MATH SOLVE

5 months ago

Q:
# Write the equation in slope intercept form for the line perpendicular to c(-4,-5) and D(4,9) passing through the midpoint of the line

Accepted Solution

A:

Slope intercept form of line passing through midpoint of CD and perpendicular to CD is [tex]\Rightarrow y=-\frac{4}{7} x+2[/tex]Solution:Need to find the slope intercept form for the line perpendicular to C(-4,-5) and D(4,9)
And passing through the midpoints of the line CD.
Lets first calculate slope of CD Let say slope of CD be represented by [tex]m_1[/tex]General formula of slope of line passing through points [tex]\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right)[/tex] is as follows:[tex]m=\frac{\left(y_{2}-y_{1}\right)}{\left(x_{2}-x_{1}\right)}[/tex][tex]\text { In case of line } \mathrm{CD} , x_{1}=-4, \quad y_{1}=-5 \text { and } x_{2}=4, y_{2}=9[/tex][tex]\text {So slope of line } \mathrm{CD} \text { that is } m_{1}=\frac{(9-(-5))}{(4-(-4))}=\frac{14}{8}=\frac{7}{4}[/tex]Let’s say slope of required line which is perpendicular to CD be [tex]m_2[/tex]As product of slope of the lines perpendicular to each other is -1
=> slope of line CD [tex]\times[/tex] slope of line perpendicular to CD = -1[tex]\begin{array}{l}{=>m_{1} \times m_{2}=-1} \\\\ {\Rightarrow \frac{7}{4} \times m_{2}=-1} \\\\ {\Rightarrow m_{2}=-\frac{4}{7}}\end{array}[/tex]Now let’s find midpoint of CD [tex]\text { Midpoint }(x, y) \text { of two points }\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right) \text { is given by }[/tex][tex]x=\frac{x_{2}+x_{1}}{2} \text { and } y=\frac{y_{2}+y_{1}}{2}[/tex][tex]\text { So in case of line } \mathrm{CD} , x_{1}=-4, y_{1}=-5 \text { and } x_{2}=4, y_{2}=9[/tex]And midpoint of CD will be as follows[tex]x=\frac{x_{2}+x_{1}}{2}=\frac{4+(-4)}{2}=0 \text { and } y=\frac{y_{2}+y_{1}}{2}=\frac{9-5}{2}=2[/tex]So midpoint of CD is ( 0 , 2 )
As it is given that line whose slope intercept form is required is perpendicular to CD and passing through midpoint of CD , we need equation of line passing through ( 0 , 2 ) and having slope as [tex]m_{2}=-\frac{4}{7}[/tex]Generic equation of line passing through [tex]\left(x_{1}, y_{1}\right)[/tex] and having slope of m is given by[tex]\left(y-y_{1}\right)=m\left(x-x_{1}\right)[/tex][tex]\text { In our case } x_{1}=0 \text { and } y_{1}=2 \text { and } m=-\frac{4}{7}[/tex]Substituting the values in generic equation of line we get[tex](y-2)=-\frac{4}{7}(x-0)[/tex]As we required final equation in slope intercept form which is y = mx + c, lets rearrange our equation is required form:[tex]\Rightarrow y=-\frac{4}{7} x+2[/tex]Hence can conclude that slope intercept form of line passing through midpoint of CD and perpendicular to CD is [tex]\Rightarrow y=-\frac{4}{7} x+2[/tex]