Question

Line v passes through point [6.6] and is perpendicular to the graph of y= 3/4x - 11 line w is parallel to line v and passes through point [-6,10] what is the equation in slope intercept form of line w please help im in a exam

Answer 1

Given:

The equation of a line is

$y=\dfrac{3}{4}x-11$

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

$y=mx+b$         ...(i)

where, m is slope and b is y-intercept.

We have,

$y=\dfrac{3}{4}x-11$      ...(ii)

On comparing (i) and (ii), we get

$m=\dfrac{3}{4}$

So, slope of given line is $\dfrac{3}{4}$.

Product of slopes of two perpendicular lines is -1.

$m_1\times m_2=-1$

$\dfrac{3}{4}\times m_2=-1$

$m_2=-\dfrac{4}{3}$

Line w is perpendicular to the given line. So, the slope of line w is $-\dfrac{4}{3}$.

Slopes of parallel line are equal.

Line v is parallel to line w. So, slope of line v is also $-\dfrac{4}{3}$.

Slope of line v is $-\dfrac{4}{3}$ and it passes thorugh (-6,10). So, the equation of line v is

$y-y_1=m(x-x_1)$

where, m is slope.

$y-10=-\dfrac{4}{3}(x-(-6))$

$y-10=-\dfrac{4}{3}(x+6)$

$y-10=-\dfrac{4}{3}x-\dfrac{4}{3}(6)$

$y-10=-\dfrac{4}{3}x-8$

Adding 10 on both sides, we get

$y=-\dfrac{4}{3}x-8+10$

$y=-\dfrac{4}{3}x+2$

Therefore the equation of line v is $y=-\dfrac{4}{3}x+2$.