Question

The points P(2, m), Q(3, -2) and R(8,3) are collinear i.e they lie on a straight line. Find the
value of m​

Answer 1

Solution :-

Given :-

The points P ( 2 , m ), Q ( 3 , -2 ) and R ( 8 , 3 ) are collinear.

That is,

$\bf \dfrac{1}{2} \times [ \ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \ ] = 0$

Here :-

$\bullet \ \sf x_1= 2 \ , \ y_1= m \\\\\bullet \ x_2= 3 \ , \ y_2=-2 \\\\\bullet \ x_3= 8 \ , \ y_3= 3$

$\longrightarrow \sf \dfrac{1}{2} \times [ \ 2(-2-3)+3(3-m) +8(m-(-2)) \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ ( 2 \times -5 ) + 9 - 3m +8(m+2) \ ] =0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ -10+9-3+8m+16 \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ 15+ 5 m \ ] = 0 \\\\\longrightarrow 15+5m = 0 \\\\\longrightarrow 5m = - 15 \\\\\longrightarrow \bf m = -3$

Value of m = -3