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Find an equation of the line passing through the points (−8,2) with the slope m= 4/3

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Answer:


y=(4)/(3)x+(38)/(3)

Explanation:

So we want to find an equation of a line passing through (-8,2) with a slope of 4/3.

To do so, we can use the point-slope form. This is:


y-y_1=m(x-x_1)

So, substitute (-8,2) for x₁ and y₁, and let m be 4/3:


y-2=(4)/(3)(x+8)

Distribute the right:


y-2=(4)/(3)x+(32)/(3)

Add 2 to both sides. Note that 2 can be written as 6/3. Thus:


y=(4)/(3)x+(32)/(3)+(6)/(3)

Simplify:


y=(4)/(3)x+(38)/(3)

And this is our equation :)

User Sunil Rao
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