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# help

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Write the standard form of the line that passes through the given points. Include your work in your final answer. Type your answer in the box provided or use the upload option to submit your solution.

(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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