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

Given:  Points $( a,\ b),\ ( 3,\ -5)$ and $( -5,\ -2)$ are collinear.

To do: To find the value of $3a+8b$.

Solution:

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:

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

 

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:

$A=\frac{1}{2}[a( -5-( -2))+3( -2-b)+( -5)( b-( -5))]$

$\Rightarrow 0=\frac{1}{2}[a( -5+2)+3( -2-b)-5( b+5)]$

$\Rightarrow 0=\frac{1}{2}[-3a-6-3b-5b-25]$

$\Rightarrow 0=\frac{1}{2}[-3a-8b-31]$

$\Rightarrow 0=-3a-8b-31$

$\Rightarrow 3a+8b=-31$

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