Two mutually perpendicular straight lines thr | vedclass

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(A) Let the origin be $A(0, 0)$. The perpendicular distance $AD$ from the origin to the line $2x + y - 5 = 0$ is given by:$AD = \left|\frac{2(0) + 1(0

Vedclass · Accepted Answer

$5$

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(A) Let the origin be $A(0, 0)$. The perpendicular distance $AD$ from the origin to the line $2x + y - 5 = 0$ is given by:$AD = \left|\frac{2(0) + 1(0) - 5}{\sqrt{2^2 + 1^2}}ight| = \frac{5}{\sqrt{5}} = \sqrt{5}$.Since the two lines through the origin are mutually perpendicular and form an isosceles triangle with the given line,the angles at the base $BC$ are $45^{\circ}$.In the right-angled triangle $ABD$,we have $	an 45^{\circ} = \frac{AD}{BD}$.Since $	an 45^{\circ} = 1$,we get $BD = AD = \sqrt{5}$.Similarly,in triangle $ADC$,$DC = AD = \sqrt{5}$.Thus,the base $BC = BD + DC = \sqrt{5} + \sqrt{5} = 2\sqrt{5}$.The area of the triangle $ABC$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes BC 	imes AD$.Area $= \frac{1}{2} 	imes (2\sqrt{5}) 	imes \sqrt{5} = 5$ square units.

Two mutually perpendicular straight lines through the origin form an isosceles triangle with the line $2x + y = 5$. Then the area of the triangle is:

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