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(B) Let the line $L$ intersect the axes at $A$ and $B$. Let $OA = |a|$ and $OB = |b|$.The equation of a line perpendicular to $5x - y = 1$ is of the f

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the line $L$ intersect the axes at $A$ and $B$. Let $OA = |a|$ and $OB = |b|$.The equation of a line perpendicular to $5x - y = 1$ is of the form $x + 5y = k$.Setting $x = 0$,we get $y = k/5$,so $B = (0, k/5)$.Setting $y = 0$,we get $x = k$,so $A = (k, 0)$.The area of the triangle formed by the line and the coordinate axes is given by $\frac{1}{2} |OA| |OB| = 5$.$\frac{1}{2} |k| |k/5| = 5$$|k^2| / 10 = 5$$k^2 = 50$$k = \pm 5\sqrt{2}$Thus,the equation of the line $L$ is $x + 5y = \pm 5\sqrt{2}$.

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

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