A line having a slope of 3/4 passes through the point (−8,4).

Write the equation of this line in slope-intercept form.

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Let the required line be y = mx+c in the slope intercept form, where m is the slope and c is the y intercept.

We determine m and c by the using given conditions.

m = 3/4, given.

So y = (3/4)x+c (1).

The line at (1) has the point (-8, 4) on it.

Therefore 4 = (3/4)*(-8)+c...(2)

(1)-(2) gives: y-4 = (3/4)(x+8).

=> y -4 = (3/4)x + 6

We multiply 4 to get integral coefficients.

4(y-4) = 3x+24

We rearrange.

3x-4y+24+16= 0

**3x-4y+40 = 0 **is the required line.

We are asked to find the equation of a line which contains (-8,4) as a point and has a slope of 3/4.

We will begin by writing the slope-intercept form.

=> y = mx + b

=> We will substitute the given point for x and y and the slope of 3/4 for m. This will allow us to find the y-intercept which is "b."

=> 4 = 3/4(-8) + b

=> 4 = -6 + b

=> 10 = b

Now we will place 3/4 into the slope-intercept form as m and 10 as b.

=> y = 3/4x + 10.

**The equation of the line is y = 3/4x + 10.**

Thank you ma'am :)

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