Question

write the equation of the line that passes through the point (-1,-6) and is perpendicular to a line that passes through the points (-2,5) and (-4,8)

Answer 1

Answer:

$y=\frac{2}{3}x-\frac{16}{3}$

Step-by-step explanation:

step 1

Find the slope of the given line

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

(-2,5) and (-4,8)

substitute the values in the formula

$m=\frac{8-5}{-4+2}$

$m=\frac{3}{-2}$

$m=-\frac{3}{2}$

step 2

Find the slope of the perpendicular line to the given line

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

so

$m_1*m_2=-1$

$m_1=-\frac{3}{2}$ ----> slope of the given line

therefore

$m_2=\frac{2}{3}$ ---> slope of the perpendicular line to the given line

step 2

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{2}{3}$

$point\ (-1,-6)$

substitute

$y+6=\frac{2}{3}(x+1)$

step 3

Convert to slope intercept form

$y=mx+b$

isolate the variable y

$y+6=\frac{2}{3}x+\frac{2}{3}$

$y=\frac{2}{3}x+\frac{2}{3}-6$

$y=\frac{2}{3}x-\frac{16}{3}$