Distance Formula Practice Problems with Step-by-Step Solutions

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

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Solution:

The formula for the distance \(D\) between two points \((a, b)\) and \((c, d)\) is:

\[ D = \sqrt{(c - a)^2 + (d - b)^2} \]

Apply the formula:

\[ D = \sqrt{(0 - 2)^2 + (6 - 3)^2} \] \[ D = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]

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

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Solution:

First, find the midpoint \(M\) of the segment joining \((2, 4)\) and \((4, 6)\).

\[ M = \left( \frac{2 + 4}{2}, \frac{4 + 6}{2} \right) \] \[ M = (3, 5) \]

Now use the distance formula between \((-1, -3)\) and \((3, 5)\).

\[ D = \sqrt{(3 - (-1))^2 + (5 - (-3))^2} \] \[ D = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5} \]

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

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Solution:

\[ 5 = \sqrt{(-3 - (-2))^2 + (x - (-3))^2} \] \[ 5 = \sqrt{(-1)^2 + (x + 3)^2} = \sqrt{1 + (x + 3)^2} \]

Square both sides:

\[ 25 = 1 + (x + 3)^2 \] \[ (x + 3)^2 = 24 \] \[ x + 3 = \pm\sqrt{24} = \pm2\sqrt{6} \] \[ x = -3 + 2\sqrt{6} \quad \text{or} \quad x = -3 - 2\sqrt{6} \]

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

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Solution:

\[ (2,5) = \left( \frac{x + (-5)}{2}, \frac{y + 6}{2} \right) \] \[ 2 = \frac{x - 5}{2} \quad \text{and} \quad 5 = \frac{y + 6}{2} \] \[ 4 = x - 5 \implies x = 9 \] \[ 10 = y + 6 \implies y = 4 \]

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

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Solution:

\[ D_1 = \sqrt{(4-0)^2 + (-9-y)^2} \] \[ D_2 = \sqrt{(0-0)^2 + (-2-y)^2} \]

Since \(D_1 = D_2\):

\[ \sqrt{16 + (-9-y)^2} = \sqrt{0 + (-2-y)^2} \] \[ 16 + (81 + 18y + y^2) = (4 + 4y + y^2) \] \[ 97 + 18y = 4 + 4y \] \[ 14y = -93 \implies y = -\frac{93}{14} \]

The equidistant point is \(\left(0, -\frac{93}{14}\right)\).

Question: Show that the triangle with vertices \((0, 1)\), \((2, 3)\), and \((2, -1)\) is a right isosceles triangle.

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Solution:

\[ D_1 = \sqrt{(2-0)^2 + (3-1)^2} = \sqrt{4+4} = \sqrt{8} \] \[ D_2 = \sqrt{(2-0)^2 + (-1-1)^2} = \sqrt{4+4} = \sqrt{8} \] \[ D_3 = \sqrt{(2-2)^2 + (-1-3)^2} = \sqrt{0+16} = 4 \]

Since \(D_1 = D_2\), the triangle is isosceles.

Check the Pythagorean theorem:

\[ (\sqrt{8})^2 + (\sqrt{8})^2 = 8 + 8 = 16 = 4^2 \]

Since \(16 = 16\), the triangle is a right isosceles triangle.

Question: Find the length of the hypotenuse of the right triangle whose vertices are \((-2, 1)\), \((1, 1)\), and \((1, 2)\).

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Solution:

\[ D_1 = \sqrt{(1 - (-2))^2 + (1 - 1)^2} = \sqrt{9 + 0} = 3 \] \[ D_2 = \sqrt{(1 - (-2))^2 + (2 - 1)^2} = \sqrt{9 + 1} = \sqrt{10} \] \[ D_3 = \sqrt{(1 - 1)^2 + (2 - 1)^2} = \sqrt{0 + 1} = 1 \]

The longest side is the hypotenuse \(D_2 = \sqrt{10}\). Verify:

\[ 3^2 + 1^2 = 9 + 1 = 10 = (\sqrt{10})^2 \]

The length of the hypotenuse is \(\sqrt{10}\).

Question: Find a relationship between \(x\) and \(y\) so that the distance between \((x, y)\) and \((-2, 4)\) is exactly 5.

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Solution:

\[ 5 = \sqrt{(-2 - x)^2 + (4 - y)^2} \]

Square both sides:

\[ 25 = (x + 2)^2 + (y - 4)^2 \]

This is the equation of a circle with center \((-2, 4)\) and radius \(5\).

Question: Find a relationship between \(x\) and \(y\) so that \((x, y)\) is equidistant from \((-3, 4)\) and \((0, -3)\).

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Solution:

\[ D_1 = \sqrt{(-3 - x)^2 + (4 - y)^2} \] \[ D_2 = \sqrt{(0 - x)^2 + (-3 - y)^2} \]

Set them equal and square both sides:

\[ (-3 - x)^2 + (4 - y)^2 = (0 - x)^2 + (-3 - y)^2 \]

Expand:

\[ 9 + 6x + x^2 + 16 - 8y + y^2 = x^2 + 9 + 6y + y^2 \] \[ 6x - 14y + 16 = 0 \]

Dividing by 2 gives:

\[ 3x - 7y + 8 = 0 \]

This is the perpendicular bisector of the segment joining \((-3, 4)\) and \((0, -3)\).

Question: Find a relationship between \(x\) and \(y\) so that the triangle with vertices \((x, y)\), \((1, 1)\), and \((5, 1)\) is a right triangle with hypotenuse defined by \((1, 1)\) and \((5, 1)\).

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Solution:

Squared hypotenuse:

\[ h^2 = (5-1)^2 + (1-1)^2 = 16 \]

Squared legs:

\[ a^2 = (x-1)^2 + (y-1)^2 \] \[ b^2 = (x-5)^2 + (y-1)^2 \]

Apply \(a^2 + b^2 = h^2\):

\[ (x-1)^2 + (y-1)^2 + (x-5)^2 + (y-1)^2 = 16 \]

Expand and combine:

\[ 2x^2 - 12x + 2y^2 - 4y + 28 = 16 \] \[ x^2 - 6x + y^2 - 2y + 6 = 0 \]

Complete the squares:

\[ (x-3)^2 + (y-1)^2 = 2^2 \]

This is a circle with center \((3, 1)\) and radius \(2\). By Thales's theorem, every point on this circle forms a right angle opposite the hypotenuse, except the endpoints \((1, 1)\) and \((5, 1)\), which give a degenerate triangle.