QUESTION IMAGE
Question
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 $\frac{3}{4}$ of the way to (6,0). (2 pt)
- find the fractional distance starting from (6,0) and going $\frac{3}{4}$ of the way to (2,0). (2 pt)
- find the fractional distance starting from (4,12) and going $\frac{1}{4}$ of the way to (16,3). (2 pt)
- find the fractional distance starting from (16,3) and going $\frac{1}{4}$ of the way to (4,12). (2 pt)
Step1: Recall distance formula
The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. When $y_1 = y_2$, the formula simplifies to $d=\vert x_2 - x_1\vert$.
Step2: Solve question 1
For points $(2,0)$ and $(6,0)$, since $y$-coordinates are the same, $d=\vert6 - 2\vert= 4$.
Step3: Solve question 2
For points $(5,0)$ and $(10,0)$, since $y$-coordinates are the same, $d=\vert10 - 5\vert = 5$.
Step4: Solve question 3
For points $(2,4)$ and $(6,12)$, use the full - distance formula: $d=\sqrt{(6 - 2)^2+(12 - 4)^2}=\sqrt{4^2+8^2}=\sqrt{16 + 64}=\sqrt{80}=4\sqrt{5}$.
Step5: Solve question 4
For points $(5,7)$ and $(10,14)$, $d=\sqrt{(10 - 5)^2+(14 - 7)^2}=\sqrt{5^2+7^2}=\sqrt{25 + 49}=\sqrt{74}$.
Step6: Solve question 5
For points $(2,-4)$ and $(-6,12)$, $d=\sqrt{(-6 - 2)^2+(12+4)^2}=\sqrt{(-8)^2+16^2}=\sqrt{64 + 256}=\sqrt{320}=8\sqrt{5}$.
Step7: Solve question 6
For points $(-5,5)$ and $(5,-5)$, $d=\sqrt{(5 + 5)^2+(-5 - 5)^2}=\sqrt{10^2+(-10)^2}=\sqrt{100 + 100}=\sqrt{200}=10\sqrt{2}$.
Step8: Solve question 7
The $x$-coordinate of the point that is $\frac{3}{4}$ of the way from $(2,0)$ to $(6,0)$ is $x=2+\frac{3}{4}(6 - 2)=2 + 3=5$, so the point is $(5,0)$.
Step9: Solve question 8
The $x$-coordinate of the point that is $\frac{3}{4}$ of the way from $(6,0)$ to $(2,0)$ is $x=6-\frac{3}{4}(6 - 2)=6 - 3 = 3$, so the point is $(3,0)$.
Step10: Solve question 9
The $x$-coordinate of the point that is $\frac{1}{4}$ of the way from $(4,12)$ to $(16,3)$ is $x=4+\frac{1}{4}(16 - 4)=4 + 3=7$, and the $y$-coordinate is $y=12+\frac{1}{4}(3 - 12)=12-\frac{9}{4}=\frac{48 - 9}{4}=\frac{39}{4}$, so the point is $(7,\frac{39}{4})$.
Step11: Solve question 10
The $x$-coordinate of the point that is $\frac{1}{4}$ of the way from $(16,3)$ to $(4,12)$ is $x=16-\frac{1}{4}(16 - 4)=16 - 3 = 13$, and the $y$-coordinate is $y=3+\frac{1}{4}(12 - 3)=3+\frac{9}{4}=\frac{12 + 9}{4}=\frac{21}{4}$, so the point is $(13,\frac{21}{4})$.
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- 4
- 5
- $4\sqrt{5}$
- $\sqrt{74}$
- $8\sqrt{5}$
- $10\sqrt{2}$
- $(5,0)$
- $(3,0)$
- $(7,\frac{39}{4})$
- $(13,\frac{21}{4})$