Question

Find the coordinates of the point $Q$ on the $x$-axis which lies on the perpendicular bisector of the line segment joining the points $A(-5,-2)$ and $B(4,-2)$.

Name the type of triangle formed by the points $Q$, $A$, and $B$.

Answer 1

Step 1: Find the midpoint of $AB$ using midpoint formula

Let $A(x_1,y_1)=(-5,-2)$

$B(x_2,y_2)=(4,-2)$

Let $R$ be the midpoint of $AB$

Using the midpoint formula

$R(x,y)=\left(\frac{(x_1+x_2)}{2},\frac{(y_2+y_1)}{2}\right)$

$\Rightarrow$ $\begin{array}{rcl}& =& \left(\frac{4-5}{2},\frac{-2-2}{2}\right)\end{array}$

$\Rightarrow$ $R(x,y)=\left(\frac{-1}{2},-2\right)$

Step 2 : Find slope of line $AB$

Using the formula for slope of line,

$m=\frac{y_2-y_1}{x_2-x_1}$

$\Rightarrow$ $m=\frac{-2+2}{4+5}$

$\Rightarrow$ $m=0$

Step 3 : Find slope of $QR$

Slopes of perpendicular lines are negative reciprocals

$\Rightarrow$ Slope of $QR=-\infty$

$\Rightarrow$ $x_r-x_q=0$

$\Rightarrow$ $x_q=x_r$ $...\left(i\right)$

Step 4 : Find the co-ordinates of point $Q$

It is given that point $Q$ lies on the on the $x$ axis

$\Rightarrow$ $Q=\left(x_q,0\right)$

From $\left(i\right)$ we get,

$x_q=\frac{-1}{2}$

$\Rightarrow$ $Q=\left(\frac{-1}{2},0\right)$

Step 5: Identify the type of triangle

The vertex of the triangle lies on the perpendicular bisector of the base.

Hence, the triangle can only be an isosceles r equilateral triangle

Applying the distance formula we get,

$AQ=\sqrt{{\left(x_q-x_a\right)}^2+{\left(y_q-y_a\right)}^2}$

$\Rightarrow$ $AQ=\sqrt{{\left(\frac{-1}{2}+5\right)}^2+{\left(0+2\right)}^2}$

$\Rightarrow$ $AQ=\frac{\sqrt{97}}{2}$

Similarly,

$AB=\sqrt{{\left(-5-4\right)}^2+{\left(-2+2\right)}^2}$

$\Rightarrow$ $AB=9$

Thus the base of the triangle is not equal to the other side of the triangle.

Hence, the triangle cannot be an equilateral triangle.

Hence, the triangle formed is an isosceles triangle

Hence, the co-ordinates of the point $Q$ are $\left(\frac{-1}{2},0\right)$ and the triangle thus formed is an isosceles triangle.