A line L is perpendicular to the line 5x - y | vedclass

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Question

(B) The given line is $5x - y = 1$. Any line perpendicular to $5x - y = 1$ is of the form $x + 5y = k$. Writing this in intercept form: $\frac{x}{k} +

Vedclass · Accepted Answer

$x + 5y = \pm 5\sqrt{2}$

Vedclass · Answer

(B) The given line is $5x - y = 1$. Any line perpendicular to $5x - y = 1$ is of the form $x + 5y = k$. Writing this in intercept form: $\frac{x}{k} + \frac{y}{k/5} = 1$. The intercepts on the coordinate axes are $a = k$ and $b = k/5$. The area of the triangle formed by the line and the coordinate axes is given by $\frac{1}{2} |ab| = 5$. Substituting the values: $\frac{1}{2} |k \cdot \frac{k}{5}| = 5$. $|k^2| = 50$,which implies $k = \pm \sqrt{50} = \pm 5\sqrt{2}$. Therefore,the equation of the line $L$ is $x + 5y = \pm 5\sqrt{2}$.

$A$ line $L$ is perpendicular to the line $5x - y = 1$ and the area of the triangle formed by the line $L$ and the coordinate axes is $5$. The equation of the line $L$ is

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