Let three points be P(-1, 0),Q(0, 0),and R(3, | vedclass

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

Question

(A) The coordinates are $P(-1, 0)$,$Q(0, 0)$,and $R(3, 3\sqrt{3})$.Line $QP$ lies on the $x$-axis,so its angle with the positive $x$-axis is $	heta_1

Vedclass · Accepted Answer

$\sqrt{3}x + y = 0$

Vedclass · Answer

(A) The coordinates are $P(-1, 0)$,$Q(0, 0)$,and $R(3, 3\sqrt{3})$.Line $QP$ lies on the $x$-axis,so its angle with the positive $x$-axis is $	heta_1 = 180^{\circ}$.Line $QR$ passes through $(0, 0)$ and $(3, 3\sqrt{3})$. Its slope is $m = \frac{3\sqrt{3} - 0}{3 - 0} = \sqrt{3}$. Thus,the angle with the positive $x$-axis is $	heta_2 = 60^{\circ}$.The angle $\angle PQR$ is the angle between these two lines,which is $180^{\circ} - 60^{\circ} = 120^{\circ}$.The angle bisector divides this angle into two $60^{\circ}$ parts. The angle of the bisector with the positive $x$-axis is $180^{\circ} - 60^{\circ} = 120^{\circ}$.The slope of the bisector is $	an(120^{\circ}) = -\sqrt{3}$.Since it passes through the origin $Q(0, 0)$,the equation is $y = -\sqrt{3}x$,which simplifies to $\sqrt{3}x + y = 0$.

Let three points be $P(-1, 0)$,$Q(0, 0)$,and $R(3, 3\sqrt{3})$. The equation of the angle bisector of $\angle PQR$ is:

Similar Questions

The equations of the angle bisectors between the lines $3x - 4y + 7 = 0$ and $12x - 5y - 8 = 0$ are:

Find the equation of the locus of points equidistant from the lines $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$.

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

The equation of the bisector of the acute angle between the lines $3x - 4y + 7 = 0$ and $12x + 5y - 2 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module