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Question

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

Vedclass · Accepted Answer

$\frac{121}{1058}$

Vedclass · Answer

(A) The equation of a line 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 $(3 + 5\lambda)x + (-4 + \lambda)y + (1 - \lambda) = 0$. Since the line makes equal non-zero intercepts on the axes,the coefficients of $x$ and $y$ must be equal in magnitude and opposite in sign (for the intercept form $\frac{x}{a} + \frac{y}{a} = 1$) or equal in sign (for $\frac{x}{a} + \frac{y}{-a} = 1$). Setting the coefficients equal: $3 + 5\lambda = -(-4 + \lambda)$ $\Rightarrow 3 + 5\lambda = 4 - \lambda$ $\Rightarrow 6\lambda = 1$ $\Rightarrow \lambda = \frac{1}{6}$. Substituting $\lambda = \frac{1}{6}$ into the equation: $(3 + \frac{5}{6})x + (-4 + \frac{1}{6})y + (1 - \frac{1}{6}) = 0$ $\Rightarrow \frac{23}{6}x - \frac{23}{6}y + \frac{5}{6} = 0$ $\Rightarrow 23x - 23y + 5 = 0$. This gives $x - y = -\frac{5}{23}$,which does not yield equal intercepts. Setting the coefficients equal: $3 + 5\lambda = -4 + \lambda$ $\Rightarrow 4\lambda = -7$ $\Rightarrow \lambda = -\frac{7}{4}$. Substituting $\lambda = -\frac{7}{4}$: $(3 - \frac{35}{4})x + (-4 - \frac{7}{4})y + (1 + \frac{7}{4}) = 0$ $\Rightarrow -\frac{23}{4}x - \frac{23}{4}y + \frac{11}{4} = 0$ $\Rightarrow 23x + 23y = 11$. In intercept form: $\frac{x}{11/23} + \frac{y}{11/23} = 1$. The intercepts are $a = \frac{11}{23}$ and $b = \frac{11}{23}$. The area of the triangle is $\frac{1}{2} |ab| = \frac{1}{2} 	imes \frac{11}{23} 	imes \frac{11}{23} = \frac{121}{1058}$ sq. units.

If a straight line is passing through the point of intersection of the lines $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ and making equal non-zero intercepts on the coordinate axes,then the area (in sq. units) of the triangle formed by this line with the coordinate axes is

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