The origin and the points where the line L_1 | vedclass

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(A) The slope of line $L_2$ is $4$. Since $L_1 \perp L_2$,the slope of $L_1$ is $m = -\frac{1}{4}$.Let the equation of line $L_1$ be $y = -\frac{1}{4}

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$10 + \sqrt{68}$

Vedclass · Answer

(A) The slope of line $L_2$ is $4$. Since $L_1 \perp L_2$,the slope of $L_1$ is $m = -\frac{1}{4}$.Let the equation of line $L_1$ be $y = -\frac{1}{4}x + k$,which can be written as $x + 4y = 4k$.Let $c = 4k$. The intercepts are $(c, 0)$ and $(0, \frac{c}{4})$.The area of the right-angled triangle $T$ formed by the origin $(0, 0)$ and these intercepts is $\frac{1}{2} 	imes |c| 	imes |\frac{c}{4}| = 8$.$|c^2| = 64$,so $c = \pm 8$.Taking $c = 8$,the vertices are $(0, 0), (8, 0),$ and $(0, 2)$.The lengths of the sides are $8, 2,$ and $\sqrt{8^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68}$.The perimeter is $8 + 2 + \sqrt{68} = 10 + \sqrt{68}$.

The origin and the points where the line $L_1$ intersects the $x$-axis and $y$-axis are vertices of a right-angled triangle $T$ whose area is $8$. Also,the line $L_1$ is perpendicular to the line $L_2: 4x - y = 3$. Then,the perimeter of triangle $T$ is:

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