Question

find the equation,in standard form, of the line passing through the point (2,-3) and (4,2)

Answer 1

The standard form:

$Ax+By=C$

The point-slope form:

$y-y_1=m(x-x_1)$

m - slope

(x₁, y₁) - point

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (2, -3) and (4, 2). Substitute:

$m=\dfrac{2-(-3)}{4-2}=\dfrac{5}{2}$

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

$y+3=\dfrac{5}{2}(x-2)$       use distributive property

$y+3=\dfrac{5}{2}x-5$            multiply both sides by 2

$2y+6=5x-10$                     subtract 6 from both sides

$2y=5x-16$         subtract 5x from both sides

$-5x+2y=-16$          change the signs

$\boxed{5x-2y=16}$

Answer 2

Answer:

5x - 2y = 16

Step-by-step explanation: