SOLUTION: can somebody help me solve this problem. i need to find the equation of the line with these characteristics. I need to express my answer using either the general form or the slop

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Question 161909This question is from textbook college algebra
: can somebody help me solve this problem. i need to find the equation of the line with these characteristics. I need to express my answer using either the general form or the slope-intercept form of the equation of the line.

x-intercept=2; containing the point (4,-5)
i forgot how to find the slope with only the, x-intercept = 2 given.
This question is from textbook college algebra

Found 2 solutions by scott8148, Electrified_Levi:
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
the x-intercept is (2,0)

slope is (-5-0)/(4-2) __ m=-5/2

using point-slope __ y-0=(-5/2)(x-2) __ y=(-5/2)x+5

Answer by Electrified_Levi(103) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, Hope I can help,
.
can somebody help me solve this problem. i need to find the equation of the line with these characteristics. I need to express my answer using either the general form or the slope-intercept form of the equation of the line.
x-intercept=2; containing the point (4,-5)
i forgot how to find the slope with only the, x-intercept = 2 given.
.
The slope intercept form of the equation is +y+=+mx+%2B+b+, where "m" is the slope, "b" is the y-intercept
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The x-intercept is the point where the line crosses the "x" axis
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x intercept is given as the point ( p, 0 ) where "p" is the "x intercept number", in our case, it would be ( 2,0 )
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Now we have another point (2,0)
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Now that we have two points we can find the slope of the line
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The two points are (2,0) and (4, -5)
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(2,0)(x1,y1) and (4, -5)(x2,y2)
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The slope equation is +%28+y2-y1%29%2F%28x2-x1%29+, we can replace the variables with numbers
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Our slope = +%28+y2-y1%29%2F%28x2-x1%29+ = +%28+%28-5%29-%280%29%29%2F%28%284%29-%282%29%29+ = +-5%2F2+
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Our slope = +-5%2F2+
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Our slope intercept form is +y+=+mx+%2B+b+, "m" is our slope, so we can replace "m" with +-5%2F2+
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Our equation is now +y+=+%28-5%2F2%29x+%2B+b+, to find "b", we will replace "x" and "y" with one of our two points ( doesn't matter which one ) ( we will use point (2,0), replace "x" with "2", "y" with "0")
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(2,0)(x,y) , +y+=+%28-5%2F2%29x+%2B+b+ = +%280%29+=+%28-5%2F2%29%282%29+%2B+b+ = +0+=+%28-5%29+%2B+b+, we will move (-5) to the left side
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+0+=+%28-5%29+%2B+b+ = +0+%2B+5+=+%28-5%29+%2B+5+%2B+b+ = ++5+=+b+ = +b+=+5+, we found "b", we can replace "b" with "5" in the original equation
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+y+=+%28-5%2F2%29x+%2B+b+ = +y+=+%28-5%2F2%29x+%2B+%285%29+ = +y+=+%28-5%2F2%29x+%2B+5+
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+y+=+%28-5%2F2%29x+%2B+5+ is the slope intercept form of the equation, for the general equation, we will multiply each side by "2"
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+y+=+%28-5%2F2%29x+%2B+5+ = +%282%29y+=+%282%29%28%28-5%2F2%29x+%2B+5%29+ = +2y+=+%282%29%28-5%2F2%29x+%2B+%282%29%285%29+ = +2y+=+%28-5%29x+%2B+10+ = +2y+=+-5x+%2B+10+, we will move (-5x) to the left side
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+2y+=+-5x+%2B+10+ = +2y+%2B+5x+=+-5x+%2B+5x+%2B+10+ = +2y+%2B+5x+=+10+, rearranging it becomes
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+2y+%2B+5x+=+10+ = +5x+%2B+2y+=+10+
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+5x+%2B+2y+=+10+ is the general/standard form, we can check our equation by replacing "x" and "y" with both of our two points
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The two points are (2,0) and (4, -5)
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(2,0)(x,y) = +5x+%2B+2y+=+10+ = +5%282%29+%2B+2%280%29+=+10+ = +10+%2B+0+=+10+ = +10+=+10+ ( True )
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(4, -5)(x,y) = +5x+%2B+2y+=+10+ = +5%284%29+%2B+2%28-5%29+=+10+ = +20+%2B+%28-10%29+=+10+ = +20+-+10+=+10+ = +10+=+10+ ( True )
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+y+=+%28-5%2F2%29x+%2B+5+ is the slope-intercept form of the equation
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+5x+%2B+2y+=+10+ is the general/standard form of the equation
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This is the line on a graph
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Hope I helped, Levi