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

The equation of line is
`y=-(2)/(3)(x)+(5)/(3)`.

Explanation:

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

`y-y_1=(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)=(3-(-1))/(-2-4)(x-4)`

On simplification we get

`y+1=(3+1)/(-6)(x-4)`

`y+1=(4)/(-6)(x-4)`

`y+1=-(2)/(3)(x-4)`

Using distributive property,

`y+1=-(2)/(3)(x)-(2)/(3)(-4)`

`y+1=-(2)/(3)(x)+(8)/(3)`

Subtract from both the sides.

`y=-(2)/(3)(x)+(8)/(3)-1`

`y=-(2)/(3)(x)+(5)/(3)`

Therefore the he equation of line is
`y=-(2)/(3)(x)+(5)/(3)`.