The joint equation of the bisectors of the an | vedclass

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

Question

(B) The lines $x=5$ (a vertical line) and $y=3$ (a horizontal line) intersect at the point $(5, 3)$.Since these lines are perpendicular,the angle bise

Vedclass · Accepted Answer

$x^2-y^2-10x+6y+16=0$

Vedclass · Answer

(B) The lines $x=5$ (a vertical line) and $y=3$ (a horizontal line) intersect at the point $(5, 3)$.Since these lines are perpendicular,the angle bisectors are lines passing through $(5, 3)$ that make angles of $45^{\circ}$ and $135^{\circ}$ with the $x$-axis.The slopes of these bisectors are $m = 	an 45^{\circ} = 1$ and $m = 	an 135^{\circ} = -1$.The equations of the bisectors are:$y - 3 = 1(x - 5) \Rightarrow y - x + 2 = 0$$y - 3 = -1(x - 5) \Rightarrow y + x - 8 = 0$The joint equation is $(y - x + 2)(y + x - 8) = 0$.Expanding this: $y^2 + xy - 8y - xy - x^2 + 8x + 2y + 2x - 16 = 0$.Simplifying,we get $-x^2 + y^2 + 10x - 6y - 16 = 0$,or $x^2 - y^2 - 10x + 6y + 16 = 0$.

The joint equation of the bisectors of the angles between the lines $x=5$ and $y=3$ is

Similar Questions

If the equation $3 x^2-k x y-3 y^2=0$ represents the bisectors of the angles between the lines $x^2-3 x y-4 y^2=0$,then the value of $k$ is

The condition for the general quadratic equation $f(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ to represent coincident lines is:

If the pair of straight lines $xy-x-y+1=0$ and the line $ax+2y-3a=0$ are concurrent,then $a$ is equal to

The equation of a pair of lines $y=px$ and $y=qx$ can be written as $(y-px)(y-qx)=0$. Then the equation of the pair of angle bisectors of the lines $x^2-4xy-5y^2=0$ is

The equation of the pair of bisectors of the angles between the pair of straight lines $3x^2+7xy+2y^2+5x+5y+2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module