A factor of 2 can be removed from these numbers to get (7, -4) and then they can be used directly in an equation for the line.*
.. 7(x-3) -4(y +3) = 0
This can be rewritten to standard form as
.. 7x -4y = 33 . . . . <=== ANSWER
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See the source link for a graph.
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* I like this modified form of point-slope equation for a line. It makes it easy to write the equations of parallel or perpendicular lines through a given point. If two points have difference values of (∆x, ∆y), then a parallel line through (h, k) will have equation
.. ∆y*(x -h) - ∆x*(y -k) = 0 . . . . note the minus sign between terms
On the other hand, a perpendicular line through the point (h, k) will have the equation
.. ∆x*(x -h) + ∆y*(y -k) = 0 . . . . note the plus sign between terms
This is the form of the equation we used above.
If you are given a line in standard form, you can use the x- and y-coefficients for ∆y and -∆x, respectively.
"Standard form" requires the leading coefficient be positive, so you may have to multiply the equations given above by -1 to put the result in standard form.
Answers & Comments
Verified answer
The midpoint is the average of the two points:
.. ((10, -7) + (-4, 1))/2 = (6/2, -6/2) = (3, -3)
The difference of the two points is
.. (10, -7) - (-4, 1) = (14, -8)
A factor of 2 can be removed from these numbers to get (7, -4) and then they can be used directly in an equation for the line.*
.. 7(x-3) -4(y +3) = 0
This can be rewritten to standard form as
.. 7x -4y = 33 . . . . <=== ANSWER
_____
See the source link for a graph.
_____
* I like this modified form of point-slope equation for a line. It makes it easy to write the equations of parallel or perpendicular lines through a given point. If two points have difference values of (∆x, ∆y), then a parallel line through (h, k) will have equation
.. ∆y*(x -h) - ∆x*(y -k) = 0 . . . . note the minus sign between terms
On the other hand, a perpendicular line through the point (h, k) will have the equation
.. ∆x*(x -h) + ∆y*(y -k) = 0 . . . . note the plus sign between terms
This is the form of the equation we used above.
If you are given a line in standard form, you can use the x- and y-coefficients for ∆y and -∆x, respectively.
"Standard form" requires the leading coefficient be positive, so you may have to multiply the equations given above by -1 to put the result in standard form.
Slope of YZ = (- 7 - 1)/(10 - (-4))
= -8/14
= - 4/7
slope of perpendicular bisector = 7/4
Mid-point of YZ = ( (10-4)/2, (-7 + 1)/2)
= (3, - 3)
Equation of perpendicular bisector
y + 3 = 7/4(x - 3)
4y + 12 = 7x - 21
7x - 4y = 33
Do your own damn homework.