# Write an equation in point slope form that passes through (5,1), (-3,4) then write an equation in standard form and in slope intercept form. You need to remember the what the point slope form of equation of the line is such that:

`y - y_1 = m(x - x_1)`

`m = (y_2 - y_1)/(x_2-x_1)`

`y - 1 = (4-1)/(-3-5)*(x - 5)`

`y - 1 = 3/(-8)*(x-5)`

`y - 1 = -3/8(x - 5)`

You...

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You need to remember the what the point slope form of equation of the line is such that:

`y - y_1 = m(x - x_1)`

`m = (y_2 - y_1)/(x_2-x_1)`

`y - 1 = (4-1)/(-3-5)*(x - 5)`

`y - 1 = 3/(-8)*(x-5)`

`y - 1 = -3/8(x - 5)`

You need to remember what is the slope intercept form of equation such that:

`y = mx + b`

Hence, you need to isolate y to the left side and to open the brackets such that:

`y = -3x/8 + 15/8 + 1`

`y = -3x/8 + (15+8)/8`

`y = -3x/8 + 23/8`

You need to write the standard form of equation of line, hence, you need to bring the terms in equation `y = -3x/8 + 23/8`  to a common denominator such that:

`8y = -3x + 23`

You need to move all terms to the left side such that:

`3x + 8y - 23 = 0`

Hence, evaluating the point slope form yields `y - 1 = -3/8(x - 5),`  evaluating the slope intercept form yields `y = -3x/8 + 23/8`  and evaluating the standard form yields `3x + 8y - 23 = 0.`