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प्रश्न
Show that the straight lines x + y – 4 = 0, 3x + 2 = 0 and 3x – 3y + 16 = 0 are concurrent.
उत्तर
The lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, a3x + b3y + c3 = 0 are concurrent if
`|(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)|` = 0
The given lines x + y – 4 = 0, 3x + 0y + 2 = 0, 3x – 3y + 16 = 0
`|(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)| = |(1,1,-4),(3,0,2),(3,-3,16)|`
= 1(0 + 6) – 1(48 – 6) – 4(-9 – 0)
= 6 – (42) + 36
= 42 – 42
= 0
The given lines are concurrent.
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