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QuestionFind the standard form of the line through points (−5,−1)(-5,-1) and (10,−7)(10,-7) given the point-slope form. Options: 2x−5y=−152 x-5 y=-15, 2x−5y=−172 x-5 y=-17, 2x+5y=−152 x+5 y=-15, 2x+5y=−172 x+5 y=-17.

Studdy Solution

STEP 1

Assumptions1. The point-slope form of the equation of the line is given as y+7=−5(x−10)y+7=-\frac{}{5}(x-10). We need to convert this equation into standard form. The standard form of a linear equation is Ax+By=CAx + By = C, where A, B, and C are integers, and A and B are not both zero.

STEP 2

First, let's distribute the −25-\frac{2}{5} on the right side of the equation.
y+7=−25x+4y+7=-\frac{2}{5}x +4

STEP 3

Next, let's isolate y on one side of the equation by subtracting7 from both sides.
y=−25x+−7y=-\frac{2}{5}x + -7

STEP 4

implify the right side of the equation.
y=−2x−3y=-\frac{2}{}x -3

STEP 5

To convert this equation into standard form, we need to eliminate the fraction by multiplying every term by5.
5y=−2x−155y=-2x -15

STEP 6

Rearrange the equation to get it in the form Ax+By=CAx + By = C.
2x+5y=−152x +5y = -15

STEP 7

In standard form, A should be positive. So, multiply the entire equation by -1 to make A positive.
−2x−5y=15-2x -5y =152x+5y=−152x +5y = -15So, the standard form of the equation for this line is 2x+5y=−152x +5y = -15.

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