QUESTION IMAGE
Question
ext 1.01 - distance/fractional distance
geometry
directions: for the extension activities, you must complete a total of 12 points. each question has a number of points assigned to it. your teacher may assign you specific questions to do, or they may let you to choose.
- find the distance between (2,0) and (6,0). (1 pt)
- find the distance between (5,0) and (10,0). (1 pt)
- find the distance between (2,4) and (6,12). (1 pt)
- find the distance between (5,7) and (10,14). (1 pt)
- find the distance between (2, - 4) and (-6,12). (1 pt)
- find the distance between (-5,5) and (5, - 5). (1 pt)
- find the fractional distance starting from (2,0) and going 3/4 of the way to (6,0). (2 pt)
- find the fractional distance starting from (6,0) and going 3/4 of the way to (2,0). (2 pt)
- find the fractional distance starting from (4,12) and going 1/4 of the way to (16,3). (2 pt)
- find the fractional distance starting from (16,3) and going 1/4 of the way to (4,12). (2 pt)
1.
Step1: Identify distance formula
For two - dimensional points \((x_1,y_1)\) and \((x_2,y_2)\), the distance \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here \(x_1 = 2,y_1=0,x_2 = 6,y_2 = 0\).
\[d=\sqrt{(6 - 2)^2+(0 - 0)^2}\]
Step2: Simplify the expression
\[d=\sqrt{4^2+0^2}=\sqrt{16}=4\]
Step1: Apply distance formula
For points \((x_1,y_1)=(5,0)\) and \((x_2,y_2)=(10,0)\), \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
\[d=\sqrt{(10 - 5)^2+(0 - 0)^2}\]
Step2: Simplify
\[d=\sqrt{5^2+0^2}=\sqrt{25}=5\]
Step1: Use distance formula
For \((x_1,y_1)=(2,4)\) and \((x_2,y_2)=(6,12)\), \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
\[d=\sqrt{(6 - 2)^2+(12 - 4)^2}\]
Step2: Calculate values inside square - root
\((6 - 2)^2=4^2 = 16\) and \((12 - 4)^2=8^2 = 64\). Then \(d=\sqrt{16 + 64}=\sqrt{80}=4\sqrt{5}\)
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