Question

Find the point on the x-axis which is equidistant from the points $(-2,5)$ and $(2,-3)$. Hence find the area of the triangle formed by these points

• A. $(-2,0);$ $10$ sq. units
• B. $(-2,5);$ $10$ sq. units
• C. $(-5,0);$ $10$ sq. units
• D. $(-5,2);$ $10$ sq. units

Answer 1

The correct option is A $(-2,0);$ $10$ sq. units

Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ can be calculated using the formula $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
Let the point on the x-axis be $(x,0)$
Distance between $(x,0)$ and $(-2,5)=\sqrt{{(-2-x)}^2+{(5-0)}^2}=\sqrt{4+x^2+4x+25}=\sqrt{x^2+4x+29}$
Distance between $(x,0)$ and $(2,-3)=\sqrt{{(2-x)}^2+{(-3-0)}^2}=\sqrt{4+x^2-4x+9}=\sqrt{x^2-4x+13}$
As the point $(x,0)$ is equidistant from the two points, both the distances calculated are equal.
$\sqrt{x^2+4x+29}=\sqrt{x^2-4x+13}$
$=>x^2+4x+29=x^2-4x+13$
$4x+4x=13-29$
$8x=-16$
$x=-2$
Thus, the point is $(-2,0)$

Area of a triangle with vertices $(x_1,y_1)$ ; $(x_2,y_2)$ and $(x_3,y_3)$ is $\left|\frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\right|$
Hence, substituting the points $(x_1,y_1)=(-2,5)$ ; $(x_2,y_2)=(2,-3)$ and $(x_3,y_3)=(-2,0)$ in the area formula, we get area of triangle $=\left|\frac{(-2)(-3+0)+\left(2\right)(0-5)+(-2)(5+3)}{2}\right|=\left|\frac{6-10-16}{2}\right|=\frac{20}{2}=10$ sq. units