If (4,3) and (1,-2) are the end points of a d | vedclass

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(D) Let the vertices of the square be $A(4,3)$ and $C(1,-2)$ as the endpoints of a diagonal. The angle between the diagonal and any side of a square i

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$x-4y-9=0$

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(D) Let the vertices of the square be $A(4,3)$ and $C(1,-2)$ as the endpoints of a diagonal. The angle between the diagonal and any side of a square is $45^{\circ}$.Slope of diagonal $AC = \frac{3 - (-2)}{4 - 1} = \frac{5}{3}$.Let $m$ be the slope of a side. Then,$	an 45^{\circ} = \left| \frac{m - 5/3}{1 + m(5/3)} ight|$.$1 = \left| \frac{3m - 5}{3 + 5m} ight|$.This gives two cases:Case $1$: $\frac{3m - 5}{3 + 5m} = 1$ $\Rightarrow 3m - 5 = 3 + 5m$ $\Rightarrow 2m = -8$ $\Rightarrow m = -4$.The equation of the side passing through $(1,-2)$ with slope $-4$ is $y - (-2) = -4(x - 1)$ $\Rightarrow y + 2 = -4x + 4$ $\Rightarrow 4x + y - 2 = 0$.Case $2$: $\frac{3m - 5}{3 + 5m} = -1$ $\Rightarrow 3m - 5 = -3 - 5m$ $\Rightarrow 8m = 2$ $\Rightarrow m = \frac{1}{4}$.The equation of the side passing through $(1,-2)$ with slope $\frac{1}{4}$ is $y - (-2) = \frac{1}{4}(x - 1)$ $\Rightarrow 4y + 8 = x - 1$ $\Rightarrow x - 4y - 9 = 0$.Comparing with the options,$x - 4y - 9 = 0$ is present.

If $(4,3)$ and $(1,-2)$ are the end points of a diagonal of a square,then the equation of one of its sides is

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