Question

Question 20
Find the ratio in which the line 2x+3y = 5 = 0 divides the line segment joining the points (8,-9) and (2,1). Also, find the coordinates of the point of division.

Answer 1

Let the line 2x+3y – 5 = 0 divides the segment joining the points A(8,-9) and B(2,1) in the ratio $\lambda :1$ at point P.
$\therefore CoordinatesofP\equiv \{\frac{2\lambda +8}{\lambda +1},\frac{\lambda -9}{\lambda +1}\}[\because Byinternaldivisionformula,thecoordinatedwillbe\{\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}\}]$
But P lies on 2x+3y – 5 = 0
$\therefore 2\left(\frac{2\lambda +8}{\lambda +1}\right)+3\left(\frac{\lambda -9}{\lambda +1}\right)-5=0\Rightarrow 2(2\lambda +8)+3(\lambda -9)-5(\lambda +1)=0\Rightarrow 4\lambda +16+3\lambda -27-5\lambda -5=0\Rightarrow 2\lambda -16=0\Rightarrow \lambda =8\Rightarrow \lambda :1=8:1$
So, the point P divides the line in the ratio 8:1.
$\therefore PointofdivisionP\equiv \{\frac{2\left(8\right)+8}{8+1},\frac{8-9}{8+1}\}=(\frac{16+8}{9},-\frac{1}{9})\equiv (\frac{24}{9},\frac{-1}{9})\equiv (\frac{8}{3},\frac{-1}{9})$
Hence, the required point of division is
$(\frac{8}{3},\frac{-1}{9})$