Given:
The equation of given line is
[tex]y=3x-4[/tex]
A line is perpendicular to the given line and passes through the point (9,5).
To find:
The equation of the perpendicular line.
Solution:
We have,
[tex]y=3x-4[/tex]
On comparing this equation with slope intercept form [tex]y=mx+b[/tex], we get
[tex]m=3[/tex]
It means, the slope of the given line is 3.
The product of slopes of two perpendicular lines is -1.
[tex]m\times m_1=-1[/tex]
[tex]3\times m_1=-1[/tex]
[tex]m_1=-\dfrac{1}{3}[/tex]
Point slope form of a line is
[tex]y-y_1=m(x-x_1)[/tex]
Where, m is the slope.
The slope of perpendicular line is [tex]-\dfrac{1}{3}[/tex] and it passes through the point (9,5). So the equation of the line is
[tex]y-5=-\dfrac{1}{3}(x-9)[/tex]
[tex]y-5=-\dfrac{1}{3}(x)-\dfrac{1}{3}(-9)[/tex]
[tex]y-5=-\dfrac{1}{3}x+3[/tex]
Adding 5 on both sides, we get
[tex]y-5+5=-\dfrac{1}{3}x+3+5[/tex]
[tex]y=-\dfrac{1}{3}x+8[/tex]
Therefore, the equation of required line is [tex]y=-\dfrac{1}{3}x+8[/tex].
