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Question
Answer the following question:
Show that the lines x − y = 6, 4x − 3y = 20 and 6x + 5y + 8 = 0 are concurrent. Also find the point of concurrence
Solution
The given lines are
x − y = 6
4x − 3y = 20
6x + 5y = − 8
These lines are concurrent, if
`|(1, -1, 6),(4, -3, 20),(6, 5, -8)|` = 0
L.H.S. = `|(1, -1, 6),(4, -3, 20),(6, 5, -8)|`
= 1 (24 − 100) − ( −1)(−32 − 120) + 6 (20 + 18)
= −76 − 152 + 228
= −228 + 228 = 0
= R.H.S.
Hence, the given lines are concurrent. To find the point of concurrence, we have to solve any two equations.
Consider the equations,
x − y = 6
4x − 3y = 20
∴ D = `|(1, -1),(4, -3)|` = −3 + 4 = 1 ≠ 0
Dx = `|(6, -1),(20, -3)|` = −18 + 20 = 2
Dy = `|(1, 6),(4, 20)|` = 20 − 24 = −4
∴ x = `"D"_x/"D" = 2/1` = 2
y = `"D"_y/"D" = (-4)/1` = −4
∴ the point of concurrence is (2, −4).
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