Question

Find the point on the x-axis which is equidistant from (2, - 5) and (- 2, 9). (2 Marks)

Answer 1

We have to find a point on x-axis. Therefore, its y-coordinate will be 0.
Let the point on x-axis be (x,0)
Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}\left(1Mark\right)$

Distance between $(x,0)$ and $(2,-5)$
$=\sqrt{(x-2{)}^2+(0-(-5){)}^2}=\sqrt{(x-2{)}^2+(5{)}^2}$
Distance between $(x,0)$ and $(-2,9)$
$=\sqrt{(x-(-2){)}^2+(0-\left(9\right){)}^2}=\sqrt{(x+2{)}^2+(9{)}^2}$
By the given condition,
$\sqrt{(x-2{)}^2+(5{)}^2}=\sqrt{(x+2{)}^2+(9{)}^2}$
$(x-2{)}^2+25=(x+2{)}^2+81$
$x^2+4-4x+25=x^2+4+4x+81$
$8x=25-81$
$8x=-56$
$x=-7\left(1Mark\right)$
Therefore, the point is $(-7,0)$.