Find an equation of the line having a given slope of 3/4 that goes through the point (-8,4).

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We are asked to find the equation of the line with slope `m=3/4` and containing the point (-8,4). Recognize that a line is determined by two points (through two points there is exactly one line.) Also, through two points we can find the slope (if it exists) by using the slope formula: `m=(y_2-y_1)/(x_2-x_1)` . A line is also determined by a point on the line and the slope.

(1) We can find the unique equation of the line in slope-intercept form (y=mx+b.) We substitute the known values for m,x, and y in order to find b. Here `4=3/4(-8)+b ==> 4=-6+b ==> b=10` . Since we know m and b we can write `y=3/4x+10` as the equation we seek.

The equation is `y=3/4x+10`

(2) We can also use the point-slope formula. It is derived from the slope formula:

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

Here `y-(4)=3/4(x-(-8)) ==> y-4=3/4(x+8)`

This form is not unique as it depends on the chosen point. Using the distributive property to eliminate the parantheses and adding 4 to both sides we get y=3/4x+10 as before.

(3) We can write an equation in standard form. From `y=3/4x+10` we get `4y=3x+10` by multiplying both sides of the equation by 4. Then

3x-4y=-10 is the standard form and 3x-4y+10=0 is the general form of the line.

(4) Once we have the y-intercept as b=10 we can solve for the x-intercept `-40/3` . Then the intercept form is `x/(-40/3)+y/10=1` or `(-3x)/40+y/10=1`

This form is unique.

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