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2 years ago

The three lines given below form a triangle. Find the coordinates of the vertices of the triangle.

Mathematics

1 answer:

LenKa [72]2 years ago

8 0

Answer:

(3, 5), (1, 2) and (5, 1)

Step-by-step explanation:

Make three systems with pairs of lines and solve them to work out the vertices.

1) Line 1 and line 2

-3x + 2y = 1
2x + y = 11

Double the second equation and subtract equations:

-3x + 2y - 2(2x + y) = 1 - 2(11)
-3x - 4x = 1 - 22
-7x = - 21
x = 3

Find y:

2*3 + y = 11
6 + y = 11
y = 11 - 6
y = 5

The point is (3, 5)

2) Line 1 and line 3

-3x + 2y = 1
x + 4y = 9

Triple the second equation and add up equations:

-3x + 2y + 3(x + 4y) = 1 + 3(9)
2y + 12y = 1 + 27
14y = 28
y = 2

Find x:

x + 4*2 = 9
x + 8 = 9
x = 1

The point is (1, 2)

3) Line 2 and line 3

2x + y = 11
x + 4y = 9

Double the second equation and subtract the equations:

2x + y - 2(x + 4y) = 11 - 2(9)
y - 8y = 11 - 18
- 7y = - 7
y = 1

Find x:

x + 4*1 = 9
x + 4 = 9
x = 5

The point is (5, 1)

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The general solution is

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Step-by-step explanation:

Step :1:-

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$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

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$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

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$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

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Answer:

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