I really need help with this problem.

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  a) line: y = x - 11

  b) k = -9

  c) AP = BP = √74

Step-by-step explanation:

There are four (4) formulas that are needed for solving this problem.

  m = (y2 -y1)/(x2 -x1) . . . . . . slope of a line between points (x1, y1) and (x2, y2)

  M = ((x1 +x2)/2, (y1 +y2)/2) . . . . midpoint of the segment between the points

  y -y1 = m(x -x1) . . . . . point-slope equation of a line given point and slope

  d = √((x2 -x1)² +(y2 -y1)²) . . . . distance between two points

In addition, you need to know that perpendicular lines have opposite reciprocal slopes.

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The perpendicular bisector of a segment between two points will be a line with a slope that is the opposite reciprocal of the slope of the segment. That line will go through the midpoint of the segment.

a) The slope of the segment between A(7, -2) and B(9, -4) is ...

  m = (-4-(-2))/(9-7) = -2/2 = -1

The midpoint of AB is ...

  M = ((7+9)/2, (-2-4)/2) = (8, -3)

The opposite reciprocal of the slope of AB is ...

  m = -1/(-1) = 1

Then the point-slope equation of the perpendicular bisector is the line with slope 1 through the point (8, -3):

  y -(-3) = 1(x -8)

  y = x -11

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b) Point P on the line has x-value of 2, so the y-value will be ...

  k = 2 -11 = -9

Point P has coordinates (2, -9).

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c) The length of segment AP is given by the distance formula:

  d = √((2 -7)² +(-9-(-2))²) = √((-5)² +(-7)²) = √74

The length of segment BP is likewise given by the distance formula:

  d = √((2 -9)² +(-9 -(-4))²) = √((-7)² +(-5)²) = √74

AP = √74 = BP   ⇒   AP ≅ BP

x = 3 ± √(5+y)

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.