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oee [108]
3 years ago
10

Suppose lines EF and GH are reflected over the line Y equals X to form the lines JK and LM which statement would be true about t

he lines
A) jk and lm will be parallel

B) jk and lm will be perpendicular

C) jk will be same slope as ef

D) lm will be negative reciprocal of gh

Mathematics
1 answer:
katrin [286]3 years ago
5 0

Answer:

A) JK and LM will be parallel to each other.

Step-by-step explanation:

On reflection on y=x line the x co-ordinate changes with y co-ordinate and y co-ordinate changes with x co-ordinate

(x,y)\rightarrow (y,x)

Points on line EF

(0,6) , (-5,-2)

On reflection of this line on y=x the new points we get for line JK are

(6,0),(-2,-5)

Points on line GH

(-4,9),(-9,1)

On reflection on y=x line the new points we get for line LM are

(9,-4),(1,-9)

Slope of line JK

m=\frac{y_2-y_1}{x2-x1}\\m=\frac{(-5)-0}{(-2)-6} \\m=\frac{-5}{-8}=\frac{5}{8}

Slope of line LM

m=\frac{y_2-y_1}{x2-x1}\\m=\frac{(-9)-(-4)}{1-9} \\m=\frac{-9+4}{-8}=\frac{-5}{-8}\\m=\frac{5}{8}

For two line to be parallel, their slopes will be same.

m_{JK} =\frac{5}{8} , m_{LM}=\frac{5}{8}

Since slopes of lines JK and LM are same therefore we can say that these are parallel to each other.

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The values of x and y are 70/22 and 30/22 respectively by using the first-order condition of differential calculus.

<h3>What is the first-order condition in differential calculus?</h3>

A first-order differential equation is represented by the equation \mathbf{ \dfrac{dy}{dx} =f (x,y) }with 2 variables x & y, including its function f(x,y) specified on a xy-plane.

Given that:

\mathbf{f(x,y) =-22x^2+22xy-11y^2+110x-40y-23}

Let us first differentiate the above equation with respect to x, we have:

\mathbf{\dfrac{\partial f(x,y) }{\partial x}  = -44x +22y -0+110-0-0=0}

\mathbf{\implies  -44x +22y+110=0}        (multiply by -1)

44x - 22y = 110    ------ (equation 1)

Now, differentiating with respect to y, we have:

\mathbf{\dfrac{\partial f(x,y) }{\partial y}  =0 +22x-22y +0-40-0=0}

\mathbf{\implies 22x-22y -40=0}

22x - 22y = 40      ----- (equation 2)

Now, we have a system of equations:

44x - 22y = 110

-                      ---- ( subtracting equation 2 from 1; elimination method)

<u> 22x - 22y = 40  </u>

<u>22x    + 0  = 70    </u>

<u />

x = 70/22

Replacing the value of x into equation (1), we have:

44x - 22y = 110

44(70/22) - 22y = 110

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140 - 110 = 22y

30 = 22y

y = 30/22

Learn more about the first-order conditions in differential calculus here;

brainly.com/question/14528981

#SPJ1

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