Question

nearest tenth.
EF: E(1, 4), F(5, 1) and GH: G(-3, 1), H(1, 6)
The length of EF is
The length of GH is about

Answer 1

Answer:

EF = 5 units

GH = 6.4 units (nearest tenth)

Step-by-step explanation:

Given:

• E = (1, 4)
• F = (5, 1)

To find the length of EF, use the distance formula.

Distance between two points

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

$\textsf{where }(x_1,y_1) \textsf{ and }(x_2,y_2)\:\textsf{are the two points}.$

Substitute the given points into the formula:

$\begin{aligned}\sf EF & = \sqrt{(x_F-x_E)^2+(y_F-y_E)^2}\\& = \sqrt{(5-1)^2+(1-4)^2\\ & = \sqrt{4^2+(-3)^2}\\ & = \sqrt{16+9}\\ & = \sqrt{25}\\ & = 5\:\sf units\end{aligned}$

Therefore, the length of EF is 5 units.

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Given:

• G = (-3, 1)
• H = (1, 6)

To find the length of GH, use the distance formula.

Distance between two points

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

$\textsf{where }(x_1,y_1) \textsf{ and }(x_2,y_2)\:\textsf{are the two points}.$

Substitute the given points into the formula:

$\begin{aligned}\sf GH & = \sqrt{(x_H-x_G)^2+(y_H-y_G)^2}\\& = \sqrt{(1-(-3))^2+(6-1)^2\\ & = \sqrt{4^2+5^2}\\ & = \sqrt{16+25}\\ & = \sqrt{41}\\ & = 6.4\:\sf units\:\:(nearest\:tenth)\end{aligned}$

Therefore, the length of GH is about 6.4 units (nearest tenth).