y-3=3(x+1) equation.

rewrite this equation in slope intercept form. & what is the y-intercept of this line? & rewrite this equation in standard form. & what is the x-intercept of this line & what is the equation in the standard form of a perpendicular line that passes through (5,1). & what is the x-intercept of the perpendicular line. & graph the original equation given. include your 3 pointss from problems 6,8,and 10, labeled a,b,&c respectively? can you show me the calculations also please!

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Rewrite `y - 3 = 3 (x+1)` in slope intercept form: distribute 3 and add three to both sides we get:

`y = (3x + 3) + 3 = 3x + 6`

`y = 3x + 6` **Slope intercept form**. **Y-intercept is (0, 6)** and slope is 3.

To rewrite this equation in **standard form (ax + by = c)**, subtract y from each side, and subtract 6 from each side, we'd get: `3x - y = -6.`

x-intercept, we let y = 0, so `3x = -6, so x = -2`

**x-intercept** = `(-2, 0)`

The line perpendicular to this line will have the opposite reciprocal slope. The original equation has a slope of 3 as defined earlier, therefore the `_|_` line has a slope of `-1/3.`

Using point slope form`y-y_(1)=m(x -x_(1))` with slope of `-1/3` and containing point (5,1) we have the equation:

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

` `Now place in standard form by multiplying both sides by -3 to get:

`-3(y - 1) = x - 5` Distribute -3 to get:

`-3y + 3 = x - 5` Rearrange to ax + by = c, **to get:**

`x + 3y = 8`

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