If the equation 16x^2 - 24xy + 9y^2 - 8x + 6y | vedclass

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(A) The given equation is $16x^2 - 24xy + 9y^2 - 8x + 6y - 35 = 0$.Notice that $16x^2 - 24xy + 9y^2 = (4x - 3y)^2$.Let $u = 4x - 3y$. Then the equatio

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$4x - 3y - 1 = 0$

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(A) The given equation is $16x^2 - 24xy + 9y^2 - 8x + 6y - 35 = 0$.Notice that $16x^2 - 24xy + 9y^2 = (4x - 3y)^2$.Let $u = 4x - 3y$. Then the equation becomes $u^2 - 2u - 35 = 0$.Factoring the quadratic: $(u - 7)(u + 5) = 0$.This gives two lines: $L_1: 4x - 3y - 7 = 0$ and $L_2: 4x - 3y + 5 = 0$.The locus of points equidistant from two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is the line $ax + by + \frac{c_1 + c_2}{2} = 0$.Here $a = 4, b = -3, c_1 = -7, c_2 = 5$.The locus is $4x - 3y + \frac{-7 + 5}{2} = 0$.$4x - 3y - 1 = 0$.

If the equation $16x^2 - 24xy + 9y^2 - 8x + 6y - 35 = 0$ represents a pair of straight lines, then the equation of the locus of points equidistant from these two lines is...

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