Find Distance Using Pythagorean Theorem - 8th Grade Math
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What is the distance between $(-4,-3)$ and $(0,0)$?
What is the distance between $(-4,-3)$ and $(0,0)$?
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$5$. $\sqrt{(0-(-4))^2+(0-(-3))^2}=\sqrt{16+9}=5$
$5$. $\sqrt{(0-(-4))^2+(0-(-3))^2}=\sqrt{16+9}=5$
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Find the distance between $(1,2)$ and $(7,5)$ in simplest radical form.
Find the distance between $(1,2)$ and $(7,5)$ in simplest radical form.
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$\sqrt{45}$. $\sqrt{6^2+3^2}=\sqrt{36+9}=\sqrt{45}$
$\sqrt{45}$. $\sqrt{6^2+3^2}=\sqrt{36+9}=\sqrt{45}$
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Find the distance between $(-3,-4)$ and $(1,8)$ in simplest radical form.
Find the distance between $(-3,-4)$ and $(1,8)$ in simplest radical form.
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$\sqrt{160}$. $\sqrt{4^2+12^2}=\sqrt{16+144}=\sqrt{160}$
$\sqrt{160}$. $\sqrt{4^2+12^2}=\sqrt{16+144}=\sqrt{160}$
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Identify the Pythagorean relationship used to find distance with legs $\Delta x$ and $\Delta y$.
Identify the Pythagorean relationship used to find distance with legs $\Delta x$ and $\Delta y$.
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$d^2=(\Delta x)^2+(\Delta y)^2$. Pythagorean theorem: hypotenuse squared equals sum of legs squared.
$d^2=(\Delta x)^2+(\Delta y)^2$. Pythagorean theorem: hypotenuse squared equals sum of legs squared.
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Find the distance between $(0,0)$ and $(6,8)$ using the Pythagorean Theorem.
Find the distance between $(0,0)$ and $(6,8)$ using the Pythagorean Theorem.
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$10$. $\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$
$10$. $\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$
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Find the distance between $(2,3)$ and $(2,11)$ in simplest form.
Find the distance between $(2,3)$ and $(2,11)$ in simplest form.
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$8$. Vertical line, so distance is $|11-3|=8$.
$8$. Vertical line, so distance is $|11-3|=8$.
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Find the distance between $(-4,7)$ and $(5,7)$ in simplest form.
Find the distance between $(-4,7)$ and $(5,7)$ in simplest form.
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$9$. Horizontal line, so distance is $|5-(-4)|=9$.
$9$. Horizontal line, so distance is $|5-(-4)|=9$.
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Find the distance between $(-1,-2)$ and $(2,2)$ in simplest radical form.
Find the distance between $(-1,-2)$ and $(2,2)$ in simplest radical form.
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$5$. $\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$
$5$. $\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$
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Find the distance between $(3,-1)$ and $(-1,2)$ in simplest radical form.
Find the distance between $(3,-1)$ and $(-1,2)$ in simplest radical form.
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$5$. $\sqrt{(-4)^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$
$5$. $\sqrt{(-4)^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$
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Find the distance between $(-2,5)$ and $(4,1)$ in simplest radical form.
Find the distance between $(-2,5)$ and $(4,1)$ in simplest radical form.
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$\sqrt{52}$. $\sqrt{6^2+(-4)^2}=\sqrt{36+16}=\sqrt{52}$
$\sqrt{52}$. $\sqrt{6^2+(-4)^2}=\sqrt{36+16}=\sqrt{52}$
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Find the distance between $(-6,2)$ and $(3,-10)$ in simplest radical form.
Find the distance between $(-6,2)$ and $(3,-10)$ in simplest radical form.
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$15$. $\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15$
$15$. $\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15$
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Find the distance between $(2,-3)$ and $(8,1)$ in simplest radical form.
Find the distance between $(2,-3)$ and $(8,1)$ in simplest radical form.
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$\sqrt{52}$. $\sqrt{6^2+4^2}=\sqrt{36+16}=\sqrt{52}$
$\sqrt{52}$. $\sqrt{6^2+4^2}=\sqrt{36+16}=\sqrt{52}$
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Identify the value of $(y_2-y_1)^2$ when $y_1=10$ and $y_2=4$.
Identify the value of $(y_2-y_1)^2$ when $y_1=10$ and $y_2=4$.
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$36$. $(4-10)^2=(-6)^2=36$
$36$. $(4-10)^2=(-6)^2=36$
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Identify the value of $(x_2-x_1)^2$ when $x_1=-3$ and $x_2=5$.
Identify the value of $(x_2-x_1)^2$ when $x_1=-3$ and $x_2=5$.
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$64$. $(5-(-3))^2=8^2=64$
$64$. $(5-(-3))^2=8^2=64$
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Which expression correctly represents the distance between $(a,b)$ and $(c,d)$?
Which expression correctly represents the distance between $(a,b)$ and $(c,d)$?
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$\sqrt{(c-a)^2+(d-b)^2}$. Standard distance formula with variables $(a,b)$ and $(c,d)$.
$\sqrt{(c-a)^2+(d-b)^2}$. Standard distance formula with variables $(a,b)$ and $(c,d)$.
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Find the distance between $(5,9)$ and $(1,0)$ in simplest radical form.
Find the distance between $(5,9)$ and $(1,0)$ in simplest radical form.
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$\sqrt{97}$. $\sqrt{4^2+9^2}=\sqrt{16+81}=\sqrt{97}$
$\sqrt{97}$. $\sqrt{4^2+9^2}=\sqrt{16+81}=\sqrt{97}$
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State the distance formula between $(x_1,y_1)$ and $(x_2,y_2)$ in the coordinate plane.
State the distance formula between $(x_1,y_1)$ and $(x_2,y_2)$ in the coordinate plane.
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$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Derived from the Pythagorean theorem with legs as horizontal and vertical distances.
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Derived from the Pythagorean theorem with legs as horizontal and vertical distances.
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Which statement is always true about distance between two points: $d$ or $-d$ can be a distance?
Which statement is always true about distance between two points: $d$ or $-d$ can be a distance?
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$d\ge 0$. Distance is always non-negative by definition.
$d\ge 0$. Distance is always non-negative by definition.
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Find and correct the error: $d=\sqrt{(x_2-x_1)+(y_2-y_1)}$ for distance between two points.
Find and correct the error: $d=\sqrt{(x_2-x_1)+(y_2-y_1)}$ for distance between two points.
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$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Missing squares; must square each difference before adding.
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Missing squares; must square each difference before adding.
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What is the horizontal distance between $(x_1,y_1)$ and $(x_2,y_2)$ on a coordinate plane?
What is the horizontal distance between $(x_1,y_1)$ and $(x_2,y_2)$ on a coordinate plane?
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$|x_2-x_1|$. The absolute value ensures distance is always positive.
$|x_2-x_1|$. The absolute value ensures distance is always positive.
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What is the vertical distance between $(x_1,y_1)$ and $(x_2,y_2)$ on a coordinate plane?
What is the vertical distance between $(x_1,y_1)$ and $(x_2,y_2)$ on a coordinate plane?
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$|y_2-y_1|$. The absolute value ensures distance is always positive.
$|y_2-y_1|$. The absolute value ensures distance is always positive.
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State the distance formula between points $(x_1,y_1)$ and $(x_2,y_2)$ on a coordinate plane.
State the distance formula between points $(x_1,y_1)$ and $(x_2,y_2)$ on a coordinate plane.
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$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Derived from Pythagorean theorem applied to coordinate differences.
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Derived from Pythagorean theorem applied to coordinate differences.
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What is the horizontal distance between $(x_1,y_1)$ and $(x_2,y_1)$?
What is the horizontal distance between $(x_1,y_1)$ and $(x_2,y_1)$?
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$|x_2-x_1|$. Points share same y-coordinate, so distance is horizontal difference.
$|x_2-x_1|$. Points share same y-coordinate, so distance is horizontal difference.
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What is the vertical distance between $(x_1,y_1)$ and $(x_1,y_2)$?
What is the vertical distance between $(x_1,y_1)$ and $(x_1,y_2)$?
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$|y_2-y_1|$. Points share same x-coordinate, so distance is vertical difference.
$|y_2-y_1|$. Points share same x-coordinate, so distance is vertical difference.
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Identify the legs of the right triangle used to find distance between $(x_1,y_1)$ and $(x_2,y_2)$.
Identify the legs of the right triangle used to find distance between $(x_1,y_1)$ and $(x_2,y_2)$.
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Legs: $|x_2-x_1|$ and $|y_2-y_1|$. Horizontal and vertical distances form perpendicular sides.
Legs: $|x_2-x_1|$ and $|y_2-y_1|$. Horizontal and vertical distances form perpendicular sides.
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