Coordinate Geometry: Distance and Midpoint
Finding lengths and middle points on the coordinate plane
Coordinate Geometry: Distance and Midpoint
Finding lengths and middle points on the coordinate plane
Math - Grade 9-12
- 1
Find the distance between A(2, 3) and B(8, 11).
Use the distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2).
The distance is 10 units because the horizontal change is 6, the vertical change is 8, and sqrt(6^2 + 8^2) = sqrt(100) = 10. - 2
Find the midpoint of the segment with endpoints C(-4, 7) and D(6, -1).
The midpoint is (1, 3) because ((-4 + 6)/2, (7 + -1)/2) = (2/2, 6/2) = (1, 3). - 3
Find the distance between P(-3, -5) and Q(9, 0). Leave your answer in simplest radical form.
Subtract the x-coordinates and y-coordinates, square the differences, add them, and take the square root.
The distance is 13 units because the horizontal change is 12, the vertical change is 5, and sqrt(12^2 + 5^2) = sqrt(169) = 13. - 4
Find the midpoint of the segment with endpoints R(10, -6) and S(-2, 4).
The midpoint is (4, -1) because ((10 + -2)/2, (-6 + 4)/2) = (8/2, -2/2) = (4, -1). - 5
The midpoint of segment AB is M(3, -2). One endpoint is A(7, 4). Find the other endpoint B.
Set up two equations using the midpoint formula, one for x and one for y.
The other endpoint is B(-1, -8). Since the midpoint averages the coordinates, (7 + x)/2 = 3 gives x = -1, and (4 + y)/2 = -2 gives y = -8. - 6
Find the distance between E(-6, 2) and F(1, -4). Round your answer to the nearest tenth.
The distance is about 9.2 units because sqrt((1 - -6)^2 + (-4 - 2)^2) = sqrt(7^2 + -6^2) = sqrt(85), and sqrt(85) is about 9.2. - 7
A line segment has endpoints G(-8, -3) and H(4, 5). Find its midpoint and its length.
This problem uses both the midpoint formula and the distance formula.
The midpoint is (-2, 1), and the length is 4sqrt(13) units. The midpoint is ((-8 + 4)/2, (-3 + 5)/2), and the distance is sqrt(12^2 + 8^2) = sqrt(208) = 4sqrt(13). - 8
A circle has center K(2, -1). Point L(8, 7) lies on the circle. Find the radius of the circle.
The radius is 10 units because the radius is the distance from the center K to point L, and sqrt((8 - 2)^2 + (7 - -1)^2) = sqrt(6^2 + 8^2) = 10. - 9
Points A(1, 2), B(7, 2), C(7, 8), and D(1, 8) form a quadrilateral. Use distances to classify the quadrilateral as a square, rectangle, or neither.
Find the length of each side and notice whether the sides are horizontal or vertical.
The quadrilateral is a square. Each side has length 6 units, since AB = 6, BC = 6, CD = 6, and DA = 6, and the sides are horizontal or vertical so the angles are right angles. - 10
The endpoints of a diameter of a circle are A(-5, 9) and B(7, -3). Find the center of the circle.
The center of the circle is (1, 3) because the center is the midpoint of the diameter, so ((-5 + 7)/2, (9 + -3)/2) = (1, 3). - 11
Point M(4, 6) is the midpoint of segment PQ. If P(-2, 10), find Q.
The midpoint coordinates are the averages of the endpoint coordinates.
Point Q is (10, 2). Using the midpoint formula, (-2 + x)/2 = 4 gives x = 10, and (10 + y)/2 = 6 gives y = 2. - 12
Triangle JKL has vertices J(-1, 1), K(5, 1), and L(2, 5). Use distances to determine whether the triangle is isosceles, scalene, or equilateral.
Find all three side lengths, then compare them.
The triangle is isosceles. JK = 6, JL = sqrt((2 - -1)^2 + (5 - 1)^2) = 5, and KL = sqrt((2 - 5)^2 + (5 - 1)^2) = 5, so two sides are equal.