The orthocenter of the triangle formed by the | vedclass

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

Question

(A) The given lines are $6x^2+xy-y^2=0$ and $x-2y=10$.Factorizing the equation $6x^2+xy-y^2=0$,we get $(2x+y)(3x-y)=0$.This gives two lines: $L_1: 3x-

Vedclass · Accepted Answer

$(2,-4)$

Vedclass · Answer

(A) The given lines are $6x^2+xy-y^2=0$ and $x-2y=10$.Factorizing the equation $6x^2+xy-y^2=0$,we get $(2x+y)(3x-y)=0$.This gives two lines: $L_1: 3x-y=0$ and $L_2: 2x+y=0$.The third line is $L_3: x-2y=10$.Observe the slopes: slope of $L_2$ is $m_2 = -2$ and slope of $L_3$ is $m_3 = 1/2$.Since $m_2 	imes m_3 = (-2) 	imes (1/2) = -1$,the lines $L_2$ and $L_3$ are perpendicular.In a right-angled triangle,the orthocenter is the vertex where the right angle is formed.Thus,the orthocenter is the point of intersection of $L_2$ and $L_3$.Solving $2x+y=0$ and $x-2y=10$:From $y = -2x$,substitute into $x-2(-2x)=10$ $\Rightarrow x+4x=10$ $\Rightarrow 5x=10$ $\Rightarrow x=2$.Then $y = -2(2) = -4$.The orthocenter is $(2,-4)$.

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

Similar Questions

The pairs of straight lines $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$ form a

If the lines represented by $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ intersect on the $x$-axis,which of the following is in general incorrect?

The area of the triangle formed by the lines $y^2 - 9xy + 18x^2 = 0$ and $y = 9$ is

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

The figure formed by the lines $x^2 + 4xy + y^2 = 0$ and $x - y = 4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module