The distance from the origin to the orthocent | vedclass

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

Question

(D) The given lines are $x+y-1=0$ and $6x^2-13xy+5y^2=0$.Factorizing the second equation:$6x^2-10xy-3xy+5y^2=0$$(2x-y)(3x-5y)=0$So,the lines are $2x-y

Vedclass · Accepted Answer

$\frac{11\sqrt{2}}{24}$

Vedclass · Answer

(D) The given lines are $x+y-1=0$ and $6x^2-13xy+5y^2=0$.Factorizing the second equation:$6x^2-10xy-3xy+5y^2=0$$(2x-y)(3x-5y)=0$So,the lines are $2x-y=0$ and $3x-5y=0$.Let the vertices of the triangle be $O(0,0)$,$A$,and $B$.Solving $2x-y=0$ and $x+y-1=0$,we get $A = (1/3, 2/3)$.Solving $3x-5y=0$ and $x+y-1=0$,we get $B = (5/8, 3/8)$.Let the orthocentre be $H(h, k)$.The altitude from $B$ to $OA$ is perpendicular to $y=2x$. Its equation is $y - 3/8 = -1/2(x - 5/8) \Rightarrow 4x + 8y - 6 = 0$.The altitude from $A$ to $OB$ is perpendicular to $y=3/5x$. Its equation is $y - 2/3 = -5/3(x - 1/3) \Rightarrow 15x + 9y - 11 = 0$.Solving these two altitude equations:$4h + 8k = 6$ and $15h + 9k = 11$.Multiplying the first by $9$ and second by $8$:$36h + 72k = 54$$120h + 72k = 88$Subtracting gives $84h = 34 \Rightarrow h = 17/42$.Substituting $h$ in $4h + 8k = 6$: $4(17/42) + 8k = 6$ $\Rightarrow 34/42 + 8k = 6$ $\Rightarrow 8k = 6 - 17/21 = 109/21$ $\Rightarrow k = 109/168$.The distance from origin to $(h, k)$ is $\sqrt{h^2+k^2} = \sqrt{(17/42)^2 + (109/168)^2} = \frac{11\sqrt{2}}{24}$.

The distance from the origin to the orthocentre of the triangle formed by the lines $x+y-1=0$ and $6x^2-13xy+5y^2=0$ is

Similar Questions

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 line $x+y=1$ meets the lines represented by the equation $y^3-6xy^2+11x^2y-6x^3=0$ at the points $P, Q, R$. If $O$ is the origin,then $(OP)^2+(OQ)^2+(OR)^2=$

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 $a+2h+b=$

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