Co-ordinate Geometry

A(-3, 2)

B(2, 1)

C(4, -2)

D(-1, -3)

E(0, -1)

(i) (5, 0)

The point whose y-coordinate = 0, lies on the x-axis.

Hence, point (5, 0) lies on x-axis.

(ii) (0, -2)

The point whose x-coordinate = 0, lies on the y-axis.

Hence, point (0, -2) lies on y-axis.

(iii) (0, 3)

The point whose x-coordinate = 0, lies on the y-axis.

Hence, point (0, 3) lies on y-axis.

(iv) (-3, 0)

The point whose y-coordinate = 0, lies on the x-axis.

Hence, point (-3, 0) lies on x-axis.

(i) A(-3, 2)

Here, x is negative and y is positive.

Hence, point A lies in the 2nd quadrant.

(ii) B(-5, -3)

As x is negative, y is negative

Hence, point B lies in the 3rd quadrant.

(iii) C(2, -7)

As x is positive and y is negative

Hence, point C lies in the 4th quadrant.

(iv) D(-2, -2)

As x is negative, y is negative

Hence, point D lies in the 3rd quadrant.

Given,

A(2, -2), B(8, 4) and C(5, 7) are three vertices of a rectangle ABCD.

As we know diagonals of a rectangle are equal and bisect each other.

Therefore,

Midpoint of AC = Midpoint of BD

Midpoint of AC = $\Big(\frac{2 + 5}{2}, \frac{-2 + 7}{2}\Big)$

= $\Big(\frac{7}{2}, \frac{5}{2}\Big)$

Let D = (x, y)

Midpoint of BD:

$\Big(\frac{8 + x}{2}, \frac{4 + y}{2}\Big)$

Now Equating both midpoints

$\frac{8 + x}{2} = \frac{7}{2}$

$\frac{4 + y}{2} = \frac{5}{2}$

Solving for x,

$\frac{8 + x}{2} = \frac{7}{2}$

8 + x = 7

x = 7 - 8

x = -1

Solving for y,

$\frac{4 + y}{2} = \frac{5}{2}$

4 + y = 5

y = 1.

Hence, coordinates of D = (-1, 1).

A(3, 2), B(0, 5) and D(0, -1) are the three vertices of a square ABCD.

As we know diagonals of square are equal and bisect each other.

Therefore,

Midpoint of AC = Midpoint of BD

Let C = (x, y)

Midpoint of AC = $\Big(\frac{3 + x}{2}, \frac{2 + y}{2}\Big)$

Midpoint of BD :

$\Big(\frac{0 + 0}{2}, \frac{5 + (-1)}{2}\Big)$

(0, 2)

Now Equating both midpoints,

$\frac{3 + x}{2}$ = 0

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

Solving for x,

$\frac{3 + x}{2}$ = 0

3 + x = 0

x = -3

Solving for y,

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

2 + y = 4

y = 2.

Hence, coordinates of C = (-3, 2).

Given, equation :

⇒ x + 4 = 0

⇒ x = -4

Given, equation :

⇒ x - 5 = 0

⇒ x = 5

Given, equation :

⇒ y + 5 = 0

⇒ y = -5

Given,

⇒ 2y - 7 = 0

⇒ 2y = 7

⇒ y = $\frac{7}{2}$.

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

Let x = 2, then y = 2

Step 2 :

Make a table for the corresponding values of x and y:

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

Step 3 :

Plot the points (-1, -1), (0, 0), (1, 1) and (2, 2) 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 = -1, then y = 1

Let x = 0, then y = 0

Let x = -2, then y = 2

Let x = 2, then y = -2

Step 2 :

Make a table for the corresponding values of x and y:

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

Step 3 :

Plot the points (-1, 1), (0, 0), (-2, 2) and (2, -2) 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 = -1, then y = -2

Let x = 0, then y = 0

Let x = 1, then y = 2

Let x = 2, then y = 4

Step 2 :

Make a table for the corresponding values of x and y: :

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

Step 3 :

Plot the points (-1, -2), (0, 0), (1, 2) and (2, 4) 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 = -1, then y = -1

Let x = 0, then y = 2

Let x = 1, then y = 5

Let x = -2, then y = -4

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
-1	-1
0	2
1	5
-2	-4

Step 3 :

Plot the points (-1, -1), (0, 2), (1, 5) and (-2, -4) 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 = 0, then y = -1

Let x = 1, then y = 1

Let x = 2, then y = 3

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
0	-1
1	1
2	3

Step 3 :

Plot the points (0, -1), (1, 1), (2, 3) on a graph paper and then draw a straight line passing through the points plotted on the graph.

Given,

y - x = 5

y = x + 5

Step 1 :

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

Let x = 0, then y = 5

Let x = -5, then y = 0

Let x = -2, then y = 3

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
0	5
-5	0
-2	3

Step 3 :

Plot the points (0, 5), (-5, 0), (-2, 3) on a graph paper and then draw a straight line passing through the points plotted on the graph.

Given,

2x + y = 6

y = 6 - 2x

Step 1 :

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

Let x = 0, then y = 6

Let x = 1, then y = 4

Let x = 2, then y = 2

Let x = 3, then y = 0

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
0	6
1	4
2	2
3	0

Step 3 :

Plot the points (0, 6), (1, 4), (2, 2) and (3, 0) on a graph paper and then draw a straight line passing through the points plotted on the graph.

Graph meets x-axis at (3, 0).

Graph meets y-axis at (0, 6).

Given,

2x - 3y = 5

3y = 2x - 5

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

Step 1 :

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

Let x = 4, then y = 1

Let x = 2.5, then y = 0

Let x = 1, then y = -1

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
4	1
2.5	0
1	-1

Step 3 :

Plot the points (4, 1), (2.5, 0), (1, -1) on a graph paper and then draw a straight line passing through the points plotted on the graph.

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

Through the point x = 4, draw a vertical straight line which meets the line 2x - 3y = 5 at point C.

Through point C, draw a horizontal line which meets the y-axis at y = 1.

Hence, the value of y = 1, when x = 4.

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

Through the point y = 3, draw a horizontal straight line which meets the line 2x - 3y = 5 at point D.

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

Hence, the value of x = 7, when y = 3.

First Equation : y = 2x + 3

Step 1 :

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

Let x = -1.5, then y = 0

Let x = 0, then y = 3

Let x = 2, then y = 7

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
-1.5	0
0	3
2	7

Step 3 :

Plot the points (-1.5, 0), (0, 3), (2, 7) on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second 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 = 0, then y = 1

Let x = 1, then y = 4

Let x = 2, then y = 7

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
0	1
1	4
2	7

Step 3 :

Plot the points (0, 1), (1, 4), (2, 7) on a graph paper and then draw a straight line passing through the points plotted on the graph.

On the same graph paper, draw the graph for each given equation.

Both the straight lines drawn meet at a point. As it is clear from the graph, co-ordinates of the common point are (2, 7).

Hence, x = 2, y = 7.

First Equation : x + y = 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 = 0

Let x = 0, then y = 1

Let x = 4, then y = -3

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
1	0
0	1
4	-3

Step 3 :

Plot the points (1, 0), (0, 1), (4, -3) on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second Equation : 3x + 2y = 6

Step 1 :

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

Let x = 0, then y = 3

Let x = 2, then y = 0

Let x = 4, then y = -3

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
0	3
2	0
4	-3

Step 3 :

Plot the points (0, 3), (2, 0), (4, -3) on a graph paper and then draw a straight line passing through the points plotted on the graph.

On the same graph paper, draw the graph for each given equation.

Both the straight lines drawn meet at a point. As it is clear from the graph, co-ordinates of the common point are (4, -3).

Hence, x = 4, y = -3.

First Equation : y - 2x = 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 = -1

Let x = 0, then y = 1

Let x = 1, then y = 3

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
-1	-1
0	1
1	3

Step 3 :

Plot the points (-1, -1), (0, 1), (1, 3) on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second Equation : 5y - x = 14

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

Let x = -4, then y = 2

Let x = 6, then y = 4

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
1	3
-4	2
6	4

Step 3 :

Plot the points (1, 3), (-4, 2), (6, 4) on a graph paper and then draw a straight line passing through the points plotted on the graph.

On the same graph paper, draw the graph for each given equation.

Both the straight lines drawn meet at a point. As it is clear from the graph, co-ordinates of the common point are (1, 3).

Hence, x = 1, y = 3.

First Equation : x + 2y = 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

Let x = 3, then y = 0

Let x = -1, then y = 2

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
1	1
3	0
-1	2

Step 3 :

Plot the points (1, 1), (3, 0), (-1, 2) on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second Equation : 4x + 3y = 2

Step 1 :

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

Let x = 0.5, then y = 0

Let x = -1, then y = 2

Let x = 2, then y = -2

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
0.5	0
-1	2
2	-2

Step 3 :

Plot the points (0.5, 0), (-1, 2), (2, -2) on a graph paper and then draw a straight line passing through the points plotted on the graph.

On the same graph paper, draw the graph for each given equation.

Both the straight lines drawn meet at a point. As it is clear from the graph, co-ordinates of the common point are (-1, 2).

Hence, x = -1, y = 2.

First Equation : 2x + 3y = 2

Step 1 :

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

Let x = 1, then y = 0

Let x = -2, then y = 2

Let x = 4, then y = -2

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
1	0
-2	2
4	-2

Step 3 :

Plot the points (1, 0), (-2, 2), (4, -2) on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second Equation : x - 2y = 8

Step 1 :

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

Let x = 0, then y = -4

Let x = 8, then y = 0

Let x = 4, then y = -2

Step 2 :

Make a table for the corresponding values of x and y: :

x	y
0	-4
8	0
4	-2

Step 3 :

Plot the points (0, -4), (8, 0), (4, -2) on a graph paper and then draw a straight line passing through the points plotted on the graph.

On the same graph paper, draw the graph for each given equation.

Both the straight lines drawn meet at a point. As it is clear from the graph, co-ordinates of the common point are (4, -2).

Hence, x = 4, y = -2.

(i) A(7, 13) and B(10, 9)

By distance formula,

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

Substituting values we get :

$$\begin{aligned}AB = \sqrt{(10 - 7)^2 + (9 - 13)^2} \\[1em] = \sqrt{(3)^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5.\end{aligned}$$

Hence, AB = 5 units.

(ii) P(-4, 7) and Q(2, -5)

By distance formula,

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

Substituting values we get :

$$\begin{aligned}PQ = \sqrt{(2 - (-4))^2 + (-5 - 7)^2}\\[1em] =\sqrt{(6)^2 + (-12)^2}\\[1em] = \sqrt{36 + 144}\\[1em] = \sqrt{180}\\[1em] = 6 \sqrt{5}\end{aligned}$$

Hence, PQ = $6\sqrt{5}$ units.

(iii) C(4, -5) and D(12, -11)

By distance formula,

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

Substituting values we get :

$$\begin{aligned}CD = \sqrt{(12 -4)^2 + (-11 + 5)^2}\\[1em] =\sqrt{(8)^2 + (-6)^2}\\[1em] =\sqrt{64 + 36}\\[1em] =\sqrt{100}\\[1em] = 10\end{aligned}$$

Hence, CD = 10 units.

(iv) E(-6, -4) and F(9, -12)

By distance formula,

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

Substituting values we get :

$$\begin{aligned}EF = \sqrt{(9 - (-6))^2 + (-12 - (-4))^2}\\[1em] = \sqrt{(15)^2 + (-8)^2}\\[1em] = \sqrt{225 + 64}\\[1em] = \sqrt{289}\\[1em] = 17\end{aligned}$$

Hence, EF = 17 units.

(i) A(6, -6)

Origin = O(0, 0)

By distance formula,

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

Substituting values we get :

$$\begin{aligned}\text{OA } = \sqrt{(6 - 0)^2 + (-6 - 0)^2} \\[1em] = \sqrt{(6)^2 + (-6)^2} \\[1em] = \sqrt{36 + 36} \\[1em] = \sqrt{72} \\[1em] = 6\sqrt{2}.\end{aligned}$$

Hence, OA = $6\sqrt{2}$ units.

(ii) B(-5, 5)

Let origin = O(0, 0)

By distance formula,

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

Substituting values we get :

$$\begin{aligned}\text{OB} = \sqrt{(-5 - 0)^2 + (5 - 0)^2}\\[1em] = \sqrt{(-5)^2 + (5)^2}\\[1em] = \sqrt{25 + 25}\\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2}.\end{aligned}$$

Hence, OB = $5\sqrt{2}$ units.

(iii) C(4, -6)

Let origin = O(0, 0)

By distance formula,

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

Substituting values we get :

$$\begin{aligned}\text{OC } = \sqrt{(4 - 0)^2 + (-6 - 0)^2}\\[1em] = \sqrt{(4)^2 + (-6)^2}\\[1em] = \sqrt{16 + 36}\\[1em] = \sqrt{52}\\[1em] = 2\sqrt{13}.\end{aligned}$$

Hence OC = $2\sqrt{13}$ units.

Let the point on x-axis which is equidistant from the points A(2, -5) and B(-2, 9) be P(x, 0).

As point is equidistant from the points A and B.

Distance of A and P = Distance of P and B

AP = PB

$\Rightarrow \sqrt{(x - 2)^2 + (0 - (-5))^2} = \sqrt{(- 2 - x)^2 + (9 - 0)^2}$

Squaring both sides, we get :

⇒ (x - 2)² + 5² = (-2 - x)² + 9²

⇒ x² + 4 - 4x + 25 = x² + 4 + 4x + 81

⇒ -4x + 25 = 4x + 81

⇒ -4x - 4x + 25 - 81 = 0

⇒ -8x - 56 = 0

⇒ -8x = 56

⇒ x = -7.

Hence, P(-7, 0) is the point on x-axis which is equidistant from the points A(2, -5) and B(-2, 9).

Let the point on y-axis which is equidistant from the points A(-3, 2) and B(5, -2) be P(0, y).

As point is equidistant from the points A and B.

Distance of A and P = Distance of P and B

AP = PB

$\Rightarrow \sqrt{(0 - (-3))^2 + (y - 2)^2} = \sqrt{(5 - 0)^2 + ((-2) - y)^2}$

Squaring both sides, we get :

⇒ 3² + (y - 2)² = 5² + (-2 -y)²

⇒ 9 + y² + 4 - 4y = 25 + y² + 4 + 4y

⇒ y² - 4y + 13 = y² + 4y + 29

⇒ y² - y² + 4y + 4y = 13 - 29

⇒ 8y = -16

⇒ y = -2.

Hence, P(0, -2) is the point on y-axis which is equidistant from the points A(-3, 2) and B(5, -2).

We will find the length of all sides of the triangle ABC by using distance formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow AB = \sqrt{(6 -3)^2 + (4 - 0)^2}\\[1em] = \sqrt{(3)^2 + (4)^2}\\[1em] = \sqrt{9 + 16} = \sqrt{25} \\[1em] = 5\\[1em] \Rightarrow BC = \sqrt{(-1 -6)^2 + (3 - 4)^2}\\[1em] = \sqrt{(-7)^2 + (-1)^2}\\[1em] = \sqrt{49 + 1} = \sqrt{50} \\[1em] = 5\sqrt{2}\\[1em] \Rightarrow AC = \sqrt{(-1 -3)^2 + (3 - 0)^2}\\[1em] = \sqrt{(-4)^2 + (3)^2}\\[1em] = \sqrt{16 + 9} = \sqrt{25} \\[1em] = 5\end{aligned}$$

⇒ AB = AC

Also,

AB² + AC² = 5² + 5²

= 25 + 25

= 50.

BC² = $(5\sqrt{2})^2$ = 50.

Since,

AB² + AC² = BC²

⇒ △ABC is right angled triangle at A.

As two sides are equal.

⇒ △ABC is an isosceles right angled triangle.

Hence, proved that the points A(3, 0), B(6, 4) and C(-1, 3) are the vertices of an isosceles right-angled triangle.

Using distance formula,

$$\begin{aligned}\Rightarrow OA = \sqrt{(3 - 0)^2 + (\sqrt{3} - 0)^2}\\[1em] = \sqrt{(3)^2 + (\sqrt{3})^2}\\[1em] = \sqrt{9 + 3} \\[1em] = \sqrt{12} \\[1em] = 2\sqrt{3} \text{ units} \\[1em] \Rightarrow AB = \sqrt{(3 - 3)^2 + (-\sqrt{3} - \sqrt{3})^2}\\[1em] = \sqrt{(0)^2 + (-2\sqrt{3})^2}\\[1em] = \sqrt{0 + 12} \\[1em] = \sqrt{12} \\[1em] = 2\sqrt{3} \text{ units}\\[1em] \Rightarrow OB = \sqrt{(3 - 0)^2 + (-\sqrt{3} - 0)^2}\\[1em] = \sqrt{(3)^2 + (-\sqrt{3})^2}\\[1em] = \sqrt{9 + 3} \\[1em] = \sqrt{12} \\[1em] = 2\sqrt{3} \text{ units}.\end{aligned}$$

Since OA = AB = OB, all the sides of the triangle are equal.

Hence, the points O(0, 0), A(3, $\sqrt{3}$) and B(3, -$\sqrt{3}$) are the vertices of an equilateral triangle.

We will find the length of all sides by using distance formula.

$$\begin{aligned}\Rightarrow AB = \sqrt{(-1 -1)^2 + (5 - 1)^2}\\[1em] = \sqrt{(-2)^2 + (4)^2}\\[1em] = \sqrt{4 + 16} = \sqrt{20} \\[1em] = 2\sqrt{5}\\[1em] \Rightarrow BC = \sqrt{(7 -(-1))^2 + (9 - 5)^2}\\[1em] = \sqrt{(8)^2 + (4)^2}\\[1em] = \sqrt{64 + 16} = \sqrt{80} \\[1em] = 4\sqrt{5}\\[1em] \Rightarrow CD = \sqrt{(9 - 7)^2 + (5 - 9)^2}\\[1em] = \sqrt{(2)^2 + (4)^2}\\[1em] = \sqrt{4 + 16} = \sqrt{20} \\[1em] = 2\sqrt{5}\\[1em] \Rightarrow AD = \sqrt{(9 - 1)^2 + (5 - 1)^2}\\[1em] = \sqrt{(8)^2 + (4)^2}\\[1em] = \sqrt{64 + 16} = \sqrt{80} \\[1em] = 4\sqrt{5}\end{aligned}$$

⇒ AB = CD and BC = AD

Now,

We will find the length of diagonals AC and BD.

$$\begin{aligned}\Rightarrow AC = \sqrt{(7 -1)^2 + (9 - 1)^2}\\[1em] = \sqrt{(6)^2 + (8)^2}\\[1em] = \sqrt{36 + 64} = \sqrt{100} \\[1em] = 10\\[1em] \Rightarrow BD = \sqrt{(9 -(-1))^2 + (5 - 5)^2}\\[1em] = \sqrt{(9 + 1)^2 + (0)^2}\\[1em] =\sqrt{(10)^2}\\[1em] = 10\end{aligned}$$

⇒ AC = BD

Since, opposite sides are equal and diagonals are also equal.

⇒ ABCD is a rectangle.

Hence, the points A(1, 1), B(-1, 5), C(7, 9) and D(9, 5) are the vertices of a rectangle ABCD.

We will find the length of all sides by using distance formula.

$$\begin{aligned}\Rightarrow AB = \sqrt{(5 - 1)^2 + (4 - 2)^2}\\[1em] = \sqrt{(4)^2 + (2)^2}\\[1em] = \sqrt{16 + 4} = \sqrt{20} \\[1em] = 2\sqrt{5}\\[1em] \Rightarrow BC = \sqrt{(3 - 5)^2 + (8 - 4)^2}\\[1em] = \sqrt{(-2)^2 + (4)^2}\\[1em] = \sqrt{4 + 16} = \sqrt{20} \\[1em] = 2\sqrt{5}\\[1em] \Rightarrow CD = \sqrt{(-1 - 3)^2 + (6 - 8)^2}\\[1em] = \sqrt{(-4)^2 + (-2)^2}\\[1em] = \sqrt{16 + 4} = \sqrt{20} \\[1em] = 2\sqrt{5}\\[1em] \Rightarrow AD = \sqrt{(-1 -1)^2 + (6 - 2)^2}\\[1em] = \sqrt{(-2)^2 + (4)^2}\\[1em] = \sqrt{4 + 16} = \sqrt{20} \\[1em] = 2\sqrt{5}\end{aligned}$$

⇒ AB = BC = CD = AD

Now,

We will find the length of diagonals AC and BD.

$$\begin{aligned}\Rightarrow AC = \sqrt{(3 -1)^2 + (8 - 2)^2}\\[1em] = \sqrt{(2)^2 + (6)^2}\\[1em] = \sqrt{4 + 36} = \sqrt{40} \\[1em] = 2\sqrt{10}\\[1em] \Rightarrow BD = \sqrt{(-1 - 5)^2 + (6 - 4)^2}\\[1em] =\sqrt{(-6)^2 + (2)^2}\\[1em] = \sqrt{36 + 4}\\[1em] =\sqrt{(40)}\\[1em] = 2\sqrt{10}\end{aligned}$$

⇒ AC = BD

Since, all sides are equal and diagonals are also equal.

⇒ ABCD is a square.

Hence, the points A(1, 2), B(5, 4), C(3, 8) and D(-1, 6) are the vertices of a square.

By distance formula,

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

Calculate the lengths :

$$\begin{aligned}\Rightarrow AB = \sqrt{(3 - 2)^2 + (4 - (-1))^2}\\[1em] = \sqrt{(1)^2 + (5)^2}\\[1em] = \sqrt{1 + 25}\\[1em] = \sqrt{26} \text{ units}\\[1em] \Rightarrow BC = \sqrt{(-2 -3)^2 + (3 - 4)^2}\\[1em] = \sqrt{(-5)^2 + (-1)^2}\\[1em] = \sqrt{25 + 1}\\[1em] = \sqrt{26} \text{ units}\\[1em] \Rightarrow CD = \sqrt{(-3 - (-2))^2 + (-2 - 3)^2}\\[1em] = \sqrt{(-1)^2 + (-5)^2}\\[1em] = \sqrt{1 + 25}\\[1em] = \sqrt{26} \text{ units}\\[1em] \Rightarrow AD = \sqrt{(-3 -2)^2 + (-2 -(-1))^2}\\[1em] = \sqrt{(-5)^2 + (-1)^2}\\[1em] = \sqrt{25 + 1}\\[1em] = \sqrt{26} \text{ units}.\end{aligned}$$

As, AB = BC = CD = AD

Hence, the points A(2, -1), B(3, 4), C(-2, 3) and D(-3, -2) are the vertices of a rhombus.

As y = 0 on x-axis, the point (2, 0) lies on x-axis.

Hence, option 3 is the correct option.

As x = 0 on y-axis, the point (0, 3) lies on y-axis.

Hence, option 2 is the correct option.

In the IV quadrant, x is positive and y is negative.

So, the point (1, -4) lies in the IV quadrant.

Hence, option 2 is the correct option.

In the III quadrant, both x and y are negative.

So, the point (-2, -3) lies in the III quadrant.

Hence, option 4 is the correct option.

A line parallel to the x-axis has the form y = constant.

So, y = 2 is parallel to x-axis.

Hence, option 2 is the correct option.

P is on x-axis, then its y coordinate is 0.

Distance is 3 units from y-axis to its right.

Thus, x-coordinate is positive.

⇒ x = 3

So, coordinates of P = (3, 0).

Hence, option 1 is the correct option.

Let O(0, 0) be the origin.

Using distance formula,

PO = $\sqrt{(0 - 4)^2 + (0 - (-3))^2}$

= $\sqrt{(-4)^2 + (3)^2}$

= $\sqrt{16 + 9} = \sqrt{25}$

= 5 units.

Hence, option 3 is the correct option.

Let the point on x-axis be P(x, 0) which is equidistant from the points A(7, 6) and B(-3, 4).

As point is equidistant from the points A and B.

Distance of A and P = Distance of P and B

AP = PB

$\Rightarrow \sqrt{(x - 7)^2 + (0 - 6)^2} = \sqrt{(- 3 - x)^2 + (4 - 0)^2}$

Squaring both sides, we get :

⇒ (x - 7)² + (-6)² = (-3 - x)² + 4²

⇒ x² + 49 - 14x + 36 = x² + 9 + 6x + 16

⇒ -14x + 85 = 6x + 25

⇒ -14x - 6x - 25 + 85 = 0

⇒ -20x + 60 = 0

⇒ 20x = 60

⇒ x = 3

∴ P(3, 0) is the point on x-axis which is equidistant from the points A(7, 6) and B(-3, 4).

Hence, option 3 is the correct option.

1. From graph,

The coordinates of S are (-2, 3).

Hence, option (a) is the correct option.

2. As abscissa is the x-coordinate of a point.

M = (3, 1).

So, the abscissa of the point M is 3.

Hence, option (b) is the correct option.

3. As Dinesh is represented by D.

Coordinates of D = (5, -2)

The perpendicular distance from the x-axis is the absolute value of the y-coordinate = 2 units.

Hence, option (c) is the correct option.

4. Rishabh is represented by R.

The coordinates of R = (-1, -1)

Sum = Abscissa + ordinate = -1 + (-1) = -2.

Hence, option (d) is the correct option.

5. Kavita's position is on the x-axis and coordinates are (-4, 0).

Therefore, Kavita is nearest to the x-axis.

Hence, option (b) is the correct option.

1. The distance of a point from the y-axis is given by the absolute value of its x-coordinate.

Coordinates of S = (12, 8)

⇒ Distance of the point S from y-axis is 12 units.

Hence, option (d) is the correct option.

2. Coordinates of D = (4, 4)

Coordinates of R = (14, 8)

We will find distance between the points D and R by using distance formula,

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

Substituting values we get :

$$\begin{aligned}DR = \sqrt{(14 - 4)^2 + (8 - 4)^2}\\[1em] = \sqrt{(10)^2 + (4)^2}\\[1em] = \sqrt{100 + 16} \\[1em] = \sqrt{116} \text{ units}\end{aligned}$$

DR = $\sqrt{116}$ units

Hence, option (b) is the correct option.

3. Coordinates of S = (12, 8)

Coordinates of Q = (14, 4)

We will find the length of diagonal SQ by using distance formula,

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

Substituting values we get :

$$\begin{aligned}SQ = \sqrt{(14 - 12)^2 + (4 - 8)^2}\\[1em] = \sqrt{(2)^2 + (-4)^2}\\[1em] = \sqrt{4 + 16}\\[1em] = \sqrt{20}\\[1em] = 2\sqrt{5}\end{aligned}$$

SQ = $2\sqrt{5}$ units

Hence, option (c) is the correct option.

4. As,

Coordinates of A = (4, 2)

Coordinates of B = (6, 2)

Coordinates of C = (6, 4)

Coordinates of D = (4, 4)

We will find out the length of all sides of ABCD by using distance formula,

$$\begin{aligned}\Rightarrow AB = \sqrt{(6 -4)^2 + (2 - 2)^2}\\[1em] = \sqrt{(2)^2 + (0)^2}\\[1em] = \sqrt{4 + 0} = \sqrt{4} \\[1em] = 2 \\[1em] \Rightarrow BC = \sqrt{(6 -6)^2 + (4 -2)^2}\\[1em] = \sqrt{(0)^2 + (2)^2}\\[1em] = \sqrt{4} \\[1em] = 2\\[1em] \Rightarrow CD = \sqrt{(4 - 6)^2 + (4 - 4)^2}\\[1em] = \sqrt{(-2)^2 + (0)^2}\\[1em] = \sqrt{4 + 0} = \sqrt{4} \\[1em] = 2\\[1em] \Rightarrow AD = \sqrt{(4 - 4)^2 + (4 - 2)^2}\\[1em] = \sqrt{(0)^2 + (2)^2}\\[1em] = \sqrt{0 + 4} = \sqrt{4} \\[1em] = 2\end{aligned}$$

⇒ AB = BC = CD = AD

Now we will check for Diagonals,

$$\begin{aligned}\Rightarrow AC = \sqrt{(6 -4)^2 + (4 - 2)^2}\\[1em] = \sqrt{(2)^2 + (2)^2}\\[1em] = \sqrt{4 + 4} = \sqrt{8} \\[1em] = 2\sqrt{2}\\[1em] \Rightarrow BD = \sqrt{(4 - 6)^2 + (4 - 2)^2}\\[1em] = \sqrt{(-2)^2 + (2)^2}\\[1em] = \sqrt{4 + 4} = \sqrt{8} \\[1em] = 2\sqrt{2}\\[1em]\end{aligned}$$

⇒ Diagonal AC and BD are also equal.

⇒ ABCD is a square.

Hence, option (a) is the correct option.

5. As,

Coordinates of P = (12, 4)

Coordinates of Q = (14, 4)

Coordinates of R = (14, 8)

Coordinates of S = (12, 8)

We will find out the length of all sides of PQRS by using distance formula,

$$\begin{aligned}\Rightarrow PQ = \sqrt{(14 - 12)^2 + (4 - 4)^2}\\[1em] = \sqrt{(2)^2 + (0)^2}\\[1em] = \sqrt{4 + 0} = \sqrt{4} \\[1em] = 2 \\[1em] \Rightarrow QR = \sqrt{(14 - 14)^2 + (8 - 4)^2}\\[1em] = \sqrt{(0)^2 + (4)^2}\\[1em] = \sqrt{16} \\[1em] = 4\\[1em] \Rightarrow RS = \sqrt{(12 - 14)^2 + (8 - 8)^2}\\[1em] = \sqrt{(-2)^2 + (0)^2}\\[1em] = \sqrt{4 + 0} = \sqrt{4} \\[1em] = 2\\[1em] \Rightarrow SP = \sqrt{(12 - 12)^2 + (8 - 4)^2}\\[1em] = \sqrt{(0)^2 + (4)^2}\\[1em] = \sqrt{0 + 16} = \sqrt{16} \\[1em] = 4\end{aligned}$$

⇒ PQ = RS and QR = SP

⇒ Opposite sides are equal.

Now we will check for Diagonals,

$$\begin{aligned}\Rightarrow PR = \sqrt{(14 - 12)^2 + (8 - 4)^2}\\[1em] = \sqrt{(2)^2 + (4)^2}\\[1em] = \sqrt{4 + 16}\\[1em] = \sqrt{20} \\[1em] = 2\sqrt{5}\\[1em] \Rightarrow SQ = \sqrt{(14 - 12)^2 + (4 - 8)^2}\\[1em] = \sqrt{(2)^2 + (-4)^2}\\[1em] = \sqrt{4 + 16}\\[1em] = \sqrt{20} \\[1em] = 2\sqrt{5}\\[1em]\end{aligned}$$

⇒ Diagonal PR and SQ are also equal.

⇒ PQRS is a rectangle.

Hence, option (b) is the correct option.

As coordinates of P are (7, 3)

So, the distance from x-axis 3 and distance from y-axis is 7.

⇒ Assertion (A) is false .

As x-coordinate of a point is called abscissa and y-coordinate is called ordinate.

⇒ Reason (R) is true.

Hence, option 2 is the correct option.

We will find distance between O and P by using distance formula,

$$\begin{aligned}OP = \sqrt{(4 - 0)^2 + (3 - 0)^2}\\[1em] =\sqrt{4^2 + 3^2}\\[1em] =\sqrt{16 + 9}\\[1em] =\sqrt{25}\\[1em] = 5\end{aligned}$$

⇒ Assertion (A) is true.

As Distance Formula for two points A and B is

AB = $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

⇒ Reason (R) is correct.

Hence, option 3 is the correct option.

As, x = 0 and y = -2

Thus, the point (0, -2) lies on y-axis.

⇒ Assertion (A) is true.

Reason (R) states that any point of the form (0, y) lies on the x-axis, which is false. In fact, any point of the form (0, y) lies on the y-axis.

⇒ Reason (R) is false.

Hence, option 1 is the correct option.

Point = (2, 8)

Distance from x-axis = 8 units

Hence, option 2 is the correct option.

A point is 4 units from the origin.

⇒ All points lie on a circle of radius 4.

⇒ In the first quadrant, there are infinitely many points on that arc.

Hence, option 4 is the correct option.

According to question :

Length of AB = 7 - (-5) = 7 + 5 = 12.

We will find the length of AB by using distance formula,

$$\begin{aligned}AB = \sqrt{(5 - 5)^2 + (-5 -7)^2}\\[1em] = \sqrt{0 + (-12)^2}\\[1em] = \sqrt{144} \\[1em] = 12 \text{ units}.\end{aligned}$$

Hence, option 1 is the correct option.

As the point (0, 5) lies on y-axis.

⇒ It does not lie in any quadrant.

Hence, option 4 is the correct option.

As, distance from x-axis is 3 units.

⇒ |y| = 3

Distance from y-axis is 4 units.

⇒ |x| = 4

Thus, the points can be (4, 3), (4, -3), (-4, 3), (-4, -3).

Hence, option 4 is the correct option.

Distance from x-axis = Absolute value of y coordinate

Smallest distance = 2

⇒ (2, 2) is the nearest point to x-axis.

Hence, option 3 is the correct option.

Centre = O(0, 0)

Radius = 4

Equation of circle :

x² + y² = radius²

x² + y² = 4²

x² + y² = 16

If point lies inside the circle,

Equation will be :

x² + y² < 16

If point lies outside the circle,

Equation will be :

x² + y² > 16

Now, we will check for each point.

1. (4, 2)

4² + 2² = 16 + 4 = 20

20 > 16

So, point is outside the circle.

2. (2, 2)

2² + 2² = 4 + 4 = 8

8 < 16

So, point is inside the circle.

3. (-5, 1)

(-5)² + 1² = 25 + 1 = 26

26 > 16

So, point is outside the circle.

4. (-1, -2)

(-1)² + (-2)² = 1 + 4 = 5

5 < 16

So, point is inside the circle.

5. (-4, 1)

(-4)² + 1² = 16 + 1 = 17

17 > 16

So, point is outside the circle.

6. (1, 1)

1² + 1² = 1 + 1 = 2

2 < 16

So, point is inside the circle.

7. (-2, 4)

(-2)² + 4² = 4 + 16 = 20

20 > 16

So, point is outside the circle.

Hence, points lying

Inside the circle :(2, 2), (-1, -2), (1, 1)

Outside the circle : (4, 2), (-5, 1), (-4, 1), (-2, 4)

From graph,

A = (-2, 5) and B = (7, -3)

As,

x-coordinate of point C = x-coordinate of point A

y-coordinate of point C = y-coordinate of point B

Coordinates of C = (-2, -3)

Now,

Area of right triangle = $\frac{1}{2} \times \text {base} \times \text{height}$

We will find base and height by using distance formula,

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

Base = CB

$$\begin{aligned}CB =\sqrt{(7 - (-2))^2 + (-3 - (-3))^2}\\[1em] =\sqrt{(7 + 2)^2 + (-3 + 3)^2}\\[1em] =\sqrt{(9)^2 + (0)^2}\\[1em] =\sqrt{81}\\[1em] = 9.\end{aligned}$$

Base = 9 units

Height = AC

$$\begin{aligned}AC =\sqrt{(-2 - (-2))^2 + (-3 - 5)^2}\\[1em] =\sqrt{(-2 + 2)^2 + (-3 - 5)^2}\\[1em] =\sqrt{(0)^2 + (-8)^2}\\[1em] =\sqrt{64}\\[1em] = 8.\end{aligned}$$

Height = 8 units

Area of right triangle ABC = $\frac{1}{2} \times \text {base} \times \text{height}$

= $\frac{1}{2} \times 9 \times 8$

= 36 sq. units

Hence, area of triangle = 36 sq. units.