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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.

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Solution.        
\text{Use distance formula}= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}
Let Q(x, 0) {\because on x-axis y coordinate is zero}
Q lies on the perpendicular bisector of the line AB i.e.
AQ = BQ
Squaring both sides
AQ2 = BQ2
(x – 5)2 + (0 + 2)2 = (x – 4)2 + (0 + 2)2
x2 + 25 – 10x + 4 = x2 + 16 – 8x + 4
29 – 10x = 20 – 8x
29 – 20 = –8x + 10x
9 = 2x
\frac{9}{2}= x
x = 4.5
\therefore The co-ordinate of Q is (4.5, 0)
\because AQ = BQ and Q lies on the perpendicular bisector of the line AB
\therefore  \triangleABQ is an isosceles triangle.

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