Name: Centroid - Meaning
Uploaded: 2019-07-19
Duration: 2 min 53 s
Description: Medians of a triangle are concurrent and their point of concurrence is from each vertex at a distance equal to 2/3 of the length of that median. Explore the...

Question

If the point

P (x, y)

is equidistant from the points

A (5, 1)

and

B (1, 5)

, then prove that

x = y

.

Accepted Answer

Given that the point $P (x, y)$ is equidistant from the points $A (5, 1)$ and $B (1, 5)$ . We have to prove that $x = y$ .

Since the point $P (x, y)$ has the same distance from the points $A (5, 1)$ and $B (1, 5)$ , we can write

$P A = P B$

$\Rightarrow \sqrt{{(5 - x)}^{2} + {(1 - x)}^{2}} = \sqrt{(1 - y)^{2} + (5 - y)^{2}}$ [The distance between two points is given by $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ , where $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are the coordinates of the two points]

$\Rightarrow (5 - x)^{2} + (1 - x)^{2} = (1 - y)^{2} + (5 - y)^{2}$

$\Rightarrow 25 - 10 x + x^{2} + 1 - 2 x + x^{2} = 1 - 2 y + y^{2} + 25 - 10 y + y^{2}$

$\Rightarrow - 10 x + x^{2} - 2 x + x^{2} = - 2 y + y^{2} + - 10 y + y^{2}$

$\Rightarrow 2 x^{2} - 12 x = 2 y^{2} - 12 y$

$\Rightarrow x^{2} - 6 x = y^{2} - 6 y$

$\Rightarrow x^{2} - 6 x + 9 = y^{2} - 6 y + 9$ [Add $9$ to both sides of the equation]

$\Rightarrow {(x - 3)}^{2} = {(y - 3)}^{2}$

$\Rightarrow x - 3 = y - 3$

$\Rightarrow x = y$ [Proved]

If the point $P (x, y)$ is equidistant from the points $A (5, 1)$ and $B (1, 5)$ , then prove that $x = y$ .

Important Questions on Coordinate Geometry

Learn and Prepare for any exam you want !

If the point P(x,y) is equidistant from the points A(5,1) and B(1,5), then prove that x=y.

Important Questions on Coordinate Geometry

Learn and Prepare for any exam you want !

If the point $P (x, y)$ is equidistant from the points $A (5, 1)$ and $B (1, 5)$ , then prove that $x = y$ .