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Slopes and y-intercepts of lines?
I forgot to take my notes home, and now I've completely blanked out and forget how to solve a few line questions. Excuses aside, here they are:
Write the equation of the like, in standard form, for each of the following:
1) A line through the point (3,6) and parallel to 2x - 3y - 4 = 0
2) Passing through the points (-2,-7) and (3,-4)
1) For this question, I know how to find the slope. You put it in the form y = mx + b and 'm' is the slope. The slope is the same for both lines because they are parallel. However, what I don't know is how to get the y-intercept of the second line. The answer is 2x - 3y + 12 = 0
2) I think I have an idea how to do this one, but I am unsure. The answer is 3x - 5y - 29 = 0
Thanks for the help!
Grade Nine, why?
4 Answers
- Anonymous1 decade agoFavorite Answer
✐Explanation✐
1. Using the y = mx + b, m is the slope, and b is the y-intercept. If you encounter equation that is not in slope-intercept form, make x = 0 and solve for y.
2. Use the slope formula.
m = (y - y1)/(x - x1)
m = (-7 + 4)/(-2 - 3)
m = -3/-5
m = 3/5
The quickest way to solve for equation is to use slope formula again instead of using the slope-intercept formula.
y - y1 = m(x - x1)
This is same slope formula; you multiply both sides by x - x1 to get that formula above.
Take one of the points and the slope. Then, substitute them. Finally, turn that equation into standard form Ax + By + C = 0
5[y + 4 = 3/5(x - 3)]
5y + 20 = 3x - 9
-3x + 5y + 29 = 0
3x - 5y - 29 = 0
Yes, your answer is correct.
Hope that helps!
Source(s): Knowledge - Anonymous1 decade ago
slope of your given line is 2/3 so the slope of your parallel line is also. Use y = mx+b, put in the slope for m and your point for x and y getting
6 = (2/3)(3) + b solving for b
6 = 2 + b
4 = b
back into y = mx + b
y = (2/3)x + 4 multiply all by 3
3y = 2x + 12
2x - 3y + 12 = 0
2) find the slope between the points and do as above.
- spiffin456Lv 71 decade ago
1. You have 2x-3y=c, just put in x=3 and y=6 to get c=-12
2. There is a formula for this, alternatively work out the gradient and then make the line go through one of the points (and check that it goes through the other one)
