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Determine, algebraically, the vertices of the triangle formed by the lines 3x – y =  3, 2x – 3y = 2 and x + 2y = 8

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Solution:
Determine, algebraically, the vertices of the triangle formed by the lines
3x – y =  3 … (1)
2x – 3y = 2 … (2)
x + 2y = 8 ... (3)

For vertices of A, solve equation (1) and (3)
3x – y = 3
x + 2y = 8 × 3
3x + 6y = 24
3x + y = –3
7y = 21
y = 3
Put y = 3 in eq. (1)
3x – 3 = 3
3x = 6
x = 2
point A is (2, 3)
For vertices of B solve equation (1) and (2)
3x – y = 3 × 3
2x – 3y = 2
9x – 3y = 9
–    +    –
–7x = –7
x = 1
Put x = 1 in equation (1)
3(1) – y = 3
3 – 3 = y
y = 0
point B is (1, 0)
For vertices of C solve (2) and (3)
2x – 3y = 2
x + 2y = 8 × 2
2x + 4y = 16
2x – 3y = 2
–    +     –
7y = 14
y = 2
Put y = 2 in eq. (2)
2x – 3(2) = 2
2x = 2 + 6

x = \frac{8}{2} = 4
Point C (4, 2)
Hence vertices of triangle ABC are A (2, 3), B (1, 0) and C (4, 2)

 

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