Question

In what ratio is the line segment joining $(2,4)$ and $(6,8)$ divided by the line $3x+4y=5?$

Answer 1

The given line is $3x+4y=5$ ..........$\left(1\right)$

Let the line $\left(1\right)$ divide the line segment joining the point $(2,4)$ and $(6,8)$ at the point $P$ in the ratio $k:1$ then the co-ordinates of point $P$ are

$(\frac{6k+2}{k+1},\frac{8k+4}{k+1})$ $[\because Sectionformulax=\frac{m{x}_2+n{x}_1}{m+n}]$

Since $P$ lies on $\left(1\right)$, we get $3\left(\frac{7k+1}{k+1}\right)+4\left(\frac{9k+3}{k+1}\right)=5$

$\Rightarrow 21k+3+36k+12=5k+5$

$\Rightarrow 57k+15=5k+5$

$\Rightarrow 57k-5k=5-15$

$\Rightarrow 52k=-10$

$\Rightarrow 26k=-5$

$\Rightarrow k=\frac{-5}{26}$

Hence, the required ratio is $\frac{-5}{26}:1$ or $5:26$ externally.