Question

Find the coordinates of the point Q on the X-axis which lies on the perpendicular bisector of the line segment joining A the points A(-5,-2) and B(4,-2), name the type of triangle formed by the point Q, A and B

Answer 1

$A\equiv (-5,-2),B\equiv (4,-2)$
Let C $\to$ mid-point of AB
$C\equiv (\frac{-5+4}{2},\frac{-2-2}{2})\equiv (-\frac{1}{2},-2)$
$\therefore C\equiv (-\frac{1}{2},-2)$
Perpendicular drawn from C to x-axis touches x-axis at Q.
Co-ordinates of $Q\equiv (-\frac{1}{2},0)$
For $\Delta ABQ,AB=\sqrt{(4+5{)}^2+(-2+2{)}^2}=9$
$AQ=\sqrt{{(-5+\frac{1}{2})}^2+(-2-0{)}^2}=4.924$
$\sqrt{{(4+\frac{1}{2})}^2+(-2+0{)}^2}=4.924$
AQ = BQ
$\therefore \Delta ABQ$ is isosceles triangle.