Answer:
Therefore, equation of the line that passes through (16,-7) and is perpendicular to the line [tex]2x-3y=12[/tex] is [tex]3x+2y=34[/tex]
Step-by-step explanation:
Given:
[tex]2x-3y=12[/tex]
To Find:
Equation of line passing through ( 16, -7) and is perpendicular to the line
[tex]2x-3y=12[/tex]
Solution:
[tex]2x-3y=12[/tex] ...........Given
[tex]\therefore y=\dfrac{2}{3}\times x-4[/tex]
Comparing with,
[tex]y=mx+c[/tex]
Where m =slope
We get
[tex]Slope = m1 = \dfrac{2}{3}[/tex]
We know that for Perpendicular lines have product slopes = -1.
[tex]m1\times m2=-1[/tex]
Substituting m1 we get m2 as
[tex]\dfrac{2}{3}\times m2=-1\\\\m2=-\dfrac{3}{2}[/tex]
Therefore the slope of the required line passing through (16 , -7) will have the slope,
[tex]m2=-\dfrac{3}{2}[/tex]
Now the equation of line in slope point form given by
[tex](y-y_{1})=m(x-x_{1})[/tex]
Substituting the point (16 , -7) and slope m2 we will get the required equation of the line,
[tex](y-(-7))=-\dfrac{3}{2}\times (x-16)\\\\2y+14=-3x+48\\3x+2y=34......Equation\ of\ line[/tex]
Therefore, equation of the line that passes through (16,-7) and is perpendicular to the line [tex]2x-3y=12[/tex] is
[tex]3x+2y=34[/tex]
