Question

Find the value of k for which point p(k,-1) q(2,1) and r(4,5) are collinear.

Answer 1

Given :

Find the value of k for which point p(k,-1) q(2,1) and r(4,5) are collinear.

To find :

Find the value of k

Solution :

By applying area of triangle formula to solve this

Area of ∆PQR

$\sf \frac{1}{2}(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$

$\sf \frac{1}{2}(k(5-1)+2(-1-5)+4(-1-1)$

$\sf \frac{1}{2}(4k+2\times(-6)+4\times(-2)$

$\sf \frac{1}{2}(4k-12-8)$

$\sf \frac{1}{2}(4k-20)$

$\sf 2k-10$

If the points are collinear, then area of triangle = 0

=> 2k - 10 = 0

=> 2k = 10

=> k = 10/2 = 5

Additional information :

★ Distance formula = $\sf \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

★ Section formula = $\sf \frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}$

★ Mid point formula = $\sf \frac{x_1+x_2}{2},\frac{y_1+y_2}{2}$