Question

What is the equation of the line that passes through (4, -1) and (-2, 3)?

Answer 1

(4,-1),(-2,3)
slope = (3 - (-1) / (-2 - 4) = -4/6 = -2/3

y = mx + b
slope(m) = -2/3
use either of ur points...(4,-1)...x = 4 and y = -1
now sub into the formula and find b, the y int
-1 = -2/3(4) + b
-1 = -8/3 + b
-1 + 8/3 = b
-3/3 + 8/3 = b
5/3 = b
so ur equation is : y = -2/3x + 5/3

Answer 2

Answer:

The equation of line is $y=-\frac{2}{3}(x)+\frac{5}{3}$.

Step-by-step explanation:

If a line passes though two points then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

The line passes through the points (4, -1) and (-2, 3), so the equation of line is

$y-(-1)=\frac{3-(-1)}{-2-4}(x-4)$

On simplification we get

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

$y+1=\frac{4}{-6}(x-4)$

$y+1=-\frac{2}{3}(x-4)$

Using distributive property,

$y+1=-\frac{2}{3}(x)-\frac{2}{3}(-4)$

$y+1=-\frac{2}{3}(x)+\frac{8}{3}$

Subtract  from both the sides.

$y=-\frac{2}{3}(x)+\frac{8}{3}-1$

$y=-\frac{2}{3}(x)+\frac{5}{3}$

Therefore the he equation of line is $y=-\frac{2}{3}(x)+\frac{5}{3}$.