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(A) The given equation is $16x^2 - 24xy + 9y^2 + 48x - 36y + 35 = 0$. We can rewrite the first three terms as $(4x - 3y)^2$. So,the equation becomes $

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(A) The given equation is $16x^2 - 24xy + 9y^2 + 48x - 36y + 35 = 0$. We can rewrite the first three terms as $(4x - 3y)^2$. So,the equation becomes $(4x - 3y)^2 + 12(4x - 3y) + 35 = 0$. Let $t = 4x - 3y$. Then $t^2 + 12t + 35 = 0$. Factoring the quadratic,we get $(t + 7)(t + 5) = 0$. Thus,the two lines are $4x - 3y + 7 = 0$ and $4x - 3y + 5 = 0$. The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$. Here,$a = 4$,$b = -3$,$c_1 = 7$,and $c_2 = 5$. $d = \frac{|7 - 5|}{\sqrt{4^2 + (-3)^2}} = \frac{2}{\sqrt{16 + 9}} = \frac{2}{\sqrt{25}} = \frac{2}{5}$ units.

The distance between the lines represented by $16x^2 - 24xy + 9y^2 + 48x - 36y + 35 = 0$ is ...... units.

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