Question

the line MN is shown on the grid. find the equation of the perpendicular bisector of line MN.

Answer 1

Answer:

$The~endpoints~of~MN~are~(-1,7)~and~(3,-1).\\So,~Midpoint~of~MN=(\frac{-1+3}{2} ,\frac{7-1}{2}) = (\frac{2}{2},\frac{6}{2})=(1,3)\\Now,\\MN~passes~through~(-1,7)~and~(3,-1).\\Slope~of~MN(m_{MN})=(\frac{y_2-y_1}{x_2-x_1})=\frac{-1-7}{3+1} = \frac{-8}{4} =-2\\Let~the~slope~of~the~line~which~is~perpendicular~bisector~of~MN~be~'m'.\\Using~the~condition~of~perpendicular~lines,\\(m_{MN})(m)=-1\\or, -2m=-1\\or, m=\frac{1}{2}$$The~perpendicular~bisector~of~MN~has~slope~\frac{1}{2}~and~it~passes~through~ midpoint~of~MN, (1,3).\\So,~its~equation~is~given~by:\\y-3=\frac{1}{2}(x-1)\\or, 2y-6 = x-1\\or, x-2y=1-6\\or, x-2y+5=0 ~is~the~required~equation.$

Answer 2

The equation of the line which is perpendicular to MN is x -2y +5=0

What is equation of line?

The equation of a line with a slope of m and a y-intercept of (0, b) is

y = mx + b.

The coordinates of M  (-1,7) and N( 3, -1)

So, the midpoints of MN are

$(\frac{-1+3}{2},\frac{ -1+7}{2}) = (1, 3)$

Now, slope of MN be:

$\frac{y_2-y_2}{x_2-x_1}$ =$\frac{-1-7}{3+1}$ = -2

slope of MN= -2

Let the slope of line which is perpendicular to MN be m'

So, m x m'=-1

    -2m' = -1

      $m=\frac{1}{2}$

So, the equation of the line be:

$y- 3 = m (x-1)$

y-3=  $\frac{1}{2}$ (x-1)

2y-6=x-1

x - 2y +5= 0

Learn more about equation of line here:

https://brainly.com/question/21511618

#SPJ2