1(xiii)
In the given diagram, ΔAOB is a right-angled triangle and C is the mid-point of AB. The coordinates of the point which is equidistant from the three vertices of ΔAOB is :
- (a)(x, y)
- (b)(y, x)
- (c)$\Big(\frac{x}{2}, \frac{y}{2}\Big)$
- (d)$\Big(\frac{2x}{3}, \frac{2y}{3}\Big)$
Show SolutionHide Solution↓
Given,
C is midpoint of AB.
We know that,
In a right-angled triangle, the point equidistant from all three vertices is the midpoint of the hypotenuse.
Using midpoint formula :
M = $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big)$
Substituting values, we get :
$$\begin{aligned}\Rightarrow C = \Big(\frac{0 + 2x}{2}, \frac{2y + 0}{2} \Big) \\[1em] \Rightarrow C = \Big(\frac{2x}{2}, \frac{2y}{2} \Big) \\[1em] \Rightarrow C = \text{(x, y)}.\end{aligned}$$
Hence, option 1 is the correct option.