I've tried my hardest on the problem, it's quite hard...

## What is the standard form of an equation of the line that passes through the points (-1,4) and (-7,-5)?

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# 3 Answers

"What is the standard form of an equation of the line that passes through the points (-1,4) and (-7,-5)?"

x--7 -7--1 -6 2

----- = ------ = ---- = --

y--5 -5-4 -9 3

Multiply both sides by 3(y+5):

3(x+7) =2(y+5)

3x + 21 = 2y + 10

3x - 2y = -11

First use the slope formula to get the slope. m = ( y

_{1 }-_{ }y_{2 }) / (x_{1 }- x_{2}) m = (-5 - 4) / (-7 - -1)

m = -9 / -6 = 9/6

Now use the point-slope form with the slope you just found and either of the points given:

Plug in m and a point

_{....}y - y_{1}= m (x- x_{1}) y - 4 = 9/6 (x - -1)

Simplify.... ANSWER: y - 4 = 9/6 (x + 1)

# Comments

Standard form is Ax + By = C or Ax + By + C = 0 depending on the author.

Hi Mikk;

(-1,4) and (-7,-5)

Let's first establish the slope. This is the change-of-y divided by the change-of-x...

(y-y

_{1})/(x-x_{1})(4--5)/(-1--7)

Subtracting a negative number is the same as adding a positive number...

(4+5)/(-1+7)

9/6=3/2

This is the slope.

Standard equation is...

Ax+By=C, neither A nor B equal zero, and A is greater than zero.

slope=-A/B

slope=3/-2. The two must be negative because A must be greater than zero.

3x-2y=C

Let's plug-in one coordinate to establish C...

(-1,4) and (-7,-5). I randomly select the first...

[(3)(-1)]-[(2)(4)]=C

-3-8=C

-11=C

Let's plug-in the other coordinate to verify C as -11...

3x-2y=C

[(3)(-7)]-[(2)(-5)]=C

-21-(-10)=C

-21+10=C

-11=C

**3x-2y=-11**

## Comments

_{1}- y_{2 }/ x_{1 }- x_{2 . }The x values go in the denominator.Comment