The line 5x + y - 1 = 0 coincides with one of | vedclass

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(C) The given equation is $5x^2 + xy - kx - 2y + 2 = 0$. Since this represents a pair of lines and one line is $5x + y - 1 = 0$,we can write the equat

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$11$

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(C) The given equation is $5x^2 + xy - kx - 2y + 2 = 0$. Since this represents a pair of lines and one line is $5x + y - 1 = 0$,we can write the equation as $(5x + y - 1)(ax + by + c) = 0$.Comparing the coefficients of $y^2$,we see the coefficient is $0$,so the second line must be of the form $ax + c = 0$.Thus,$(5x + y - 1)(ax + c) = 5ax^2 + axy + (5c - a)x + cy - c = 0$.Comparing this with $5x^2 + xy - kx - 2y + 2 = 0$:From the $xy$ term,$a = 1$.From the constant term,$-c = 2$,so $c = -2$.From the $y$ term,$c = -2$ (which matches).From the $x$ term,$-k = 5c - a$.Substituting the values: $-k = 5(-2) - 1 = -10 - 1 = -11$.Therefore,$k = 11$.

The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0$. Then the value of $k$ is:

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