# If P(9a -2, – b) divides line segment joining

Question:

If P(9a -2, – b) divides line segment joining A(3a + 1,-3) and B(8a, 5) in the ratio 3 : 1, then find the values of a and b.

Solution:

Let P(9a – 2, – b) divides AS internally in the ratio 3:1.

By section formula,

$9 a-2=\frac{3(8 a)+1(3 a+1)}{3+1}$

$[\because$ internal section formula, the coordinates of point $P$ divides the line segment joining point $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ in the ratio $m_{1}: m_{2}$ internally is $\left.\left(\frac{m_{2} x_{1}+m_{1} x_{2}}{m_{1}+m_{2}}, \frac{m_{2} y_{1}+m_{1} y_{2}}{m_{1}+m_{2}}\right)\right]$

and $-b=\frac{3(5)+1(-3)}{3+1}$

$\Rightarrow$ $9 a-2=\frac{24 a+3 a+1}{4}$

and $-b=\frac{15-3}{4}$

$\Rightarrow$ $-b=\frac{15-3}{4}$

and $-b=\frac{12}{4}$

$\Rightarrow$ $9 a-2=\frac{27 a+1}{4}$

and $-b=\frac{12}{4}$

$\Rightarrow$ $36 a-8=27 a+1$

and $b=-3$

$\Rightarrow \quad 36 a-27 a-8-1=0$

$\Rightarrow \quad 9 a-9=0$

$\therefore$ $a=1$

Hence, the required values of a and b are 1 and – 3.