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Question 351732: a line passes through a point (4,-2) and is parallel to the line containing (-1,4) and (2,-3), what is the standard form?
Found 2 solutions by Alan3354, ewatrrr:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! ----------------------
A line and a point example.
Write in standard form the eqation of a line that satisfies the given conditions. Perpendicular to 9x+3y=36, through (1,2)
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Find the slope of the line. Do that by putting the equation in slope-intercept form, y = mx + b. That means solve for y.
9x+3y = 36
3y= - 9x + 36
y = -3x + 12
The slope, m = -3
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The slope of lines parallel is the same.
The slope of lines perpendicular is the negative inverse, m = +1/3
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Use y = mx + b and the point (1,2) to find b.
2 = (1/3)*1 + b
b = 5/3
The equation is y = (1/3)x + 5/3 (slope-intercept form)
x - 3y = -5 (standard form)
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Answer by ewatrrr(24785)  (Show Source):
You can put this solution on YOUR website! Hi,
*Note paralel lines have the same slop
.Using the point-slope form
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finding the slope
(-1,4) and (2,-3),
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m = -7/3 (the slope of both lines) As the slope is negative, we know the line(s) slope to the left
.
Using the standard slope-intercept form y = mx + b where m is the slope and b is the y- intercept(the value of y when the line passes thru the y-axis)
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for the New line
y = (-7/3)x + b
.
Substituting Pt(4,-2) to solve for b
-2 = -28/3 + b
-6/3 + 28/3 = b
22/3 = b
.
New line is:
y = (-7/3)x + 22/3
.
Standar form would be
3y + 7x = 22
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