Given: Lines y and z are parallel, and ABC forms a triangle. Prove: m25 + m_2 + m26 = 180° M
1 answer:
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Write the equations in matrix,
![\left[\begin{array}{ccc}5&-1&1\\1&2&-1\\2&3&-3\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\5\\5\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%261%5C%5C1%262%26-1%5C%5C2%263%26-3%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C5%5C%5C5%5Cend%7Barray%7D%5Cright%5D%20)
Using row transformation,
R₂ <---> R₃
![\left[\begin{array}{ccc}5&-1&1\\2&3&-3\\1&2&-1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\5\\5\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%261%5C%5C2%263%26-3%5C%5C1%262%26-1%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C5%5C%5C5%5Cend%7Barray%7D%5Cright%5D%20)
Using,
R₂ ---> R₂ - 2R₃
![\left[\begin{array}{ccc}5&-1&1\\0&-1&-1\\1&2&-1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\-5\\5\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%261%5C%5C0%26-1%26-1%5C%5C1%262%26-1%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C-5%5C%5C5%5Cend%7Barray%7D%5Cright%5D%20)
Using,
R₂ --- > (-1)R₂
![\left[\begin{array}{ccc}5&-1&1\\0&1&1\\1&2&-1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\5\\5\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%261%5C%5C0%261%261%5C%5C1%262%26-1%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C5%5C%5C5%5Cend%7Barray%7D%5Cright%5D%20)
Using row transformation,
R₂ <----> R₃
![\left[\begin{array}{ccc}5&-1&1\\1&2&-1\\0&1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\5\\5\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%261%5C%5C1%262%26-1%5C%5C0%261%261%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C5%5C%5C5%5Cend%7Barray%7D%5Cright%5D%20)
Using,
R₂ ---> R₂ - R₁/5
![\left[\begin{array}{ccc}5&-1&1\\0&11/5&-6/5\\0&1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\21/5\\5\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%261%5C%5C0%2611%2F5%26-6%2F5%5C%5C0%261%261%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C21%2F5%5C%5C5%5Cend%7Barray%7D%5Cright%5D%20)
Using,
R₃ ---> R₃ - 5R₂/11
![\left[\begin{array}{ccc}5&-1&1\\0&11/5&-6/5\\0&0&17/11\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\21/5\\34/11\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%261%5C%5C0%2611%2F5%26-6%2F5%5C%5C0%260%2617%2F11%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C21%2F5%5C%5C34%2F11%5Cend%7Barray%7D%5Cright%5D%20)
∴ 5x-y+z = 4 ====(i)
11y-6z = 21 === (ii)
17z=34 === (iii)
from iii,
z=2.
Plug z=2 in ii to get y,
∴y=3.
Plug y and z values in i to get x,
∴x=1
Therefore the solution to the system of equations is (1,3,2)
Using row transformation,
R₂ <---> R₃
Using,
R₂ ---> R₂ - 2R₃
Using,
R₂ --- > (-1)R₂
Using row transformation,
R₂ <----> R₃
Using,
R₂ ---> R₂ - R₁/5
Using,
R₃ ---> R₃ - 5R₂/11
∴ 5x-y+z = 4 ====(i)
11y-6z = 21 === (ii)
17z=34 === (iii)
from iii,
z=2.
Plug z=2 in ii to get y,
∴y=3.
Plug y and z values in i to get x,
∴x=1
Therefore the solution to the system of equations is (1,3,2)