Question

Fine the equation, in standard form, of the line passing through the points (3,-4) and (5,1)

Answer 1

The point-slope form:

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

We have the points (3, -4) and (5, 1). Substitute:

$m=\dfrac{1-(-4)}{5-3}=\dfrac{5}{2}\\\\y-(-4)=\dfrac{5}{2}(x-3)$

The standard form: $Ax+By=C$

$y+4=\dfrac{5}{2}(x-3)$           multiply both sides by 2

$2y+8=5(x+3)$           use distributive property

$2y+8=5x+15$            subtract 2y from both sides

$8=5x-2y+15$          subtract 15 from both sides

$-7=5x-2y$

Answer: 5x - 2y = -7