Question

For which value of 'k' the points (7,-2),(5,1),(3,k) are collinear ?

• A. -4
• B. 4
• C. -8
• D. 8

Answer 1

Answer:

B. 4

Since the points are collinear ,

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

Take $(7,-2) as (x_1,y_1), (5,1) as (x_2,y_2) and (3,k) as(x_3,y_3)$

we have,

$7(1-k)+5(k+2)+3(-2-1)=0$

$7-7k+5k-9=0$

$2k=8$

$k=8/2=4$