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# 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 $.

The line segment joining the points \( A(-5,-2) \) and \( B(4,-2) \).

To do:

We have to 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 \( B(4,-2) \) and name the type of triangle formed by the points \( Q, A \) and \( B \).

Solution:

Let the coordinates of the point $Q$ be $(x, 0)$.

Point $Q$ is equidistant from $( A(5,2)$ and $B(4,2)$

This implies,

$AQ=BQ$

Using distance formula,

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$A Q=\sqrt{(x-5)^{2}+(0-2)^{2}}$

$B Q=\sqrt{(x-4)^{2}+(0-2)^{2}}$

Therefore,

$\sqrt{(x-5)^{2}+(0-2)^{2}}=\sqrt{(x-4)^{2}+(0-2)^{2}}$

Squaring on both sides, we get,

$(x-5)^2+(-2)^2=(x-4)^2+(-2)^2$

$x^2+25-2(x)(5)+4=x^2-2(x)(4)+4^2+4$

$25-10x=-8x+16$

$10x-8x=25-16$

$2x=9$

$x=\frac{9}{2}$

The co-ordinates of the point $Q$ are $(\frac{9}{2},0)$.

Here,

$A Q=B Q$ and $Q$ lies on the perpendicular bisector of $A B$

This implies,

Triangle $\mathrm{ABQ}$ is an isosceles triangle.

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