Question

Please help urgent math photo attached ​

Answer 1

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'

__

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

__

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

__

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)

_____

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)

__

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.