Mathematics, 02.07.2019 00:00 msdsh03

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

Answers

Answer from: tami5

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

5x - 2y = -7

Answer from: Quest

There is no graph but what i can say is that 1 is the slope and that -2 is the y- intercept

Answer from: Quest

answer: just subtact the units and then add the non negative : happy to

step-by-step explanation:

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