Question

Find the equation of the straight line passing through the point (2, 1) and bisecting the portion of the straight line 3x - 5y = 15 lying between the axes.

Answer 1

$\text{The equation of the line in intercept form is}\frac{x}{a}+\frac{y}{b}=1$

$\text{The line passes through (2, 1)}$

$\therefore \frac{2}{a}+\frac{1}{b}=1...\left(i\right)$

$\text{Let the line}3x-5y=15\text{intersect the x-axis and the y-axis at A and B respectively.}$

$Atx=0\text{we have,}$

$0-5y=15$

$\Rightarrow y=-3$

$Aty=0,\text{we have}$

$3x-0=15$

$\Rightarrow x=5$

$\therefore A=(0,-3)andB=(5,0)$

$\text{The midpoint of AB is}(\frac{5}{2},-\frac{3}{2})$

$\text{Clearly, the point}(\frac{5}{2},-\frac{3}{2})\text{lies on the line}\frac{x}{a}+\frac{y}{b}=1$

$\therefore \frac{5}{2a}-\frac{3}{2b}=1...\left(ii\right)$

$\text{Using}\frac{3}{2}\times eq.\left(i\right)+eq.\left(ii\right)\text{we get,}$

$\frac{3}{a}+\frac{5}{2a}=\frac{3}{2}+1$

$\Rightarrow a=\frac{11}{5}$

$Fora=\frac{11}{5}\text{we have,}$

$\frac{10}{11}+\frac{1}{b}=1$

$\Rightarrow b=1$

$\text{Hence, the equation of the required line is}$

$\frac{5x}{11}+\frac{y}{11}=1$

$\Rightarrow 5x+y=11$