Question

Find the value of p for which a(6,-3),b(2,3),c(4,p)are collinear

Answer 1

• If the Given points are collinear then the area of the ∆ formed by these points = 0

• Area of in co-ordinate geometry;

$\frac{1}{2} |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|$

But for the Given situation;

$\frac{1}{2} |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| = 0 \\ \\ |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| = 0$

Given points;

• A(6,-3)
• B(2,3)
• C(4,p)

Thus using the formula;

• 6(3-p) + 2(p+3) + 4(-3-3) = 0
• 18 - 6p + 2p + 6 + 4(-6) = 0
• ( -6p +2p ) + (18 + 6 - 24)=0
• -4p + (24 - 24) = 0
• -4p = 0
• p = 0

#answerwithquality

#BAL