Find the value of k for which the lines 3x-4y = 7, 4x-5y = 11 and 2x+3y+k = 0 are concurrent.
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We can say, (3x-4y)/(4x-5y) = 7/11
=> 11(3x-4y) = 7(4x-5y)
=> 33x -44y = 28x -35y
=> 5x = 9y
=> x/y = 9/5
Therefore x= 9 and y =5
From the condition we get,
2x + 3y + k = 0
=> 2×9 + 3×5 + k = 0
=> 18+15+k=0
=> k = -33
Hope this helps.
=> 11(3x-4y) = 7(4x-5y)
=> 33x -44y = 28x -35y
=> 5x = 9y
=> x/y = 9/5
Therefore x= 9 and y =5
From the condition we get,
2x + 3y + k = 0
=> 2×9 + 3×5 + k = 0
=> 18+15+k=0
=> k = -33
Hope this helps.
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