How do you find the magnitude of YZ given Y(-4,12) and Z(1,19)?

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Answer 1 · 2016-09-18T02:36:07

Use the distance formula to find $∣ ∣ ∣ - - \to Y Z ∣ ∣ ∣ = \sqrt{74} \approx 8.6$ $|\vec(YZ) |=sqrt(74)~~8.6$

Use the distance formula.

$d = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$(x_{1}, y_{1}) = (- 4, 12)$ $(x_1,y_1)=(-4,12)$ and $(x_{2}, y_{2}) = (1, 19)$ $(x_2,y_2)=(1,19)$

$∣ ∣ ∣ - - \to Y Z ∣ ∣ ∣ = \sqrt{{(1 - - 4)}^{2} + {(19 - 12)}^{2}}$ $|\vec(YZ) |= sqrt((1- -4)^2+(19-12)^2$

$∣ ∣ ∣ - - \to Y Z ∣ ∣ ∣ = \sqrt{5^{2} + 7^{2}}$ $|\vec(YZ) |=sqrt(5^2+7^2)$

$∣ ∣ ∣ - - \to Y Z ∣ ∣ ∣ = \sqrt{25 + 49}$ $|\vec(YZ) |=sqrt(25+49)$

$∣ ∣ ∣ - - \to Y Z ∣ ∣ ∣ = \sqrt{74} \approx 8.6$ $|\vec(YZ) |=sqrt(74)~~8.6$