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Kruka [31]
2 years ago
9

27.) Find the sum of the length of line segments EF and CD

Mathematics
1 answer:
Lelechka [254]2 years ago
6 0

Answer:

5√10

Step-by-step explanation:

The distance between two points A(x_1,y_1) and B(x_2,y_2) on the coordinate plane is given by:

|AB|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2 }

From the image attached the coordinates of the points are E(-1, -1), F(8, -4), C(-7, -4) and D(-1, 2).

Hence the lengths of the line segment EF and CD are:

|EF|=\sqrt{(8-(-1))^2+(-4-(-1))^2} =\sqrt{90}=3\sqrt{10} \\\\CD=\sqrt{(-1-(-7))^2+(-2-(-4))^2}=\sqrt{40}  =2\sqrt{10} \\\\Therefore:\\\\|EF|+|CD|=3\sqrt{10} +2\sqrt{10} =5\sqrt{10}

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<u>Solution:</u>

Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).

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\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}

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Equation of line passing through (x_1 , y_1) and having lope of m is given by

\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)

\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}

Substituting the values we get,

\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}

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