Find the equation of the line that bisects th | vedclass

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

Question

(C) The slope of the line is $m = 	an(45^{\circ}) = 1$.The midpoint of the line segment joining $(5, 3)$ and $(4, 4)$ is given by:$M = \left( \frac{5

Vedclass · Accepted Answer

$x - y - 1 = 0$

Vedclass · Answer

(C) The slope of the line is $m = 	an(45^{\circ}) = 1$.The midpoint of the line segment joining $(5, 3)$ and $(4, 4)$ is given by:$M = \left( \frac{5+4}{2}, \frac{3+4}{2} ight) = \left( \frac{9}{2}, \frac{7}{2} ight)$.The equation of the line passing through $(x_1, y_1) = \left( \frac{9}{2}, \frac{7}{2} ight)$ with slope $m = 1$ is:$y - y_1 = m(x - x_1)$$y - \frac{7}{2} = 1 \left( x - \frac{9}{2} ight)$$y - \frac{7}{2} = x - \frac{9}{2}$$x - y - \frac{9}{2} + \frac{7}{2} = 0$$x - y - 1 = 0$.

Find the equation of the line that bisects the line segment joining the points $(5, 3)$ and $(4, 4)$ and makes an angle of $45^{\circ}$ with the positive direction of the $x$-axis.

Similar Questions

The equation of a line passing through $(p \cos \alpha, p \sin \alpha)$ and making an angle $(90^\circ + \alpha)$ with the positive direction of the $X$-axis is:

Find the equation of the line whose perpendicular distance from the origin is $4$ units and the angle which the normal makes with the positive direction of the $x$-axis is $15^{\circ}$.

Given that lines $L_1: y=m_a x, L_2: y=m_b x$ and $L_3: y=m_c x$ make equal intercepts on the line $x+y=1$,then

If a line intercepted between the coordinate axes is trisected at a point $A(4, 3)$,which is nearer to the $x$-axis,then its equation is

The equation of the line passing through the point $(-3, 2)$ and parallel to the $x$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module