Question

P(-5,a),Q(b,7),R(1,-3)points are collinear such that PQ=QR find the value of a & b

Answer 1

P(-5,a),Q(b,7),R(1,-3)points are collinear such that PQ=QR. Find the value of a & b.

Solution:

There are two cases: green line and red line. The second case (red line) is impossible because $PQ \neq QR$. Thus we have only one case (green line). Further

$$\angle PQC = \angle RQD = \alpha$$

because these angles are vertical ones. And we have

$$\begin{array}{l} \angle CPQ = \pi - \angle PCQ - \angle PQC = \pi - \frac{\pi}{2} - \alpha = \frac{\pi}{2} - \alpha; \\ \angle DRQ = \pi - \angle RDQ - \angle PQD = \pi - \frac{\pi}{2} - \alpha = \frac{\pi}{2} - \alpha. \end{array}$$

Thus $\angle CPQ \cong \angle DRQ$. If $PQ = QR$ then $\Delta RDQ \cong \Delta PCQ$ (by Angle-Side-Angle (ASA) Congruence). So

$$\begin{array}{l} QC = DQ, \\ b - (-5) = 1 - b, \\ b + 5 = 1 - b, \\ 2b = -4, \end{array}$$ $$b = -2$$

Also we have

$$\begin{array}{l} DR = CP, \\ 7 - (-3) = a - 7, \\ 7 + 3 = a - 7, \end{array}$$ $$a = 17$$

Answer:

$$a = 17, \quad b = -2.$$