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 $B$.

(N/A) The points are $A(-5, -2)$ and $B(4, -2)$.
Since the $y$-coordinates of both points are the same $(-2)$,the line segment $AB$ is a horizontal line parallel to the $x$-axis.
The perpendicular bisector of a horizontal line is a vertical line passing through the midpoint of the segment.
The midpoint $R$ of $AB$ is given by $\left(\frac{-5+4}{2}, \frac{-2-2}{2}\right) = \left(-\frac{1}{2}, -2\right)$.
The perpendicular bisector is the vertical line $x = -\frac{1}{2}$.
The point $Q$ lies on the $x$-axis,so its $y$-coordinate is $0$. Since it lies on the perpendicular bisector $x = -\frac{1}{2}$,the coordinates of $Q$ are $\left(-\frac{1}{2}, 0\right)$.
To identify the triangle $QAB$,we calculate the lengths of the sides:
$AB = \sqrt{(4 - (-5))^2 + (-2 - (-2))^2} = \sqrt{9^2 + 0^2} = 9$.
$QA = \sqrt{(-5 - (-0.5))^2 + (-2 - 0)^2} = \sqrt{(-4.5)^2 + (-2)^2} = \sqrt{20.25 + 4} = \sqrt{24.25}$.
$QB = \sqrt{(4 - (-0.5))^2 + (-2 - 0)^2} = \sqrt{(4.5)^2 + (-2)^2} = \sqrt{20.25 + 4} = \sqrt{24.25}$.
Since $QA = QB$,the triangle $QAB$ is an isosceles triangle.