Question 732415: Find the equation with the given properties and sketch the graph-passes through the point of intersection of 2x+y-3=0 and x-y-3=0 and through the point (4,3). I am a bit lost on this: I've solved for each y intercepts (y=-2x-3 and y=x-3). Do I take the intersection of these two (2,-1) and do something like y=-x+1? This was done by looking at the graph, not performing the math...which I am unable to figure out-help.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find the equation with the given properties and sketch the graph-passes through the point of intersection of 2x+y-3=0 and x-y-3=0 and through the point (4,3). I am a bit lost on this: I've solved for each y intercepts (y=-2x+3 and y=x-3).
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So: -2x+3 = x-3
3x = 6
x = 2
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And:
y = -2*2+3 = -1
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You want the equation of the line thru (2,-1) and (4,3)
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slope = (3--1)/(4-2) = 4/2 = 2
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Form: y = mx + b
Solve for "b":
3 = 2*4 + b
b = -5
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Equation:
y = 2x-5
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Cheers,
Stan H.
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