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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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...
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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.
Plugging in:
y - 4 = .75(x - -8)
Solve for y and distribute the slope.
y - 4 = .75x + 6
y = .75x + 10
Thank you ma'am :)
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.
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