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1) Find the equation of the line with a slope of 7 and a y intercept of -1(in slope intercept form) compared with 2) the equation of the line that passes through (1,5) and has a slope of 3(in point slope form)
Also find the slope of a line containing points (-2,8) and 3,5).
All help is very appreciated, if you could show/tell me how you get the results please and thank you.
Is #1 y=7x-1 & #2 y-5=3(x-1)
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The standard form of the equation `y=mx+c` is the result we seek.
When using the point-slope form you need the formula
`y - y_(1) = m(x - x_(1))` .
The `x_(1) and y_(1)` represent the points at 1) (0;-1) and at 2) (1;5) Note that the y-intercept is what tells us that the x value is zero at 1). `m` represents the slope. Plug in these values to the equation:
1) `y-(-1) = 7(x-0)`
`therefore y+1= 7x`
`therefore y = 7x - 1` so yes your answer is correct.
Simarly for 2)`y-(5) = 3(x-1)`
`therefore y=3x -3 +5`
`therefore y=3x+2` so yes your answer for 2) is correct just not completed.
Please repost your remaining question separately as eNotes rules do not allow multiple questions
Ans: y = 7x - 1 compared to y=3x+2
`1) y=mx+q` so m=7
Subduing it at the passage for the point: `P(0;-1)` :
`-1= 0 xx x +q` `rArr q=-1`
`2)` `y=mx+q` m=3 `Q(1;5)`
`5=3xx 1+q` `rArr q=2`
Red line : `y=7x-1`
Blue line: `y=3x+2`
`3)` We have to use:
So that we've found the slope: `y=-3/5 x+q`
Subduing the passage in `(-2;8)`
`8=-3/5 xx(-2)+q` `q=34/5`
Note, if we subdue it at passage in `(3;5)` :
`5= -3/5 xx3+q` `q=34/5`
The value of q doesn't depend by the point the straight line pass. as we are awaitng.
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