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

  • A
    $\left(\frac{1}{4}, \frac{1}{4}\right)$
  • B
    $\left(\frac{1}{3}, \frac{1}{3}\right)$
  • C
    $\left(\frac{1}{2}, \frac{1}{2}\right)$
  • D
    $(1, 1)$

Explore More

Similar Questions

$A$ pair of straight lines passes through the point $(1,1)$. One of the lines makes an angle $\theta$ with the positive direction of the $X$-axis and the other makes the same angle with the positive direction of the $Y$-axis. If the equation of the pair of straight lines is $x^2-(a+2)xy+y^2+a(x+y-1)=0$,$a \neq -2$,then the value of $\theta$ is

If the equation of the pair of straight lines passing through the point $(1,1)$ and perpendicular to the pair of lines $3x^2+11xy-4y^2=0$ is $ax^2+2hxy+by^2+2gx+2fy+12=0$,then $2(a-h+b-g+f-12)=$

If the centroid of the triangle formed by the lines $2y^2+5xy-3x^2=0$ and $x+y=k$ is $(\frac{1}{18}, \frac{11}{18})$,then the value of $k$ equals $..........$

The area (in square units) of the quadrilateral formed by the two pairs of lines $l^2x^2 - m^2y^2 - n(lx + my) = 0$ and $l^2x^2 - m^2y^2 + n(lx - my) = 0$ is

If two pairs of straight lines with combined equations $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$ form a square,then the combined equation of its diagonals is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo