Two mutually perpendicular straight lines thr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$5$

Vedclass · Answer

(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:

Similar Questions

Let $A(-2,-1)$,$B(1,0)$,$C(\alpha, \beta)$,and $D(\gamma, \delta)$ be the vertices of a parallelogram $ABCD$. If the point $C$ lies on $2x-y=5$ and the point $D$ lies on $3x-2y=6$,then the value of $|\alpha+\beta+\gamma+\delta|$ is equal to . . . . . . .

The diagonals of a parallelogram $ABCD$ are along the lines $x+3y=4$ and $6x-2y=7$. Then $ABCD$ must be a

The circumcentre of the triangle formed by the lines $xy+2x+2y+4=0$ and $x+y+2=0$ is

Which one of the following statements is true?

For what value of $k$ are the points $(k, 2 - 2k)$,$(1 - k, 2k)$,and $(-4 - k, 6 - 2k)$ collinear?

Vedclass Test Series

Exam Paper Generator

Online Exam Module