Question

if the points (-5,1), (p,5) and (10,7) are collinear, then the value of p will be

Answer 1

Answer:

Step-by-step explanation:

please see the attachment below.

Just apply the formula of area of

ABC=0

Hope it helps.

Answer 2

$\frak{Here} \begin{cases} & \sf{(x_1 , y_1) = \bf{(-5, 1)}} \\ & \sf{(x_2 , y_2) = \bf{(p,5)}} \\ & \sf{(x_3 , y_3) = \bf{(10,7)}}\end{cases}$

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

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

We know that,

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

Therefore,

⠀⠀⠀

$:\implies\sf \dfrac{1}{2} \bigg[ -5(5 (- 7)) + p(7 - 1) + 10(1 - 5) \bigg] = 0$

$:\implies\sf \bigg[ -5(-2) + p(6) + 10(-4) \bigg] = 0$

$:\implies\sf \bigg[ 10 + 6p - 40\bigg] = 0$

$:\implies\sf 6p = 40-10$

$:\implies\sf 6p = 30$

$:\implies\sf p = {\cancel\dfrac{30}{6}}$

$:\implies{\boxed{\frak{\purple{p = 5}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;the\;value\;of\;p\;is\; \bf{5}.}}}$