Answer:
K = (-5, 14)
Step-by-step explanation:
For writing equations of parallel and perpendicular lines, it is convenient to start with equations in general form:
ax +by +c = 0
Rewriting the given equations, we have ...
y = 1/2x -1
x -2y -2 = 0 . . . . . line d
y = -3x +2
3x +y -2 = 0 . . . . . line d'
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A parallel line will have the same coefficients, but a constant appropriate to the required point on the line.
Parallel to d' through A(-1, 2):
3(x -(-1)) +(y -2) = 0
3x +y +1 = 0 . . . . . . . . simplify
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A perpendicular line will have the coefficients swapped and one of them negated. Again, the constant is appropriate to the required point.
Perpendicular to d through B(3, -2):
2(x -3) +(y -(-2)) = 0
2x +y -4 = 0 . . . . . . . . simplify
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The solution to this system of equations can be found by subtracting the second equation from the first:
(3x +y +1) -(2x +y -4) = (0) -(0)
x +5 = 0
x = -5
Substituting into the second equation gives ...
2(-5) +y -4 = 0
y -14 = 0
y = 14
The coordinates of point K are ...
K = (-5, 14)
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Additional comment
The general line through the origin ...
ax +by = 0
can be translated to go through point (h, k) by the usual method for translating functions. The variable x is replaced by (x -h), and the variable y is replaced by (y -k). Using the general form equation means we can keep the variables and constants all on the same side of the equal sign, saving a bit of work and confusion.
a(x -h) +b(y -k) = 0 . . . . . . line through point (h, k)
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The graph is drawn by a geometry program capable of drawing parallel and perpendicular lines through given points. It confirms the equations and th solution.
