Question

Find the equation of the straight line which passes through the point of intersection of the straight lines 3x - 4y + 1 = 0 and 5x + y - 1 = 0 and cuts off equal intercepts from the axis.

• A. 32x + 32y + 11 = 0
• B. 23x + 23y = 11
• C. 9x + 18y + 5 = 0
• D. None of these

Answer 1

The correct option is B 23x + 23y = 11

The equation of any line passing through the point of intersection of the lines 3x - 4y + 1 = 0 and 5x + y - 1 = 0 is (3x - 4y + 1) + k(5x+ y-1) = 0......(1)
For intercept of this line with the x-axis, y = 0
$\therefore$ 3x+ 1 + k (5x - 1) = 0
$\Rightarrow x=\frac{k-1}{5k+3}$
For intercept of the line (1) on the y -axis, x = 0
$\therefore -4y+1+k(y-1)=0\Rightarrow y=\frac{k-1}{k-4}$
Since the intercepts on the axes are equal.
$\therefore$ $\frac{k-1}{5k+3}=\frac{k-1}{k-4}\Rightarrow k=1,ork=-\frac{7}{4}$
But k $\ne$ 1, because if k = 1, the line (1) becomes 8x - 3y = 0 which passes through the origin and therefore cannot make non-zero intercepts on the axis.
$\therefore k=-\frac{7}{4}$ and from (1), we get
$3x-4y+1-\frac{7}{4}(5x+y-1)=0$
$\Rightarrow 23x+23y=11$, which is the required equation of the line.