Question

If the points $(a,b),(3,-5)$ and $(-5,-2)$ are collinear. Then find the value of $3a+8b$

• A. $20$
• B. $-31$
• C. $10$
• D. None

Answer 1

The correct option is B $-31$

If three points are collinear then the area of the triangle is zero.

Area of the triangle passing through the points $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is given by:

$A=\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$

Here the points are $(a,b)$, $(3,-5)$ and $(-5,-2)$ and it is given that these points are collinear therefore the area is zero that is:

$A=\frac{1}{2}\left|a\right(-5-(-2))+3(-2-b)+(-5\left)\right(b-(-5)\left)\right|\Rightarrow 0=\frac{1}{2}\left|a\right(-5+2)+3(-2-b)-5(b+5\left)\right|\Rightarrow 0=\frac{1}{2}|-3a-6-3b-5b-25|\Rightarrow 0=\frac{1}{2}|-3a-8b-31|\Rightarrow 0=-3a-8b-31\Rightarrow 3a+8b=-31$

Hence, $3a+8b=-31$.