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 line $y = 2x + c$ passes through a point that is equidistant from both the axes and lies in the first quadrant $(x > 0, y > 0)$. Then the value of $c$ is...

The $y$-intercept of a line is twice its $x$-intercept. If the line passes through the point $(1, 2)$,find its equation.

The equation of a straight line passing through $(-3, 2)$ and cutting an intercept equal in magnitude but opposite in sign from the axes is given by:

The line $(3x - y + 5) + \lambda (2x - 3y - 4) = 0$ will be parallel to the $y$-axis,if $\lambda$ =

If the line $\frac{x}{a} + \frac{y}{b} = 1$ passes through the points $(2, -3)$ and $(4, -5)$,then $(a, b) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module