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  • Find the value of p for which the points (– 5, 1), (1, p) and (4, – 2) are collinear.      

Find the value of p for which the points (– 5, 1), (1, p) and (4, – 2) are collinear.

 

 
 
 
 
 

Answers (1)

Given \rightarrow A= \left ( x_{1},y_{1} \right )= \left ( -5,1 \right )
                B= \left ( x_{2},y_{2} \right )= \left ( 1,p \right )
                C= \left ( x_{3},y_{3} \right )= \left ( 4,-2 \right )
The points are colinear if the triangle formed by these point has Zero Area
\Rightarrow Area\, \triangle ABC= 0
\Rightarrow \frac{1}{2}\left [ x_{1}\left ( y_{2}-y_{3} \right )+k_{2}\left ( y_{3} -y_{1}\right )+x_{3}\left ( y_{1}-y_{2} \right ) \right ]= 0
\Rightarrow \frac{1}{2}\left [ -5\left ( p- \left ( -2 \right )\right ) +1\left ( -2-1 \right )+4\left ( 1-p \right )\right ]= 0
\Rightarrow -5p-5\times 2+1\times \left ( -3 \right )+4\times 1-p\times 4= 0
\Rightarrow -9p-10-3+4= 0
\Rightarrow -9p-9= 0
\Rightarrow p= -1

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Safeer PP

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