The area (in sq. units) of the triangle forme | vedclass

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

Question

(D) The area of the triangle formed by the pair of lines $ax^2+2hxy+by^2=0$ and the line $lx+my+n=0$ is given by the formula $	ext{Area} = \frac{n^2\

Vedclass · Accepted Answer

$\frac{1}{2\sqrt{5}}$

Vedclass · Answer

(D) The area of the triangle formed by the pair of lines $ax^2+2hxy+by^2=0$ and the line $lx+my+n=0$ is given by the formula $	ext{Area} = \frac{n^2\sqrt{h^2-ab}}{|am^2-2hlm+bl^2|}$.Here,the pair of lines is $x^2-3xy+y^2=0$,so $a=1, h=-\frac{3}{2}, b=1$.The third line is $x+y+1=0$,so $l=1, m=1, n=1$.Substituting these values into the formula:$	ext{Area} = \frac{1^2\sqrt{(-\frac{3}{2})^2-(1)(1)}}{|(1)(1)^2-2(-\frac{3}{2})(1)(1)+(1)(1)^2|}$$	ext{Area} = \frac{\sqrt{\frac{9}{4}-1}}{|1+3+1|} = \frac{\sqrt{\frac{5}{4}}}{5} = \frac{\frac{\sqrt{5}}{2}}{5} = \frac{\sqrt{5}}{10} = \frac{1}{2\sqrt{5}}$ sq. units.

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

Similar Questions

The distance between the lines represented by the equation $4x^2 + 4xy + y^2 - 6x - 3y - 4 = 0$ is

The equation of the bisectors of the angles between the lines joining the origin to the points of intersection of the curve $x^2+xy+y^2+x+3y+1=0$ and the line $x+y+2=0$ is

If $\theta$ is the acute angle between the lines joining the origin to the points of intersection of the curve $x^2+xy+y^2+x+3y+1=0$ and the straight line $x+y+2=0$,then $\cos \theta=$

The angle between the lines joining the origin to the points of intersection of the line $y = 3x + 2$ and the curve $x^{2} + 2xy + 3y^{2} + 4x + 8y - 11 = 0$ is:

The condition that the lines joining the origin to the points of intersection of the two curves $x^2+y^2+gx+c=0$ and $x^2+y^2+2fy-c=0$ are at right angles is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module