Question

Find an equation of the line that has the given slope and contains the given point. m=7/3 (1,6)

Answer 1

The point-slope form:

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

We have the point (1, 6) and the slope m = 7/3. Substitute:

$y-6=\dfrac{7}{3}(x-1)$      use distributive property

$y-6=\dfrac{7}{3}x-\dfrac{7}{3}$          add 6 to both sides

$y=\dfrac{7}{3}x+\dfrac{11}{3}$        multiply both sides by 3

$3y=7x+11$           subtract 7x from both sides

$-7x+3y=11$         change the signs

$7x-3y=-11$

Answer:

point-slope form: $y-6=\dfrac{7}{3}(x-1)$

slope-intercept form: $y=\dfrac{7}{3}x+\dfrac{11}{3}$

standard form: $7x-3y=-11$