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Question

(A) The equation of the family of lines passing through the intersection of $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ is given by $(3x - 4y + 1) + \lambd

Vedclass · Accepted Answer

$23x + 23y - 11 = 0$

Vedclass · Answer

(A) The equation of the family of lines passing through the intersection of $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ is given by $(3x - 4y + 1) + \lambda(5x + y - 1) = 0$. Rearranging the terms,we get $(3 + 5\lambda)x + (\lambda - 4)y + (1 - \lambda) = 0$. For the line to cut off equal intercepts on the axes,the coefficients of $x$ and $y$ must be equal in magnitude or the line must pass through the origin. If the intercepts are equal,then the slope of the line must be $-1$ (for intercepts $a, a$) or $1$ (for intercepts $a, -a$). Case $1$: Slope $m = -\frac{3 + 5\lambda}{\lambda - 4} = -1$ $\Rightarrow 3 + 5\lambda = \lambda - 4$ $\Rightarrow 4\lambda = -7$ $\Rightarrow \lambda = -\frac{7}{4}$. Substituting $\lambda = -\frac{7}{4}$ into the equation: $(3 - \frac{35}{4})x + (-\frac{7}{4} - 4)y + (1 + \frac{7}{4}) = 0$ $\Rightarrow -23x - 23y + 11 = 0$ $\Rightarrow 23x + 23y - 11 = 0$. Case $2$: The line passes through the origin $(1 - \lambda = 0 \Rightarrow \lambda = 1)$,which gives $8x - 3y = 0$ (intercepts are $0, 0$,which are equal). Since $23x + 23y - 11 = 0$ is one of the options,option $A$ is correct.

The equation of the line passing through the intersection of $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ which cuts off equal intercepts on the axes is given by

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