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(B) The equation of any line passing through the intersection of the given lines is given by $(3x - 4y + 1) + k(5x + y - 1) = 0$.This simplifies to $(

Vedclass · Accepted Answer

$23x + 23y = 11$

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(B) The equation of any line passing through the intersection of the given lines is given by $(3x - 4y + 1) + k(5x + y - 1) = 0$.This simplifies to $(3 + 5k)x + (k - 4)y + (1 - k) = 0$.The intercepts on the axes are given by $x$-intercept $= -\frac{1 - k}{3 + 5k} = \frac{k - 1}{3 + 5k}$ and $y$-intercept $= -\frac{1 - k}{k - 4} = \frac{k - 1}{k - 4}$.Since the intercepts are equal,we have $\frac{k - 1}{3 + 5k} = \frac{k - 1}{k - 4}$.This implies either $k - 1 = 0$ or $\frac{1}{3 + 5k} = \frac{1}{k - 4}$.If $k = 1$,the equation becomes $8x - 3y = 0$,which passes through the origin (intercepts are zero).If $k - 4 = 3 + 5k$,then $4k = -7$,so $k = -7/4$.Substituting $k = -7/4$ into the equation: $4(3x - 4y + 1) - 7(5x + y - 1) = 0$.$12x - 16y + 4 - 35x - 7y + 7 = 0$.$-23x - 23y + 11 = 0$,which simplifies to $23x + 23y = 11$.

What is the equation of the line passing through the intersection of the lines $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ and having equal intercepts on the axes?

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