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-5=3x-3`

`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`

so: `y=7x-1`

`2)` `y=mx+q` m=3 `Q(1;5)`

`5=3xx 1+q` `rArr q=2`

`y=3x+2`

Red line : `y=7x-1`

Blue line: `y=3x+2`

`3)` We have to use:

`(y-y_0)=m(x-x_0)`

So that:

`8-5=m(-2-3)`

`-3/5=m`

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`

`y=1/5(-3x+34)`

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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