Question 360551
What is the equation of a line with points (3,4) and (11,18). After that I need the intersection of that line and line y=-3/2x +12.
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Slope of (3,4) and (11,18)
m = (18-4)/(11-3) =14/8 = 7/4
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Using one point (3,4) and the slope 7/4, plug into "point-slope" form:
y - y1 = m(x - x1)
y - 4 = 7/4(x - 3)
y - 4 = (7/4)x - 21/4
y = (7/4)x - 21/4 + 4
y = (7/4)x - 21/4 + 16/4
y = (7/4)x - 5/4
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Intersection:
(7/4)x - 5/4 = -3/2x +12
multiply both sides by 4:
7x - 5 = -6x + 48
13x - 5 = 48
13x = 53
x = 53/13
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find y by plugging x into:
y=-3/2x +12
y=-3/2(53/13) +12
2y = -3(53/13) +24
26y = -3(53) +312
26y = -159 +312
26y = 153
y = 153/26
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intersects at (53/13, 153/26)