Question

write the equation of the line described below in slope - intercept form. find the equation of the line that is parallel to the line (y = x - 6) and passes through the point ((-8,10)). write the equation in slope - intercept form. show your work here enter your answer write the equation of the line described below in slope - intercept form. find the equation of the line that is perpendicular to the line (y = 3x + 1) and passes through the point ((2,-4)).

Explanation:

Step1: Determine the slope for the parallel line

Parallel lines have equal slopes. The slope of the line $y = x - 6$ is $m = 1$.

Step2: Use the point - slope form to find the equation

The point - slope form is $y - y_1=m(x - x_1)$. Substitute $m = 1$, $x_1=-8$ and $y_1 = 10$:
$y - 10=1\times(x + 8)$

Step3: Convert to slope - intercept form

Expand and simplify:
$y-10=x + 8$
$y=x+18$

Step4: Determine the slope for the perpendicular line

The slope of the line $y = 3x+1$ is $m_1 = 3$. For a line perpendicular to it, the slope $m_2$ satisfies $m_1m_2=-1$. So $m_2=-\frac{1}{3}$.

Step5: Use the point - slope form for the perpendicular line

Substitute $m_2 = -\frac{1}{3}$, $x_1 = 2$ and $y_1=-4$ into the point - slope form $y - y_1=m(x - x_1)$:
$y+4=-\frac{1}{3}(x - 2)$

Step6: Convert to slope - intercept form

Expand: $y+4=-\frac{1}{3}x+\frac{2}{3}$
Subtract 4 from both sides: $y=-\frac{1}{3}x+\frac{2}{3}-4=-\frac{1}{3}x-\frac{10}{3}$

Answer:

The equation of the parallel line is $y=x + 18$.
The equation of the perpendicular line is $y=-\frac{1}{3}x-\frac{10}{3}$