Distance Problems with Solutions

The concept of the distance between two points is an important one in mathematics. A set of distance problems with detailed solutions (at the bottom of this page) are presented.

Problems

Problem 1:

Find the distance between the points (2, 3) and (0, 6).

Problem 2:

Find the distance between point (-1, -3) and the midpoint of the line segment joining (2, 4) and (4, 6).

Problem 3:

Find x so that the distance between the points (-2, -3) and (-3, x) is equal to 5.

Problem 4:

Find x and y if (2, 5) is the midpoint of points (x, y) and (-5, 6).

Problem 5:

Find the point (0, y) that is equidistant from (4, -9) and (0, -2).

Problem 6:

Show that the triangle that has (0, 1), (2, 3) and (2, -1) as vertices is right isosceles.

Problem 7:

Find the length of the hypotenuse of the right triangle whose vertices are given by the points (-2, 1), (1, 1) and (1, 2).

Problem 8:

Find a relationship between x and y so that the distance between the points (x, y) and (-2, 4) is equal to 5.

Problem 9:

Find a relationship between x and y so that (x, y) is equidistant from the two points (-3, 4) and (0, -3).

Problem 10:

Find a relationship between x and y so that the triangle whose vertices are given by (x, y), (1, 1) and (5, 1) is a right triangle with the hypotenuse defined by the points (1, 1) and (5, 1).

Solutions to the Above Problems

Solution to Problem 1:

The formula for the distance \(D\) between two points \((a, b)\) and \((c, d)\) is given by \[ D = \sqrt{(c - a)^2 + (d - b)^2} \] Apply the formula given above to find the distance \(D\) between the points \((2, 3)\) and \((0, 6)\) as follows \[ D = \sqrt{(0 - 2)^2 + (6 - 3)^2} \] \[ = \sqrt{13} \]

Solution to Problem 2:

We first find the coordinates of the midpoint \(M\) of the segment joining \((2, 4)\) and \((4, 6)\). \[ M = \left( \dfrac{2 + 4}{2}, \dfrac{4 + 6}{2} \right) \] \[ M = (3, 5) \] We now use the distance formula to find the distance between the points \((-1, -3)\) and \((3, 5)\). \[ D = \sqrt{(3 - (-1))^2 + (5 - (-3))^2} \] \[ D = \sqrt{80} = 4\sqrt{5} \]

Solution to Problem 3:

Use the distance formula to write an equation in \(x\): \[ 5 = \sqrt{\,(-3 - (-2))^2 + (x - (-3))^2} \] Simplify the expression under the square root: \[ 5 = \sqrt{\,1 + (x + 3)^2} \] Square both sides: \[ 25 = 1 + (x + 3)^2 \] Solve for \(x\): \[ (x + 3)^2 = 24 \] Two solutions: \[ x = -3 + 2\sqrt{6} \quad \text{or} \quad x = -3 - 2\sqrt{6} \]

Solution to Problem 4:

Use midpoint formula. \[ (2,5) = \left( \dfrac{x + (-5)}{2}, \; \dfrac{y + 6}{2} \right) \] Equate the coordinates: \[ 2 = \dfrac{x - 5}{2}, \qquad 5 = \dfrac{y + 6}{2} \] Solve for \(x\) and \(y\): \[ x = 9, \qquad y = 4 \]

Solution to Problem 5:

Use distance formula to find the distance \(D_1\) from \((0, y)\) to \((4, -9)\) and the distance \(D_2\) from \((0, y)\) to \((0, -2)\). \[ D_1 = \sqrt{(4-0)^2 + (-9-y)^2} \] \[ D_2 = \sqrt{(0-0)^2 + (-2-y)^2} \] The two distances are equal, hence \[ \sqrt{(4-0)^2 + (-9-y)^2} = \sqrt{(0-0)^2 + (-2-y)^2} \] Square both sides and simplify: \[ 16 + (-9-y)^2 = (-2-y)^2 \] \[ 16 + 81 + 18y + y^2 = 4 + y^2 + 4y \] Simplify and solve for \(y\): \[ y = -\dfrac{93}{14} \] The point equidistant from the two given points is \[ (0, -\dfrac{93}{14}) \]

Solution to Problem 6:

Let us first define the distances \(D_1, D_2\) and \(D_3\) as follows \[ D_1 = \text{distance from } (0,1) \text{ to } (2,3) \] \[ D_2 = \text{distance from } (0,1) \text{ to } (2,-1) \] \[ D_3 = \text{distance from } (2,3) \text{ to } (2,-1) \] We now use the distance formula \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] to find the above distances. \[ D_1 = \sqrt{(2-0)^2 + (3-1)^2} = \sqrt{4+4} = \sqrt{8} \] \[ D_2 = \sqrt{(2-0)^2 + (-1-1)^2} = \sqrt{4+4} = \sqrt{8} \] \[ D_3 = \sqrt{(2-2)^2 + (-1-3)^2} = \sqrt{0+16} = 4 \] Note that two sides of the triangle have equal lengths, so the triangle is isosceles. For the triangle to be a right triangle, \(D_3\), the largest of the three distances, must be the length of the hypotenuse, and all three sides must satisfy the Pythagorean theorem. Let us find the sum of the squares of \(D_1\) and \(D_2\) and compare it with the square of \(D_3\). \[ (D_1)^2 + (D_2)^2 = (\sqrt{8})^2 + (\sqrt{8})^2 = 8 + 8 = 16 \] \[ (D_3)^2 = 4^2 = 16 \] We can write \[ (D_1)^2 + (D_2)^2 = (D_3)^2 \] Thus, \(D_1, D_2\) and \(D_3\) satisfy the Pythagorean theorem, and therefore the triangle whose vertices are the three points given above is a right and isosceles triangle.

Solution to Problem 7:

Let \(D_1\) be the distance from \((-2, 1)\) to \((1, 1)\), \(D_2\) be the distance from \((-2, 1)\) to \((1, 2)\), and \(D_3\) the distance from \((1, 1)\) to \((1, 2)\). We now use the distance formula to find these distances. \[ D_1 = \sqrt{(1 - (-2))^2 + (1 - 1)^2} = 3 \] \[ D_2 = \sqrt{(1 - (-2))^2 + (2 - 1)^2} = \sqrt{10} \] \[ D_3 = \sqrt{(1 - 1)^2 + (2 - 1)^2} = 1 \] If the triangle defined by the three vertices is a right triangle, the hypotenuse is the side with the largest size, which is \(D_2 = \sqrt{10}\). We now use the Pythagorean theorem to check that it is a right triangle with \(D_2\) the hypotenuse and \(D_1\) and \(D_3\) the sides of the triangle. \[ (D_1)^2 + (D_3)^2 = 3^2 + 1^2 = 10 \] \[ (D_2)^2 = \left(\sqrt{10}\right)^2 = 10 \] According to the above, we can write \[ (D_1)^2 + (D_3)^2 = (D_2)^2 \] Thus, the triangle defined by the given points is a right triangle, and the size of its hypotenuse is \(\sqrt{10}\).

Solution to Problem 8:

We apply the distance formula. \[ 5 = \sqrt{(-2 - x)^2 + (4 - y)^2} \] Square both sides: \[ 25 = (-2 - x)^2 + (4 - y)^2 \] The above relationship between \(x\) and \(y\) is the equation of a circle. As an exercise, explain why.

Solution to Problem 9:

Let \(D_1\) be the distance from \((x, y)\) to \((-3, 4)\) and find it. \[ D_1 = \sqrt{(-3 - x)^2 + (4 - y)^2} \] Let \(D_2\) be the distance from \((x, y)\) to \((0, -3)\) and find it. \[ D_2 = \sqrt{(0 - x)^2 + (-3 - y)^2} \] The two distances are equal, hence \[ \sqrt{(-3 - x)^2 + (4 - y)^2} = \sqrt{(0 - x)^2 + (-3 - y)^2} \] We now square both sides and simplify: \[ (-3 - x)^2 + (4 - y)^2 = (0 - x)^2 + (-3 - y)^2 \] \[ 9 + x^2 + 6x + 16 - 8y + y^2 = x^2 + 9 + y^2 + 6y \] \[ 6x - 14y + 16 = 0 \] The above relationship between \(x\) and \(y\) is the equation of a line, which is the perpendicular bisector of the line segment defined by the point

Solution to Problem 10:

Let us use the distance formula to find the length of the hypotenuse \(h\). \[ h = \sqrt{(5-1)^2 + (1-1)^2} = 4 \] We now use the distance formula to find the lengths of the two other sides \(a\) and \(b\) of the triangle. \[ a = \sqrt{(x-1)^2 + (y-1)^2} \] \[ b = \sqrt{(x-5)^2 + (y-1)^2} \] The Pythagorean theorem gives \[ 4^2 = (x-1)^2 + (y-1)^2 + (x-5)^2 + (y-1)^2 \] Expand the squares, simplify, and complete the squares to rewrite the above relationship between \(x\) and \(y\) as follows: \[ (x-3)^2 + (y-1)^2 = 2^2 \] The above relationship between \(x\) and \(y\) is the equation of a circle because it is of the form \[ (x-h)^2 + (y-k)^2 = r^2 \] where \((h,k)\) is the center and \(r\) is the radius of the circle.