1.5: The Plane
- Page ID
- 228
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The Distance Formula
Definition: Distance
Recall that for two points \((a,b)\) and \((c,d)\) in a plane, the distance is found by the formula
\[\text{Distance}=\sqrt{(c-a)^2+(d-b)^2}.\]
Example \(\PageIndex{1}\)
Find the distance between the points \((1,1)\) and \((-4,3)\).
Solution
\[\begin{align*} \text{Distance} &=\sqrt{(-4-1)^2+(3-1)^2} \\[4pt] &=\sqrt{25+4}\\ [5pt] &=\sqrt{29}. \end{align*}\]
The Midpoint Formula
Definition: Midpoint
For points \((a,b)\) and \((c,d)\) the midpoint of the line segment formed by these points has coordinates:
\[M=\left(\dfrac{a+c}{2},\dfrac{b+d}{2}\right). \]
Example \(\PageIndex{2}\)
Suppose that you have a boat at one side of the lake with coordinates \((3,4)\) and your friend has a boat at the other side of the lake with coordinates \((18,22)\). If you want to meet half way, at what coordinates should you meet?
Solution:
\[\begin{align*} M &= \left(\dfrac{3+18}{2}, \dfrac{4+22}{2}\right) \\[4pt] &=(10.5,13). \end{align*}\]
Exercises
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Show that the points \((-5,14)\), \((1,4)\), and \((11,10)\) are vertices of an isosceles triangle.
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Show that the triangle with vertices \((1,1)\), \((-1,-1)\), and \((\sqrt{3},-\sqrt{3})\) are vertices of a right triangle.
Graphing on a Calculator
We will graph the equations:
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\(y = 2x - 3\) (Use graph then y(x) =)
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\(y = 5x^2 + 4\)
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\(y = |x + 1|\) (To find absolute value, use catalog then hit enter)
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\(y = 2x + \{-1,0,1,2,3,5\}\) (find the curly braces "{" and "}" use the list feature)
Contributors
- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.