept of the line = -6.

Given, point P(-5, 4)

Abscissa (x-coordinate): -5

Ordinate (y-coordinate): 4

Hence, option 2 is the correct option.

As we know that the distance of a point from the X-axis is the absolute value of its y-coordinate.

So, if perpendicular distance of a point from the x-axis is 5 units. Then the y-coordinate can be 5 or -5.

Hence, option 4 is the correct option.

If a point lies on the y‑axis, its x‑coordinate is 0.

If it’s at a distance of 2 units from the x‑axis, that means its y-coordinate is either +2 or −2 (since distance ignores sign).

Its co-ordinates are (0, 2) or (0, -2).

Hence, option 4 is the correct option.

The points are shown on the graph below:

Steps of construction :

1. Plot points A(2, 0), B(-3, 0) and C(-3, -5) on graph.

2. Measure AB.

3. Mark point D such that it is at a distance AB from points A and C.

4. Join AB, BC, CD and DA.

From graph,

The co-ordinates of fourth vertex D is (2, -5).

Hence, option 2 is the correct option.

Given equilateral triangle OAB.

OA = OB = AB = 8 units.

Draw BD ⊥ OA.

In an equilateral triangle, a perpendicular drawn from one of the vertices to the opposite side bisects the side.

∴ OD = $\frac{1}{2}$ x OA = ​$\frac{1}{2}$ x 8 = 4.

In right angle triangle OBD,

By pythagoras theorem,

⇒ OB² = OD² + BD²

⇒ 8² = 4² + BD²

⇒ 64 = 16 + BD²

⇒ BD² = 64 - 16

⇒ BD² = 48

⇒ BD = $\sqrt{48}$

⇒ BD = $4\sqrt{3}$

From graph,

Co-ordinates of O = (0, 0)

Co-ordinates of A = (8, 0)

As, OD = 4 units and BD = $4\sqrt{3}$ units.

Co-ordinates of B = (4, $4\sqrt{3}$).

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Vertex B lies in the fourth quadrant. The square's sides are along the coordinate axes, so one vertex, O, is at the origin (0, 0).

The side length is 4 units.

With O at (0, 0), the other two vertices on the axes must be A (4, 0) on the positive x-axis and C (0, -4) on the negative y-axis.

The fourth quadrant is where the x-coordinates are positive and the y-coordinates are negative.

Vertex B is located at the point where the line extending 4 units to the right from C (0, -4) intersects with the line extending 4 units down from A (4, 0), which is (4, -4).

⇒ B = (4, –4).

∴ Statement 1 is true, and statement 2 is false.

Hence, option 3 is the correct option.

The coordinates of R(0, 2) and P(0, -2).

Using distance formula,

Distance between two points = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

PR = $\sqrt{(0 - 0)^2 + (-2 - 2)^2} = \sqrt{(-4)^2} = \sqrt{16} = 4$ units.

Given equilateral triangle PQR.

∴ PQ = QR = PR = 4 units.

In an equilateral triangle, a perpendicular drawn from one of the vertices to the opposite side bisects the side.

∴ OP = $\frac{1}{2}$ x PR = ​$\frac{1}{2}$ x 4 = 2 units.

In right angle triangle OPQ,

By pythagoras theorem,

⇒ QP² = OP² + OQ²

⇒ OQ² = QP² - OP²

⇒ OQ² = 4² - 2²

⇒ OQ² = 16 - 4

⇒ OQ² = 12

⇒ OQ = $\sqrt{12}$

⇒ OQ = $2\sqrt{3}$

Since, OQ = $2\sqrt{3}$ units and Q lies on x-axis.

Co-ordinates of Q = ($2\sqrt{3}$, 0).

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Given, (2x - 3y, 8) = (2, x + 2y)

Thus,

2x - 3y = 2 ........(1)

x + 2y = 8 ........(2)

So, reason (R) is true.

Multiplying equation (2) by 2, we get :

⇒ 2(x + 2y) = 2 × 8

⇒ 2x + 4y = 16 ........(3)

Subtracting equation (1) from (3), we get :

⇒ 2x + 4y - (2x - 3y) = 16 - 2

⇒ 2x - 2x + 4y + 3y = 14

⇒ 7y = 14

⇒ y = $\frac{14}{7}$ = 2.

Substituting the value y = 2 in first equation,

⇒ 2x - 3 x 2 = 2

⇒ 2x - 6 = 2

⇒ 2x = 2 + 6

⇒ 2x = 8

⇒ x = $\frac{8}{2}$ = 4.

Hence, the value of x = 4 and y = 2.

∴ A is false, but R is true.

Hence, option 2 is the correct option.

(3, 5), (1, 1) and (0, -1)

As shown in the figure, all three points lie on the same straight line.

Hence, the points are collinear.

(-2, -1), (-1, -4) and (-4, 1)

As shown in the figure, all three points do not lie on the same straight line.

Hence, the points are not collinear.

Given point : A (5, -7)

AM is perpendicular to the x-axis.

AN is perpendicular to the y-axis.

The graph intersects the x-axis at M(5, 0).

The graph intersects the y-axis at N(0, -7).

Hence, the coordinates of the points are M = (5, 0) and N = (0, -7).

Plot the points A = (3, 4), B = (-2, 4) and C = (-2, -1) on the graph paper. Join point A with B and B with C.

From the graph, it is clear that the horizontal distance between the points A (3, 4) and B (-2, 4) is 5 units and the vertical distance between the points B (-2, 4) and C (-2, -1) is 5 units. Therefore, the vertical distance between the points A (3, 4) and D must be 5 units and the horizontal distance between the points C (-2, -1) and D must be 5 units.

Now, complete the square ABCD and read the coordinates of point D, as shown on the graph, D = (3, -1).

Area of the square ABCD = AD x CD

= 5 x 5 square units

= 25 square units

Hence, the coordinates of D = (3, -1) and the area of the square = 25 square units.

Given that in rectangle OABC, point O is the origin and OA = 10 units along x-axis, therefore we have O(0, 0) and A(10, 0).

It is also given that AB = 8 units. Since AB is perpendicular to OA (as OABC is a rectangle), the line through A is a vertical line and it meets B at B(10, 8).

Through point B, draw a horizontal line (parallel to x-axis) that meets the y-axis at y = 8. This gives the coordinates of point C: (0, 8).

Thus, the vertices of rectangle OABC are: O(0, 0), A(10, 0), B(10, 8) and C(0, 8)

Hence, the coordinates of the vertices are A = (10, 0), B = (10, 8) and C = (0, 8).

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then (-1) + 2y - 3 = 0 ⇒ y = 2

Let x = 0, then 0 + 2y - 3 = 0 ⇒ y = 1.5

Let x = 1, then 1 + 2y - 3 = 0 ⇒ y = 1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	2	1.5	1

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line AB passing through the points plotted on the graph.

(i) To find x₁, the value of x, when y = 3:

Through the point y = 3, draw a horizontal straight line which meets the line AB at point C. Through point C, draw a vertical line which meets the x-axis at x = -3.

Hence, the value of x, when y = 3 is -3 , i.e, x₁ = -3.

(ii) To find x₂, the value of x, when y = -2:

Through the point y = -2, draw a horizontal straight line which meets the line AB at point D. Now, through point D, draw a vertical line which meets the y-axis at x = 7.

Hence, the value of x, when y = -2 is 7 , i.e, x₂ = 7.

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -2, then 3 $\times$ (-2) - 4y = 12 ⇒ y = -4.5

Let x = 1, then 3 $\times$ 1 - 4y = 12 ⇒ y = -2.2

Let x = 2, then 3 $\times$ 2 - 4y = 12 ⇒ y = -1.5

Let x = 4, then 3 $\times$ 4 - 4y = 12 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-2	1	2	4
y	-4.5	-2.2	-1.5	0

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

(i) To find y₁, the value of y, when x = 4:

Through the point x = 4, draw a horizontal line which meets the y-axis at y = 0.

Hence, the value of y, when x = 4 is 0 , i.e, y₁ = 0.

(ii) To find y₂, the value of y, when x = 0:

Through the point x = 0, draw a vertical line which meets the y-axis at x = -3.

Hence, the value of y, when x = 0 is -3 , i.e, y₂ = -3.

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -4, then $\frac{-4}{4} + \frac{y}{5} = 1$ ⇒ y = 10

Let x = 0, then $\frac{0}{4} + \frac{y}{5} = 1$ ⇒ y = 5

Let x = 4, then $\frac{4}{4} + \frac{y}{5} = 1$ ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	- 4	0	4
y	10	5	0

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line AB passing through the points plotted on the graph.

(i) To find x₁, the value of x, when y = 10:

Through the point y = 10, draw a horizontal straight line which meets the line AB at point C.

Through point C, draw a vertical line which meets the x-axis at x = -4.

Hence, the value of x, when y = 10 is -4 , i.e, x₁ = -4.

(ii) To find y₁, the value of y, when x = 8:

Through the point x = 8, draw a vertical line which meets the line AB at point D.

Now, through point D, draw a horizontal line which meets the y-axis at y = -5.

Hence, the value of y, when x = 8 is -5 , i.e, y₁ = -5.

First equation : x + y = 2

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then (-1) + y = 2 ⇒ y = 3

Let x = 0, then 0 + y = 2 ⇒ y = 2

Let x = 1, then 1 + y = 2 ⇒ y = 1

Let x = 3, then 3 + y = 2 ⇒ y = -1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1	3
y	3	2	1	-1

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation : x - 2y = 5

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then (-1) - 2y = 5 ⇒ y = -3

Let x = 0, then 0 - 2y = 5 ⇒ y = -2.5

Let x = 1, then 1 - 2y = 5 ⇒ y = -2

Let x = 3, then 3 - 2y = 5 ⇒ y = -1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1	3
y	-3	-2.5	-2	-1

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Third equation : $\frac{x}{3} + y = 0$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -3, then $\frac{-3}{3} + y = 0$ ⇒ y = 1

Let x = 0, then $\frac{0}{3} + y = 0$ ⇒ y = 0

Let x = 3, then $\frac{3}{3} + y = 0$ ⇒ y = -1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-3	0	3
y	1	0	-1

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Hence, the three straight lines pass through the same point.

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = (-1), then (-1) + y = 5 ⇒ y = 6

Let x = 0, then 0 + y = 5 ⇒ y = 5

Let x = 1, then 1 + y = 5 ⇒ y = 4

Let x = 5, then 5 + y = 5 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1	5
y	6	5	4	0

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

From the graph, inclination of the line = θ = 135° and y-intercept of the line = OB = 5.

Hence, inclination = 135° and y-intercept of the line = 5.

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -1, then 2 $\times$ (-1) + y = 5 ⇒ y = 7

Let x = 0, then 2 $\times$ 0 + y = 5 ⇒ y = 5

Let x = 1, then 2 $\times$ 1 + y = 5 ⇒ y = 3

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	-1	0	1
y	7	5	3

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 4, then 4 $\times$ (4) - y = 5 ⇒ y = 11

Let x = 3.5, then 4 $\times$ 3.5 - y = 5 ⇒ y = 9

Let x = 1, then 4 $\times$ 1 - y = 5 ⇒ y = -1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

x	4	3.5	3
y	11	9	7

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line AB passing through the points plotted on the graph.

(i) To find x₁, the value of x, when y = 3:

Through the point y = 3, draw a horizontal straight line which meets the line AB at point C.

Through point C, draw a vertical line which meets the x-axis at x = 2.

Hence, the value of x, when y = 3 is 2 , i.e, x₁ = 2.

(ii) To find y₁, the value of y, when x = 3:

Through the point x = 3, draw a vertical line which meets the line AB at point D.

Through point D, draw a horizontal line which meets the y-axis at y = 7.

Hence, the value of y, when x = 3 is 7 , i.e, y₁ = 7.