Question

find the value of p for which the points ( -5,1) (1,p ) , (4,-2) are collinear ​

Answer 1

EXPLANATION.

→ points are → ( -5,1) , ( 1,p) , ( 4,-2) are collinear.

→ To find the value of p.

→ if point are collinear → ∆ = 0

$\sf : \implies \: x_{1} = - 5 \: and \: \: y_{1} = 1 \\ \\ \sf : \implies \: \: x_{2} = 1 \: \: \:and \: \: y_{2} = p \\ \\ \sf : \implies \: \: x_{3} = 4 \: \: and \: \: y_{3} = - 2$

$\sf : \implies \: \dfrac{1}{2} [ x_{1}( y_{2} - y_{3}) + x_{2}( y_{3} - y_{1} ) + x_{3}( y_{1} - y_{2} )] \\ \\ \sf : \implies \: [ x_{1}( y_{2} - y_{3}) + x_{2}( y_{3} - y_{1} ) + x_{3}( y_{1} - y_{2} )] \: = 0$

→ [ -5 ( p - ( -2)) + 1 ( -2 - 1 ) + 4 ( 1 - p) ] = 0

→ [ -5 ( p + 2 ) + 1 ( -3 ) + 4 ( 1 - p) ] = 0

→ [ -5p - 10 - 3 + 4 - 4p ] = 0

→ [ -9p - 9 ] = 0

→ p = -1

Answer 2

$\begin{array}{|c|c|c|c|c|}1& - 5&1&1 \\ \frac{1}{2} &1&k&1 \\ &4& - 2&1 \end{array} = 0$

$\left| \begin{array}{cc} - 5 & 1 & 1 \\ 1 & k & 1 \\ 4 & - 2 & 1 \end{array} \right|$

$\blue{ - 5 \Big(k - ( - 2) \Big) - 1(1 - 4) + 1( - 2 - 4k) = 0}$

$\red{ - 5 \: (k + 2) \: - 1( - 3) + 1( - 2 + 4k) = 0}$

$\red{ - 5k \: - 10 + 3 - 2 - 4k = 0}$

$- 9k - 9 = 0$

$-9k = 9$

$k = (-9/9)$

$\boxed{ \leadsto \red{ \bf k = - 1} }$