Question

The point (-5,1) , (1,p) and (4,-2) are collinear if the value of p is​

Answer 1

$\frak{Here} \begin{cases} & \sf{(x_1 , y_1) = \bf{(-5, 1)}} \\ & \sf{(x_2 , y_2) = \bf{(1,p)}} \\ & \sf{(x_3 , y_3) = \bf{(4,-2)}}\end{cases}$

The Given points are collinear which means the area of the triangle formed by the collinear points is 0.

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

We know that,

$\star\;{\boxed{\sf{\purple{Area_{ \triangle} = \dfrac{1}{2} \bigg[ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \bigg]}}}}$

Therefore,

⠀⠀⠀

$:\implies\sf \dfrac{1}{2} \bigg[ -5(p - (- 2)) + 1(-2 - 1) + 4(1 - p) \bigg] = 0$

$:\implies\sf \dfrac{1}{2} \bigg[ -5(p + 2) + 1(-2 - 1) + 4(1 - p) \bigg] = 0$

$:\implies\sf \dfrac{1}{2} \bigg[ -5p + 10 - 3 + 4 - 4p\bigg] = 0$

$:\implies\sf \dfrac{1}{2} \bigg[- 5p - 9 - 4p \bigg] = 0$

$:\implies\sf \dfrac{1}{2} \bigg[- 5p - 4p - 9 \bigg] = 0$

$:\implies\sf \dfrac{1}{2} \bigg[- 9p - 9 \bigg] = 0$

$:\implies\sf - 9p - 9 = 0$

$:\implies\sf - 9(p + 1) = 0$

$:\implies\sf p + 1 = 0$

$:\implies{\boxed{\frak{\pink{p = - 1}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;the\;value\;of\;p\;is\; \bf{- 1}.}}}$

Answer 2

Step-by-step explanation:

The points (-5, 1), (1, p) and (4, -2) are collinear if the value of p is