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# help

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(7, -3) and (4, -8)

Guest Sep 6, 2017
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First, we must understand what standard form of a line is. Standard form of a line is written like $$Ax+By=C$$ such that A,B, and C are all integers, and A must be positive. First, we must calculate the slope of the line that passes through theses coordinates.

 $$m=\frac{y_2-y_1}{x_2-x_1}$$ As a refresher, this is the equation to figure out the slope of two coordinates. $$m=\frac{-8-(-3)}{4-7}$$ Now, we just simplify the numerator and denominator. $$m=\frac{-8+3}{4-7}$$ $$m=\frac{-5}{-3}=\frac{5}{3}$$

The next step is to utilize point-slope form, which is $$y-y_1=m(x-x_1)$$ where $$(x_1,y_1)$$ is a point on the line. Of course, we already know that (7,-3) and (4,-8) both lie of the line. Therefore, plug in one fot he coordinates. Once converted into point-slope, we must then convert into standard form. This is what is demonstrated in the next step.

 $$y+3=\frac{5}{3}(x-7)$$ Let's multiply all sides by 3 to get rid of the fraction early. $$3y+9=5(x-7)$$ Distribute the 5 to both terms in the parentheses. $$3y+9=5x-35$$ Subtract 9 from both sides. $$3y=5x-44$$ Subtract 5x on both sides. $$-5x+3y=-44$$ We aren't done yet! The coefficient of the x-term must be positive. Therefore, divide by -1 on both sides. $$5x-3y=44$$ This is standard form now, so we are done!
TheXSquaredFactor  Sep 6, 2017

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