Question 1057109
Find the distance between the lines with the equations 3x+2y=14 and y=-3/2x-1/2

a.) Write the y-intercept of the first line, 3x+2y=14 as an ordered pair: (  ,  ) This will be the (x1,y1) you will use in the distance formula.
y-int is (0,7)
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b.) Use (x1,y1) to write the equation of your perpendicular segment in slope intercept form.
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The slope of the given line is -3/2.  The slope of lines perpendicular is +2/3.
Use y-y1 = m*(x-x1) where m = slope and (x1,y1) is the point (0,7)
y - 7 = (2/3)*x
y = 2x/3 + 7
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c.) Give the point (x2,y2) where your perpendicular segment intersects with the second line, y=-3/2x-1/2
y = -3x/2 - 1/2
y = 2x/3 + 7
Find the intersection.
Since they both = y,
-3x/2 - 1/2 = 2x/3 + 7
Multiply by 6
-9x - 3 = 4x + 42
x = -45/13
y = 61/13
--> (-45/13,61/13) is the intersection.
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d.) Find the distance between (x1,y1) and (x2,y2). Round to the nearest hundredth if necessary.
d = sqrt(diffy^2 + diffx^2) = sqrt(30^2 + 45^2)/13
d = sqrt(2925)/13 = 15/sqrt(13)
Same answer.
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Use (0,7) and the distance formula to find the distance from the point to the line.
y=-3/2x-1/2
3x + 2y + 1 = 0 and (0,7)
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d = |Ax + By + C|/sqrt(A^2 + B^2)
d = |3*0 + 2*7 + 1|/sqrt(3^2+2^2)
d = 15/sqrt(13)
d =~ 4.16
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The other steps are not necessary.
If you need to do them, email via the TY note.