Question

What is the equation, in point-slope form, for a line that goes through (8, −4) and has a slope of −5/6 ?

Answer 1

Point-slope form, y-y1 = m(x-x1)

(y-(-4)) = -5/6 (x-8)

y+4 = -5x/6+20/3

y = -5x/6+20/3 - 4

y = -5x/6 + 8/3

Answer 2

Answer:

$y-(-4)=\frac{-5}{4}(x-8)$

Step-by-step explanation:

Given point = (8,-4)

Slope = $\frac{-5}{6}$

General equation of point slope form : $y-y_1=m(x-x_1)$

$(x_1,y_1)=(8,-4)$

$m=\frac{-5}{6}$

Substitute the values in general equation :

$y-(-4)=\frac{-5}{4}(x-8)$

$y+4=\frac{-5}{4}(x-8)$

$y+4=\frac{-5}{4}x+10$

$y=\frac{-5}{4}x+6$

Hence the equation, in point-slope form, for a line that goes through (8, −4) and has a slope of −5/6  is $y-(-4)=\frac{-5}{4}(x-8)$