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schepotkina [342]
2 years ago
6

Triangle ABC has vertices A(-3,0) B(0,6) C(4,6). Find the equations of the three medians of triangle ABC

Mathematics
1 answer:
Mariana [72]2 years ago
6 0

Answer:

1.y=-6x+6, y=\frac{6}{5} x+\frac{18}{5}, y=\frac{6}{11} x+\frac{42}{11}

2.y=-\frac{1}{2} x+2\frac{1}{4} , x=2, y=-\frac{7}{6}x+\frac{43}{12}

Step-by-step explanation:

A(−3, 0), B(0, 6), and C(4, 6)

using the midpoint formula: (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) we can find all of their midpoint which is necessary to complete question 1 and 2

Mid AB=(-1.5, 3), Mid BC=(2, 6), Mid AC=(0.5,3)

Next, we have to find the slope of from the vertex and the midpoint of each side. We use the slope formula: \frac{y_2-y_1}{x_2-x_1}.

You would get -6, 6/5, and 6/11.

Next you substitute in the coordinates for the equation of each for example: y=-6x, you switch in the set of cords AB which is (-1.5, 3). 3=-6(-1.5)+z and find that z=6, so the equation is y=-6x+6. As for the rest, repeat the same process until you find all 3 cords. You should get an answer of y=-6x+6, y=\frac{6}{5} x+\frac{18}{5}, y=\frac{6}{11} x+\frac{42}{11}

For question 2,

The line has to be perpendicular bisector to the midpoint, signifying that it has to split 2 angles into congruent angles and a side into half. In order to do that we need to first find the opposite reciprocal of each side's slope. Which means we need to once again use the slope formula and get the opposite reciprocal by -\frac{1}{slope}. For all the opposite reciprocal slopes you get will be incorporated into the final formulas. That gives us the slopes of -1/2, 0 and -7/6. Then do the same process of matching cords with these incomplete equations and you can find the final equations of y=-\frac{1}{2} x+2\frac{1}{4} , x=2, y=-\frac{7}{6}x+\frac{43}{12}

I hope that helped with your question!! :)

Also I'm an RSM student too and I didn't know how to do them before so I searched it up.

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If you are looking for the product, then:

Use foil to get: sec²(1) - sec²(-csc²) -1(1) -1(-csc²)

= sec² + sec²csc² - 1 + csc²

= sec²csc² + sec² + csc² - 1

= sec²csc² + 1 - 1 (NOTE: sec² + csc² = 1 is an identity)

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Answer: sec²csc²

***************************************************

If you are looking for the zeroes, then:

Using the zero product property, set each factor equal to zero and solve.

<u>First factor:</u>

sec²Θ - 1 = 0

sec²Θ = 1

secΘ = 1, -1

remember that secΘ is \frac{1}{cos}

\frac{1}{cos} = 1 \frac{1}{cos} = -1

cross multiply to get:

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use the unit circle (or a calculator) to find that Θ = 0 and π

<u>Second factor:</u>

1 - csc²Θ = 0

1 = csc²Θ

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remember that cscΘ is \frac{1}{sin}

\frac{1}{sin} = 1 \frac{1}{sin} = -1

cross multiply to get:

sinΘ = 1 sinΘ = -1

use the unit circle (or a calculator) to find that Θ = \frac{\pi}{2} and \frac{3\pi}{2}

Answer: 0, π, \frac{\pi}{2} , \frac{3\pi}{2}

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The variable x would represent the number/value we don't know, and in this case, we don't know what number is raised to the third power. This being said, x would represent that number.

The question, although worded a bit confusingly, asks to add 155, the number (x) to the exponent of 3, and 14. Mathematically, this would be 155 + x^{3} + 14.

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