A pair of lines drawn through the origin form | vedclass

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

Question

(A) Let the triangle be $	riangle ABC$ with the right angle at the origin $A(0, 0)$. The base of the triangle lies on the line $2x + 3y = 6$.Since th

Vedclass · Accepted Answer

$\frac{36}{13}$

Vedclass · Answer

(A) Let the triangle be $	riangle ABC$ with the right angle at the origin $A(0, 0)$. The base of the triangle lies on the line $2x + 3y = 6$.Since the triangle is a right-angled isosceles triangle with the right angle at the origin,the two lines passing through the origin make an angle of $45^{\circ}$ with the line $2x + 3y = 6$.The perpendicular distance $p$ from the origin $A(0, 0)$ to the line $2x + 3y - 6 = 0$ is given by:$p = \frac{|2(0) + 3(0) - 6|}{\sqrt{2^2 + 3^2}} = \frac{6}{\sqrt{13}}$.In a right-angled isosceles triangle,the altitude from the right-angle vertex to the hypotenuse bisects the hypotenuse and is equal to half the length of the hypotenuse.Thus,the length of the altitude $AL = p = \frac{6}{\sqrt{13}}$.The base $BC$ of the triangle is $2p$ because the altitude $AL$ divides the triangle into two smaller right-angled isosceles triangles with legs of length $p$.Area of $	riangle ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2p) 	imes p = p^2$.Area $= \left(\frac{6}{\sqrt{13}}ight)^2 = \frac{36}{13}$ sq. units.

$A$ pair of lines drawn through the origin forms a right-angled isosceles triangle with the line $2x + 3y = 6$,having the right angle at the origin. The area (in sq. units) of the triangle thus formed is

Similar Questions

The area (in sq. units) of the triangle formed by the lines $x^2-3xy+y^2=0$ and $x+y+1=0$ is

If $f(x, y) = 0$ is the combined equation of the lines joining the origin to the points where the line $4x - 6y - 2 = 0$ meets the curve $3x^2 - 4xy + 5y^2 - 2x + y - 6 = 0$,then $\frac{f(1, -1)}{f(-1, -1)} = $

If the lines joining the origin to the points of intersection of $y=mx+1$ and $x^2+y^2=1$ are perpendicular,then .........

The slope of a line which cuts off intercepts of equal lengths on the axes is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module