The equation of the line which bisects the ob | vedclass

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

Question

(B) The equations of the angle bisectors are given by $\frac{x - 2y + 4}{\sqrt{1^2 + (-2)^2}} = \pm \frac{4x - 3y + 2}{\sqrt{4^2 + (-3)^2}}$.This simp

Vedclass · Accepted Answer

$(4 + \sqrt{5})x - (3 + 2\sqrt{5})y + (2 + 4\sqrt{5}) = 0$

Vedclass · Answer

(B) The equations of the angle bisectors are given by $\frac{x - 2y + 4}{\sqrt{1^2 + (-2)^2}} = \pm \frac{4x - 3y + 2}{\sqrt{4^2 + (-3)^2}}$.This simplifies to $\frac{x - 2y + 4}{\sqrt{5}} = \pm \frac{4x - 3y + 2}{5}$,or $\sqrt{5}(x - 2y + 4) = \pm (4x - 3y + 2)$.Case $1$ (Positive sign): $\sqrt{5}x - 2\sqrt{5}y + 4\sqrt{5} = 4x - 3y + 2$,which rearranges to $(4 - \sqrt{5})x - (3 - 2\sqrt{5})y + (2 - 4\sqrt{5}) = 0$.Case $2$ (Negative sign): $\sqrt{5}x - 2\sqrt{5}y + 4\sqrt{5} = -4x + 3y - 2$,which rearranges to $(4 + \sqrt{5})x - (3 + 2\sqrt{5})y + (2 + 4\sqrt{5}) = 0$.To identify the obtuse bisector,check the sign of $a_1a_2 + b_1b_2$. Here $a_1a_2 + b_1b_2 = (1)(4) + (-2)(-3) = 4 + 6 = 10 > 0$. Since the sign is positive,the bisector corresponding to the negative sign in the formula $\frac{L_1}{\sqrt{a_1^2+b_1^2}} = -\frac{L_2}{\sqrt{a_2^2+b_2^2}}$ is the obtuse bisector.Thus,the equation is $(4 + \sqrt{5})x - (3 + 2\sqrt{5})y + (2 + 4\sqrt{5}) = 0$.

The equation of the line which bisects the obtuse angle between the lines $x - 2y + 4 = 0$ and $4x - 3y + 2 = 0$ is:

Similar Questions

If the lengths of the perpendiculars drawn from a point $(a, b)$ to the lines $2x + 3y + 4 = 0$ and $3x - 2y + 4 = 0$ are equal,then the point $(a, b)$ lies on the line

If $P(-1, 0)$,$Q(0, 0)$,and $R(3, 3\sqrt{3})$ are three points,then the equation of the bisector of the $\angle PQR$ is

The sides of a rhombus $ABCD$ are parallel to the lines $x - y + 2 = 0$ and $7x - y + 3 = 0$. If the diagonals of the rhombus intersect at $P(1, 2)$ and the vertex $A$ (different from the origin) is on the $y$-axis,then the ordinate of $A$ is

The equation of the perpendicular bisector of the line segment joining the points $(1, 2)$ and $(-2, 0)$ is:

The equation of the bisector of the acute angle between the lines $2x - y + 4 = 0$ and $x - 2y - 1 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module