Find the coordinates of the point \(Q\) on the x–axis which lies on the perpendicular bisector of the segment joining \(A(-5,-2)\) and \(B(4,-2)\). Also, name the type of triangle formed by \(Q, A, B\).
\(Q = \left(-\dfrac{1}{2}, 0\right)\)
Triangle \(QAB\) is isosceles (\(QA=QB\)).
Step 1: Let \(Q(x,0)\) be a point on the x-axis. Since \(Q\) lies on the perpendicular bisector of \(AB\), it is equidistant from \(A\) and \(B\). Hence,
\(QA = QB\) or \(QA^2 = QB^2\).
Step 2: Write \(QA^2\) and \(QB^2\) using the distance formula.
\(A(-5,-2)\), \(B(4,-2)\), \(Q(x,0)\)
\(QA^2 = (x-(-5))^2 + (0-(-2))^2 = (x+5)^2 + (-2)^2\)
\(QB^2 = (x-4)^2 + (0-(-2))^2 = (x-4)^2 + (-2)^2\)
Step 3: Equate \(QA^2\) and \(QB^2\) and simplify.
\((x+5)^2 + (-2)^2 = (x-4)^2 + (-2)^2\)
\((x+5)^2 + 4 = (x-4)^2 + 4\)
Cancel 4 from both sides:
\((x+5)^2 = (x-4)^2\)
Step 4: Expand and solve for \(x\).
\(x^2 + 10x + 25 = x^2 - 8x + 16\)
\(10x + 8x = 16 - 25\)
\(18x = -9\)
\(x = -\dfrac{9}{18} = -\dfrac{1}{2}\)
Hence, \(Q = \left(-\dfrac{1}{2}, 0\right)\).
Step 5: Now find \(QA\) and \(QB\) to name the triangle.
\(QA^2 = \left(-\dfrac{1}{2}+5\right)^2 + (0+2)^2 = \left(\dfrac{9}{2}\right)^2 + 2^2 = \dfrac{81}{4} + 4 = \dfrac{81+16}{4} = \dfrac{97}{4}\)
\(\Rightarrow QA = \sqrt{\dfrac{97}{4}} = \dfrac{\sqrt{97}}{2}\) units.
\(QB^2 = \left(-\dfrac{1}{2}-4\right)^2 + (0+2)^2 = \left(-\dfrac{9}{2}\right)^2 + 2^2 = \dfrac{81}{4} + 4 = \dfrac{97}{4}\)
\(\Rightarrow QB = \dfrac{\sqrt{97}}{2}\) units.
Step 6: Since \(QA = QB\), triangle \(\triangle QAB\) has two equal sides.
\(AB = \sqrt{(4-(-5))^2 + (-2-(-2))^2} = \sqrt{9^2+0} = 9\) units.
Therefore, \(\triangle QAB\) is an isosceles triangle.