How to find the equation of the line passing through the points (-2/3,1)and (-2,1/2) and how to....

write the equation in point-slope form, slope-intercept form, and standard form?

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The slope of the line will be (-2 - (-2/3)) / ((1/2) - 1) = 8/3.

Point-slope form using (-2,1/2) as the point: y - 1/2 = (8/3)(x + 2)

Slope-intercept from form using (-2,1/2) as the point: y = (8/3)x + 35/6

Standard form using (-2,1/2) as the point: (8/3)x - y = -35/6 OR 16x - 6y = -35.

The formula for the slope is ( y2 - y1 ) / (x2 - x1 ) , so the slope commuted here in this answer needs to be inverted to be correct.

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First, find the gradient:

(1/2-1)(/-2-(-2/3)) = 3/8

Use y-b = m(x-a) and (-2/3, 1)

y-1 = 3/8(x-(-2/3))

y-1 =3/8(x+2/3) in point slope form

8(y-1) = 3(x+2/3)

8y-8 = 3x+2

3x-8y+2+8

3x-8y+10 = 0 in form of general equation of a line

If 8y-8 = 3x+2

8y = 3x+2+8

8y = 3x+10

y = 3/8x+10/8

y = 3/8x+5/4 in form y = mx+c