Conic Sections: Background You’ll Need 2

  • Use the distance formula

Distance Formula

This is an image of a triangle on an x, y coordinate plane. The x and y axes range from 0 to 7. The points (x sub 1, y sub 1); (x sub 2, y sub 1); and (x sub 2, y sub 2) are labeled and connected to form a triangle. Along the base of the triangle, the following equation is displayed: the absolute value of x sub 2 minus x sub 1 equals a. The hypotenuse of the triangle is labeled: d = c. The remaining side is labeled: the absolute value of y sub 2 minus y sub 1 equals b.
Triangle on a coordinate plane with all sides and vertices labeled

Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem, [latex]{a}^{2}+{b}^{2}={c}^{2}[/latex], is based on a right triangle where [latex]a[/latex] and [latex]b[/latex] are the lengths of the legs adjacent to the right angle, and [latex]c[/latex] is the length of the hypotenuse.

The relationship of sides [latex]|{x}_{2}-{x}_{1}|[/latex] and [latex]|{y}_{2}-{y}_{1}|[/latex] to side d is the same as that of sides a and b to side c. We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, [latex]|-3|=3[/latex]. ) The symbols [latex]|{x}_{2}-{x}_{1}|[/latex] and [latex]|{y}_{2}-{y}_{1}|[/latex] indicate that the lengths of the sides of the triangle are positive. To find the length c, take the square root of both sides of the Pythagorean Theorem.

[latex]{c}^{2}={a}^{2}+{b}^{2}\rightarrow c=\sqrt{{a}^{2}+{b}^{2}}[/latex]

It follows that the distance formula is given as

[latex]{d}^{2}={\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}\to d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}[/latex]

Distance Formula

The distance formula is a mathematical equation used to determine the exact distance between two points ([latex]\left({x}_{1},{y}_{1}\right)[/latex] and [latex]\left({x}_{2},{y}_{2}\right)[/latex]) on a coordinate plane.

[latex]\text{Distance}: d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}[/latex]

Use the distance formula to find the distance between the points [latex](-5, -3)[/latex] and [latex](7,2)[/latex].[latex]\begin{align*} d &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ d &= \sqrt{(72 - (-5))^2 + (2 - (-3))^2} \\ d &= \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} \\ d &= 13 \end{align*}[/latex]
Find the distance between the points [latex]\left(-3,-1\right)[/latex] and [latex]\left(2,3\right)[/latex].