Question

Find the equation of the line passing through the point of intersection of the lines $4x-7y-3=0and2x-3y+1=0$ that has equal intercepts on the axes.

Answer 1

$\text{Given equation of lines are}$

$4x+7y-3=0...\left(i\right)$

$and2x-3y+1=0...\left(ii\right)$

$\text{On solving Eqs. (i) and (ii),}$

$\text{From Eq. (ii),}$

$3y=2x+1$

$\Rightarrow y=\frac{2}{3}x+\frac{1}{3}$

$\text{On putting in Eq. (i), we get}$

$4x+7(\frac{2}{3}x+\frac{1}{3})-3=0$

$\Rightarrow 4x+\frac{14}{3}x+\frac{7}{3}-3=0$

$\Rightarrow \frac{12x+14x}{3}+\frac{7-9}{3}=0$

$\Rightarrow \frac{26x-2}{3}=0$

$\Rightarrow x=\frac{2}{26}\Rightarrow x=\frac{1}{13}$

$\text{On putting the value of x in Eq. (i), we get}$

$\frac{4}{13}+7y-3=0$

$\Rightarrow 7y=3-\frac{4}{13}$

$\Rightarrow 7y=\frac{39-4}{13}$

$\Rightarrow 7y=\frac{35}{13}\Rightarrow y=\frac{5}{13}$

$\text{Point is}(\frac{1}{13},\frac{5}{13})$

$\text{Now, equation of line in intercept form is}$

$\frac{x}{a}+\frac{y}{b}=1$

$\text{Since, line (iii) has equal intercepts on the axis i.e., a = b}$

$\Rightarrow \frac{x}{a}+\frac{y}{b}=1$

$x+y=a$

$\text{Above line passes through the point}$

$(\frac{1}{13},\frac{5}{13})\text{i.e.,point will satisfy it}$

$\frac{1}{13}+\frac{5}{13}=a$

$a=\frac{6}{13}$

$\text{Hence, required equation of line (iv) becomes}$

$x+y=\frac{6}{13}$

$13x+13y=6$