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Mathematics 12 Online
OpenStudy (thedj4jc):

What is the equation of the line, in slope-intercept form, that passes through (3, -1) and (-1, 5)? y = -3/2x + 7/2 2y = -3x + 7 2x + 3y - 7 = 0

OpenStudy (studygurl14):

Hello @thedj4JC ! Do you know the slope formula?

OpenStudy (thedj4jc):

No, I don't

OpenStudy (studygurl14):

The slope formula is a very useful tool that helps you find the slope of a line. If you know the slope and a point the line passes through, you can find the equation of the line. So, very handy. If a line passes through two points \((x_1,y_1)\) and \((x_2,y_2)\), the slope of that line equals... \(\Large\frac{y_2-y_1}{x_2-x_1}\)

OpenStudy (studygurl14):

Remember that yours points are (3.-1) and (-1.5). Can you find the slope of the line?

OpenStudy (thedj4jc):

It'd be -1-5/3-(-1)? -6/4?

OpenStudy (thedj4jc):

-1.5??

OpenStudy (studygurl14):

Yeah, -6/4 is correct. nice job. But you should reduce it.

OpenStudy (studygurl14):

-6/4 = -3/2

OpenStudy (studygurl14):

So, next step. Now you want to find the equation of the line. Slope-intercept form of a line: y = mx + b m = slope So, you get y = (-3/2)x + b But we still don't know what b is. Luckily, we know that point (3, -1) goes through the line. Remember that a coordinate is in form (x,y). So, that means you can plug in 3 for x and -1 for y into the equation and solve for b. can you do that?

OpenStudy (thedj4jc):

To solve for b, just substrate mx ((-3/2)3) from y (-1)?

OpenStudy (studygurl14):

y = mx + b We know the slope y = (-3/2)x + b We have point (3, -1) (-1) = (-3/2)(3) + b Now solve for b

OpenStudy (thedj4jc):

(-3/2)(3) = - 4 1/2 -1 - (-4 1/2) = 7/2

OpenStudy (studygurl14):

Nice job. Now just plug in the slope and the b-value back in... y=mx + b m = -3/2, b = 7/2 \(\Large y=-frac{3}{2}x+\frac{7}{2}\) Now you've got your equation

OpenStudy (studygurl14):

Sorry, let me do that again

OpenStudy (studygurl14):

\(\Large y=- \frac{3}{2}x+\frac{7}{2}\)

OpenStudy (thedj4jc):

Thank you!!!!

OpenStudy (studygurl14):

You're welcome! :)

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