The area of the triangle formed by the lines | vedclass

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

Question

(A) The given equations are $3x + y + 15 = 0$ $(i)$ and $3x^2 + 12xy - 13y^2 = 0$ (ii).Equation (ii) represents a pair of lines passing through the or

Vedclass · Accepted Answer

$\frac{15\sqrt{3}}{2}$

Vedclass · Answer

(A) The given equations are $3x + y + 15 = 0$ $(i)$ and $3x^2 + 12xy - 13y^2 = 0$ (ii).Equation (ii) represents a pair of lines passing through the origin $(0, 0)$.Let the lines be $y = m_1x$ and $y = m_2x$. Substituting $y = mx$ into (ii),we get $3 + 12m - 13m^2 = 0$,or $13m^2 - 12m - 3 = 0$.The roots are $m = \frac{12 \pm \sqrt{144 - 4(13)(-3)}}{26} = \frac{12 \pm \sqrt{144 + 156}}{26} = \frac{12 \pm \sqrt{300}}{26} = \frac{12 \pm 10\sqrt{3}}{26} = \frac{6 \pm 5\sqrt{3}}{13}$.The vertices of the triangle are $O(0, 0)$,$A$,and $B$,where $A$ and $B$ are the intersection points of $3x + y + 15 = 0$ with the lines $y = m_1x$ and $y = m_2x$.The area of a triangle formed by $ax^2 + 2hxy + by^2 = 0$ and $lx + my + n = 0$ is given by $\frac{n^2 \sqrt{h^2 - ab}}{|am^2 - 2hlm + bl^2|}$.Here $a = 3, h = 6, b = -13, l = 3, m = 1, n = 15$.Area $= \frac{15^2 \sqrt{6^2 - 3(-13)}}{|3(1)^2 - 2(6)(3)(1) + (-13)(3)^2|} = \frac{225 \sqrt{36 + 39}}{|3 - 36 - 117|} = \frac{225 \sqrt{75}}{|-150|} = \frac{225 	imes 5\sqrt{3}}{150} = \frac{1125\sqrt{3}}{150} = \frac{15\sqrt{3}}{2}$.

The area of the triangle formed by the lines represented by $3x + y + 15 = 0$ and $3x^2 + 12xy - 13y^2 = 0$ is

Similar Questions

The centroid of the triangle formed by the lines $x+3y=10$ and $6x^2+xy-y^2=0$ is

One line of the pair of lines $x^2+xy-2y^2=0$ is perpendicular to one line of the pair of lines $3y^2-5xy-2x^2=0$. If the combined equation of the two lines other than those two perpendicular lines is $ax^2+2hxy+by^2=0$,then $a+2h+b=$

If the slopes of both the lines given by $x^2 + 2hxy + 6y^2 = 0$ are positive and the angle between these lines is $\operatorname{Tan}^{-1}\left(\frac{1}{7}\right)$,then the product of the perpendiculars from the point $(1, 1)$ to these lines is:

The lines represented by the equations $23x^2 - 48xy + 3y^2 = 0$ and $2x + 3y + 4 = 0$ form

The triangle formed by the lines $2x^2+xy-6y^2=0$ and $x+y-1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module