Question
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
Let the line 2x + 3y - 5 = 0 divides the line segment joining the points A (8, -9) and B (2, 1) in the ratio $\lambda:1$ at point P.
$\therefore$ Coordinates of $\text{P}\equiv\left\{\frac{2\lambda+8}{\lambda+1},\frac{\lambda-9}{\lambda+1}\right\}$
$\bigg[\because$ internal division $\left\{\frac{\text{m}_2\text{x}_1+\text{m}_1\text{x}_2}{\text{m}_1+\text{m}_1},\frac{\text{m}_2\text{y}_1+\text{m}_1\text{y}_2}{\text{m}_1+\text{m}_2}\Big)\right\}\bigg]$
But P lies on $2\text{x} + 3\text{y} - 5 = 0$
$\therefore2\Big(\frac{2\lambda+8}{\lambda+1}\Big)+3\Big(\frac{\lambda-9}{\lambda+1}\Big)-5=0$
$\Rightarrow2(2\lambda+8)+3(\lambda-9)-5(\lambda+1)=0$
$\Rightarrow4\lambda+16+3\lambda-27-5\lambda-5=0$
$\Rightarrow2\lambda-16=0$
$\Rightarrow\lambda=8$
$\Rightarrow\lambda:1$
$\Rightarrow8:1$
So, the point P divides the lies the line in the ratio 8 : 1
$\therefore$ Point of division $\text{P}\equiv\left\{\frac{2(8)+8}{8+1},\frac{8-9}{8+1}\right\}$
$\text{P}\equiv\Big(\frac{16+8}{9},-\frac{1}{9}\Big)$
$\text{P}\equiv\Big(\frac{24}{9},\frac{-1}{9}\Big)$
$\text{P}\equiv\Big(\frac{8}{3},\frac{-1}{9}\Big)$
Hence, the requied point of division is $\Big(\frac{8}{3},\frac{-1}{9}\Big).$
$\therefore$ Coordinates of $\text{P}\equiv\left\{\frac{2\lambda+8}{\lambda+1},\frac{\lambda-9}{\lambda+1}\right\}$
$\bigg[\because$ internal division $\left\{\frac{\text{m}_2\text{x}_1+\text{m}_1\text{x}_2}{\text{m}_1+\text{m}_1},\frac{\text{m}_2\text{y}_1+\text{m}_1\text{y}_2}{\text{m}_1+\text{m}_2}\Big)\right\}\bigg]$
But P lies on $2\text{x} + 3\text{y} - 5 = 0$
$\therefore2\Big(\frac{2\lambda+8}{\lambda+1}\Big)+3\Big(\frac{\lambda-9}{\lambda+1}\Big)-5=0$
$\Rightarrow2(2\lambda+8)+3(\lambda-9)-5(\lambda+1)=0$
$\Rightarrow4\lambda+16+3\lambda-27-5\lambda-5=0$
$\Rightarrow2\lambda-16=0$
$\Rightarrow\lambda=8$
$\Rightarrow\lambda:1$
$\Rightarrow8:1$
So, the point P divides the lies the line in the ratio 8 : 1
$\therefore$ Point of division $\text{P}\equiv\left\{\frac{2(8)+8}{8+1},\frac{8-9}{8+1}\right\}$
$\text{P}\equiv\Big(\frac{16+8}{9},-\frac{1}{9}\Big)$
$\text{P}\equiv\Big(\frac{24}{9},\frac{-1}{9}\Big)$
$\text{P}\equiv\Big(\frac{8}{3},\frac{-1}{9}\Big)$
Hence, the requied point of division is $\Big(\frac{8}{3},\frac{-1}{9}\Big).$
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