A straight line L is perpendicular to the lin | vedclass

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(A) The given line is $5x - y = 1$,which can be written as $y = 5x - 1$. The slope of this line is $m_1 = 5$. Since line $L$ is perpendicular to this

Vedclass · Accepted Answer

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

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(A) The given line is $5x - y = 1$,which can be written as $y = 5x - 1$. The slope of this line is $m_1 = 5$. Since line $L$ is perpendicular to this line,its slope $m$ must satisfy $m 	imes 5 = -1$,so $m = -1/5$. The equation of line $L$ in slope-intercept form is $y = -\frac{1}{5}x + c$,or $x + 5y = 5c$. Let $k = 5c$,so the equation is $x + 5y = k$. The intercepts of this line are $x = k$ (when $y=0$) and $y = k/5$ (when $x=0$). The area of the triangle formed with coordinate axes is $\frac{1}{2} |x_{int} 	imes y_{int}| = 5$. $\frac{1}{2} |k 	imes \frac{k}{5}| = 5 \implies |k^2| = 50 \implies k = \pm \sqrt{50} = \pm 5 \sqrt{2}$. Thus,the equation of line $L$ is $x + 5y = \pm 5 \sqrt{2}$.

$A$ straight 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 $5$ square units. The equation of the line $L$ can be

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