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Question 49763This question is from textbook PreCalculus with limits agraphing approach
: In Exercises 43-52, find the general form of the equation of the line that passes through the points. Use a graphing utility to sketch the line
44.(4,3),((-4,-4)
This question is from textbook PreCalculus with limits agraphing approach
Answer by tutorcecilia(2152)   (Show Source):
You can put this solution on YOUR website! (4,3),(-4,-4) [Given points]
m =(y2-y1)/(x2-x1) [To find the slope, use the slope formula since two points are given]
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m =(-4-3)(-4-4) [Plug-in the values]
m = -7/-8 = 7/8 [The slope of the line]
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Ax+By=C [General Form of the equation of the line]
(y-y1) = m(x-x1) [Use the point-slope formula since you now have the slope and two points]
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(y-3) = (7/8)(x-4) [Plug-in the values for the (m) and one of the points (4,3)]
y-3 = (7/8)x - (7/8)(4) [Simplify]
y-3 = (7/8)x -7/2 [Isolate all of the variables on one side of the equation]
y-3+3 = (7/8)x-(7/2)+3
y = (7/8)x-(1/2)
y -(7/8)x = (7/8)x-(7/8)x -(1/2)
y -(7/8)x = -1/2 [Re-arrange the terms]
-7/8x + y = -1/2 [General Form of the equation of the line][Final answer]
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Check by plugging the values of the points back into the General Form]
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