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

Open in App

Solution

$\begin{matrix} P (a, b + c) = (x_{1}, y_{1}) Q = (b, c + a) = (x_{1}, y_{2}) R = (c, a + b) = (x_{3}, y_{3}) m_{p q} = s l o p e o f P Q = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{c + a - b - c}{b - a} = \frac{a - b}{- (a - b)} = - 1 m_{Q R =} s l o p e o f Q R = \frac{y_{3} - y_{2}}{x_{3} - x_{2}} = \frac{a + b - c - a}{c - b} = \frac{b - c}{- (b - c)} = - 1 S l o p e o f P Q = S l o p e o f Q R s o, p o i n t s A, B, C a r e c o l l i n e a r . \end{matrix}$

Suggest Corrections

Similar questions

\frac{1}{x} + \frac{1}{y} = \frac{1}{z} .

p^{a} = q^{b} = r^{c} = s^{d}

p, q, r, s

a, b, c, d

A (a, b + c), B (b, c + a)

C (c, a + b)

Join BYJU'S Learning Program