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garik1379 [7]
3 years ago
10

on the coordinate plane, what is the length of the line segment that connects points at (5, 4) and (−3, −1) ? Enter your answer

in the box. Round to the nearest hundredth.
Mathematics
1 answer:
gtnhenbr [62]3 years ago
5 0
Distance between points
D=\sqrt{(y2-y2)^2+(x2-x1)^2}
<span>
points are (5, 4) and (−3, −1)
</span>
D=√(-1-4)^2+(-3-5)^2
D=√25+64
D=√89
D=9.433 units
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\boxed{\mathsf{A} \triangle = \red{\dfrac{67}{2}u.a}}

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Let's follow up with the solution. Considering a triangle with the vertices \mathsf{A(x_A, y_A)}, \mathsf{B(x_B, y_B)} and \mathsf{C(x_C, y_C)}, have a look at the representation in the cartesian plan.

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\mathsf{A} \triangle =  \dfrac{\left| \begin{array}{ccc}  \mathsf{x_A} & \mathsf{y_A }& 1 \\  \mathsf{x_B} &  \mathsf{ y_B} & 1 \\ \mathsf{ x_C} &  \mathsf{ y_C} & 1 \end{array} \right|}{2}

So, applying this knowledge we're going to have,

\mathsf{A} \triangle =  \dfrac{\left| \begin{array}{ccc}  3 & -7 & 1 \\ 6 &  4 & 1 \\ -2 &  -3 & 1 \end{array} \right|}{2}

\mathsf{A} \triangle =  \dfrac{1}{2}\left[  \left.\begin{array}{ccc}   3 & -7 & 1 \\ 6 &  4 & 1 \\ -2&  -3 & 1 \end{array}  \right| \begin{array}{cc} 3 & -7 \\ 6 & 4 \\ -2 & -3 \end{array} \right]

\mathsf{A} \triangle = \dfrac{12 + 14 - 18 - (-8 - 9 - 42)}{2}

\red{\mathsf{A} \triangle = \dfrac{67}{2} = 33,5u.a}

Hope you enjoy it, see ya!)

\green{\mathsf{FROM}}: Mozambique, Maputo – Matola City – T-3

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