Respuesta :
Given:
The equation of a line is
[tex]y=\dfrac{3}{4}x-11[/tex]
Line v passes through point (6,6) and it is perpendicular to the given line.
Line w passes through point (-6,10) and it is parallel to the line v.
To find:
The equation in slope intercept form of line w.
Solution:
Slope intercept form of a line is
[tex]y=mx+b[/tex] ...(i)
where, m is slope and b is y-intercept.
We have,
[tex]y=\dfrac{3}{4}x-11[/tex] ...(ii)
On comparing (i) and (ii), we get
[tex]m=\dfrac{3}{4}[/tex]
So, slope of given line is [tex]\dfrac{3}{4}[/tex].
Product of slopes of two perpendicular lines is -1.
[tex]m_1\times m_2=-1[/tex]
[tex]\dfrac{3}{4}\times m_2=-1[/tex]
[tex]m_2=-\dfrac{4}{3}[/tex]
Line w is perpendicular to the given line. So, the slope of line w is [tex]-\dfrac{4}{3}[/tex].
Slopes of parallel line are equal.
Line v is parallel to line w. So, slope of line v is also [tex]-\dfrac{4}{3}[/tex].
Slope of line v is [tex]-\dfrac{4}{3}[/tex] and it passes thorugh (-6,10). So, the equation of line v is
[tex]y-y_1=m(x-x_1)[/tex]
where, m is slope.
[tex]y-10=-\dfrac{4}{3}(x-(-6))[/tex]
[tex]y-10=-\dfrac{4}{3}(x+6)[/tex]
[tex]y-10=-\dfrac{4}{3}x-\dfrac{4}{3}(6)[/tex]
[tex]y-10=-\dfrac{4}{3}x-8[/tex]
Adding 10 on both sides, we get
[tex]y=-\dfrac{4}{3}x-8+10[/tex]
[tex]y=-\dfrac{4}{3}x+2[/tex]
Therefore the equation of line v is [tex]y=-\dfrac{4}{3}x+2[/tex].
