Question

Can't solve can u help

Answer 1

Given:

The equation of line p is

$y=-\dfrac{1}{3}x-3$

Line p and q are parallel.

To find:

The equation of line q.

Solution:

The slope intercept form of a line is

$y=mx+b$

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

The equation of line p is

$y=-\dfrac{1}{3}x-3$

The slope of the line is $-\dfrac{1}{3}$.

We know that the slopes of parallel lines are equal.

Line p and q are parallel. So,

Slope of line q = $-\dfrac{1}{3}$

Line q passes through (6,-4) with slope $-\dfrac{1}{3}$, so the equation of the line is

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

Where, m is the slope.

$(y-(-4))=-\dfrac{1}{3}(x-6)$

$y+4=-\dfrac{1}{3}x+2$

$y=-\dfrac{1}{3}x+2-4$

$y=-\dfrac{1}{3}x-2$

Therefore, the equation of line q is $y=-\dfrac{1}{3}x-2$.