Question

Find the equation of the line passing through the point of intersection of the lines 4x + 7y - 3 = 0 and 2x - 3y + 1 = 0 that has equal intercepts on the axis.

Answer 1

The equation of given lines are

4x + 7y - 3 = 0 and 2x - 3y + 1 = 0.

Now the equation of any line through intersection of these lines is

4x + 7y - 3 + k (2x - 3y + 1) = 0 . . . (i)

$\Rightarrow$ (4 + 2k) x + (7 - 3k)y = 3 - k

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

$\Rightarrow \frac{x}{\frac{3-k}{4+2k}}+\frac{y}{\frac{3-k}{7-3k}}=1$

It is given that $\frac{3-k}{4+2k}=\frac{3-k}{7-3k}$

$\Rightarrow (3-k)[\frac{1}{4+2k}-\frac{1}{7-3k}]=0$

$\Rightarrow 3-k=0$

or $\frac{1}{4+2k}-\frac{1}{7-3k}=0\Rightarrow 3=k$

or 7 - 3k - 4 - 2k = 0

$\Rightarrow$ k = 3 or -5k = -3

$\Rightarrow$ k = 3 or $k=\frac{3}{5}$

Putting k = 3 in(i,), we have

4x + 7y - 3 + 3(2x - 3y+ 1) = 0

$\Rightarrow$ 4x + 7y - 3 + 6x - 9y + 3 = 0

$\Rightarrow$ 10x - 2y = 0 $\Rightarrow$ = 5x - y = 0

Putting $k=\frac{3}{5}$ in (i), we have

$4x+7y-3+\frac{3}{5}(2x-3y+1)=0$

$\Rightarrow$ 20x + 35y - 15 + 6x - 9y + 3 = 0

$\Rightarrow$ 26x + 26y - 12 = 0

$\Rightarrow$ 13x + 13y - 6 = 0