The coordinates of the orthocentre of the tri | vedclass

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

Question

(B) The given pair of lines is $2x^2 - 2y^2 + 3xy + 3x + y + 1 = 0$.Factorizing the expression: $2x^2 + 3xy - 2y^2 + 3x + y + 1 = 0$.$(x + 2y + 1)(2x

Vedclass · Accepted Answer

$\left( -\frac{3}{5}, -\frac{1}{5} \right)$

Vedclass · Answer

(B) The given pair of lines is $2x^2 - 2y^2 + 3xy + 3x + y + 1 = 0$.Factorizing the expression: $2x^2 + 3xy - 2y^2 + 3x + y + 1 = 0$.$(x + 2y + 1)(2x - y + 1) = 0$.Thus,the sides of the triangle are:$L_1: x + 2y + 1 = 0$$L_2: 2x - y + 1 = 0$$L_3: 3x + 2y + 1 = 0$Since the product of the slopes of $L_1$ and $L_2$ is $(-\frac{1}{2}) 	imes (2) = -1$,the lines $L_1$ and $L_2$ are perpendicular.In a right-angled triangle,the orthocentre is the vertex where the right angle is formed.Solving $x + 2y + 1 = 0$ and $2x - y + 1 = 0$:From $y = 2x + 1$,substitute into $x + 2(2x + 1) + 1 = 0$ $\Rightarrow 5x + 3 = 0$ $\Rightarrow x = -\frac{3}{5}$.Then $y = 2(-\frac{3}{5}) + 1 = -\frac{6}{5} + \frac{5}{5} = -\frac{1}{5}$.The orthocentre is $\left( -\frac{3}{5}, -\frac{1}{5} ight)$.

The coordinates of the orthocentre of the triangle formed by the lines $2x^2 - 2y^2 + 3xy + 3x + y + 1 = 0$ and $3x + 2y + 1 = 0$ are:

Similar Questions

The triangle formed by the lines $2x^2 - 3xy - 2y^2 = 0$ and $3x - y = 7$ is...

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

The combined equation of the diagonals of the parallelogram formed by the lines $(7 x^2-4 x y+8 y^2)^2+(4 x-8 y-32)(7 x^2-4 x y+8 y^2)=0$ is

The equation of a line which makes an angle of $45^{\circ}$ with each of the pair of lines $xy-x-y+1=0$ is

If $2x^2+3xy-2y^2=0$ represents two sides of a parallelogram and $3x+y+1=0$ is one of its diagonals,then the other diagonal is

Vedclass Test Series

Exam Paper Generator

Online Exam Module