Answer:
Line LM is perpendicular to line AB. It's equation is 5y - 6x = 15
Step-by-step explanation:
Given: Slope of line, AB [tex]\frac{-5}{6}[/tex]
To find: A line which is perpendicular to given line AB
We know that if two lines are perpendicular than product of their slope is -1.
Let slope of required line is m then by using given condition we get,
[tex]m\times\frac{-5}{6}\,=\,-1[/tex]
[tex]m\,=\,\frac{6}{5}[/tex]
Now we check slope of each and every line and matches with value of m.
using two point we find slope.
formula for slope,
[tex]Slope\,=\,\frac{y_2-y_1}{x_2-x_1}[/tex]
Coordinates of Given points are P( -5 , 4 ) , Q( 0 , -2 ) , J( -6 , 1 ) , K( 0 , -4 ) ,
L( -5 , -3 ) , M( 0 , 3 ) , N( -6 , -5 ) and O( 0 , 0 )
Slope of line PQ = [tex]\frac{-2-4}{0-(-5)}\:=\:\frac{-6}{5)}[/tex]
Slope line JL = [tex]\frac{-4-1}{0-(-6)}\:=\:\frac{-5}{6)}[/tex]
Slope line LM = [tex]\frac{3-(-3)}{0-(-5)}\:=\:\frac{6}{5)}[/tex]
Slope line NO = [tex]\frac{0-(-5)}{0-(-6)}\:=\:\frac{5}{6)}[/tex]
Thus, By comparing with above slope.
LM is our required line which is perpendicular to given Line AB.
For equation we use point-slope form,
Equation line LM
[tex](y-y_1)=m\times(x-x_1)[/tex]
[tex](y-3)=\frac{6}{5}\times(x-0)[/tex]
[tex]5\times(y-3)=6\times(x-0)[/tex]
[tex]5y-15=6x[/tex]
5y - 6x = 15
Therefore, Line LM is perpendicular to line AB. It's equation is 5y - 6x = 15
