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(A) The vertex $(3, 0)$ does not lie on the diagonal $x = 2y$. Let the slope of the side passing through $(3, 0)$ be $m$. The equation of the side is

Vedclass · Accepted Answer

$y - 3x + 9 = 0, 3y + x - 3 = 0$

Vedclass · Answer

(A) The vertex $(3, 0)$ does not lie on the diagonal $x = 2y$. Let the slope of the side passing through $(3, 0)$ be $m$. The equation of the side is $y - 0 = m(x - 3)$,or $mx - y - 3m = 0$.Since the angle between the diagonal $x - 2y = 0$ (slope $1/2$) and the side is $\pi/4$,we have:$	an(\pi/4) = |\frac{m - 1/2}{1 + m(1/2)}| = 1$$|\frac{2m - 1}{2 + m}| = 1$Case $1$: $\frac{2m - 1}{2 + m} = 1$ $\Rightarrow 2m - 1 = m + 2$ $\Rightarrow m = 3$.Case $2$: $\frac{2m - 1}{2 + m} = -1$ $\Rightarrow 2m - 1 = -m - 2$ $\Rightarrow 3m = -1$ $\Rightarrow m = -1/3$.Substituting $m = 3$ into $y = m(x - 3)$,we get $y = 3x - 9 \Rightarrow y - 3x + 9 = 0$.Substituting $m = -1/3$ into $y = m(x - 3)$,we get $y = -1/3(x - 3)$ $\Rightarrow 3y = -x + 3$ $\Rightarrow 3y + x - 3 = 0$.Thus,the equations are $y - 3x + 9 = 0$ and $3y + x - 3 = 0$.

If one of the diagonals of a square is along the line $x = 2y$ and one of its vertices is $(3, 0)$,then the equations of the sides passing through this vertex are:

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