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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
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 :)
slope 3/4 passes through the point (−8,4)
The y-intercept form is y = mx + b where m is slope and b is the y-intercept.
Any easy path to this solution is to first right in point-slope form and solve for y.
point-slope formula: y - y1 = m(x - x1) where (x1, y1) is a point on your line and m is slope.
y - 4 = .75(x - -8)
Solve for y and distribute the slope.
y - 4 = .75x + 6
y = .75x + 10
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