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Geometry Grade 9-12 Answer Key

Geometry: Coordinate Geometry

Using coordinates to analyze distance, slope, midpoint, and equations of lines

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Geometry: Coordinate Geometry

Using coordinates to analyze distance, slope, midpoint, and equations of lines

Geometry - Grade 9-12

Instructions: Read each problem carefully. Show your work in the space provided. Give exact answers when possible, and round only when the problem asks you to.
  1. 1

    Find the distance between A(2, -3) and B(8, 5).

    Use the distance formula with the changes in x and y.

    The distance is 10 units. Using the distance formula, d = sqrt((8 - 2)^2 + (5 - (-3))^2) = sqrt(6^2 + 8^2) = sqrt(100) = 10.
  2. 2

    Find the midpoint of the segment with endpoints P(-4, 7) and Q(6, -1).

    The midpoint is (1, 3). The x-coordinate is (-4 + 6) / 2 = 1, and the y-coordinate is (7 + (-1)) / 2 = 3.
  3. 3

    Find the slope of the line passing through C(-2, 4) and D(3, -6).

    Slope equals change in y divided by change in x.

    The slope is -2. The change in y is -6 - 4 = -10, and the change in x is 3 - (-2) = 5, so the slope is -10 / 5 = -2.
  4. 4

    Write the equation of the line with slope 3 that passes through the point (2, -5). Give your answer in slope-intercept form.

    The equation is y = 3x - 11. Substituting (2, -5) into y = 3x + b gives -5 = 6 + b, so b = -11.
  5. 5

    Determine whether the line through A(1, 2) and B(5, 10) is parallel to the line through C(-3, 4) and D(2, 14). Explain your answer.

    Parallel nonvertical lines have equal slopes.

    The lines are parallel because they have the same slope. The slope of AB is (10 - 2) / (5 - 1) = 8 / 4 = 2, and the slope of CD is (14 - 4) / (2 - (-3)) = 10 / 5 = 2.
  6. 6

    Determine whether the line through E(-1, 3) and F(5, 6) is perpendicular to the line through G(2, -4) and H(5, -10). Explain your answer.

    Perpendicular nonvertical lines have slopes whose product is -1.

    The lines are perpendicular. The slope of EF is (6 - 3) / (5 - (-1)) = 3 / 6 = 1/2, and the slope of GH is (-10 - (-4)) / (5 - 2) = -6 / 3 = -2. Since 1/2 and -2 are negative reciprocals, the lines are perpendicular.
  7. 7

    A triangle has vertices A(0, 0), B(6, 0), and C(6, 8). Find the lengths of all three sides and classify the triangle by its side lengths.

    The side lengths are AB = 6, BC = 8, and AC = 10. Since all three side lengths are different, the triangle is scalene. It is also a right triangle because 6^2 + 8^2 = 10^2.
  8. 8

    Find the coordinates of the point that divides the segment from A(2, 1) to B(8, 13) in a 1:2 ratio, measured from A to B.

    A 1:2 ratio means the whole segment is split into 3 equal parts.

    The point is (4, 5). A 1:2 ratio from A to B means the point is one third of the way from A to B. The change from A to B is (6, 12), and one third of that is (2, 4). Adding this to A gives (4, 5).
  9. 9

    Write the equation of the perpendicular bisector of the segment with endpoints M(-2, 4) and N(4, -2).

    Find the midpoint first, then use the negative reciprocal of the segment's slope.

    The equation is y = x. The midpoint of MN is (1, 1). The slope of MN is (-2 - 4) / (4 - (-2)) = -6 / 6 = -1, so the perpendicular slope is 1. A line with slope 1 through (1, 1) is y - 1 = 1(x - 1), which simplifies to y = x.
  10. 10

    A quadrilateral has vertices A(-3, 1), B(1, 4), C(5, 1), and D(1, -2). Use slopes or distances to determine what type of quadrilateral it is.

    The quadrilateral is a rhombus. All four side lengths are 5 units: AB, BC, CD, and DA each have distance sqrt(4^2 + 3^2) = 5. Opposite sides are parallel because their slopes match in pairs.
  11. 11

    Find the area of the triangle with vertices A(1, 1), B(7, 1), and C(4, 6).

    Use the horizontal base AB and find the vertical height from C to that base.

    The area is 15 square units. The base from A to B is 6 units because it is horizontal, and the height from C to the line y = 1 is 5 units. The area is (1/2)(6)(5) = 15.
  12. 12

    The equation of a circle is (x - 3)^2 + (y + 2)^2 = 49. Identify the center and radius.

    The center is (3, -2), and the radius is 7. The standard form is (x - h)^2 + (y - k)^2 = r^2, so h = 3, k = -2, and r = sqrt(49) = 7.
  13. 13

    Write the equation of the circle with center (-4, 5) that passes through the point (2, 13).

    Find the radius using the distance formula, then use the standard circle equation.

    The equation is (x + 4)^2 + (y - 5)^2 = 100. The radius is the distance from (-4, 5) to (2, 13), which is sqrt(6^2 + 8^2) = 10, so r^2 = 100.
  14. 14

    A line has equation 2x - 3y = 12. Find its x-intercept and y-intercept.

    The x-intercept is (6, 0), and the y-intercept is (0, -4). To find the x-intercept, set y = 0 and solve 2x = 12. To find the y-intercept, set x = 0 and solve -3y = 12.
  15. 15

    A rectangle has vertices A(-2, -1), B(4, -1), C(4, 3), and D(-2, 3). Find the perimeter and area of the rectangle.

    Because the sides are horizontal and vertical, subtract x-values for width and y-values for height.

    The perimeter is 20 units, and the area is 24 square units. The width is 4 - (-2) = 6 units, and the height is 3 - (-1) = 4 units. The perimeter is 2(6 + 4) = 20, and the area is 6 times 4 = 24.
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