JEE Main 2014 · 19 Apr · Shift 1 · Mathematics Q79

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MathematicsSingle correct+4 / -1Question 79 of 90

If 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, then the distance of line L from the line $x+5y=0$ is :

\dfrac7{\sqrt5}

\dfrac5{\sqrt{13}}

\dfrac7{\sqrt{13}}

\dfrac5{\sqrt7}

Answer and worked solution

Official answer: B

L is $x+5y=c$ . Its intercepts are $c$ and $\frac c5$ , so the area is $\frac{c^2}{10}=5$ and $|c|=5\sqrt2$ .

Distance from $x+5y=0$ :
$\frac{5\sqrt2}{\sqrt{26}}=\frac5{\sqrt{13}}.$

Topics:Distance of a point from a line Equations of a line

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JEE Main 2014 · 19 Apr · Shift 1, Mathematics question 79