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Show that the points P(a, b + c), Q(b, c + a) and R(c, a + b) are collinear.

Straight Line Eq

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Answer

By formula,

Slope = y2−y1x2−x1\dfrac{y2 - y1}{x2 - x1}

Slope of PQ=c+a−(b+c)b−a=c+a−b−cb−a=a−bb−a=−(b−a)b−a=−1.Slope of QR=a+b−(c+a)c−b=a+b−c−ac−b=b−cc−b=−(c−b)c−b=−1.\text{Slope of PQ} = \dfrac{c + a - (b + c)}{b - a} \\[1em] = \dfrac{c + a - b - c}{b - a} \\[1em] = \dfrac{a - b}{b - a} \\[1em] = \dfrac{-(b - a)}{b - a} = -1. \\[1em] \text{Slope of QR} = \dfrac{a + b - (c + a)}{c - b} \\[1em] = \dfrac{a + b - c - a}{c - b} \\[1em] = \dfrac{b - c}{c - b} \\[1em] = \dfrac{-(c - b)}{c - b} = -1.

Since, Slope of PQ = QR.

Hence, proved that P, Q and R are collinear.

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