L equiv 7x - y + 8 = 0 is one of the diagonal | vedclass

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(A) Let the vertices of the square be $A(-4, 5)$ and $B(3, 4)$. The diagonal $L$ is $7x - y + 8 = 0$. Since $A$ and $B$ are vertices of the square,the

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$(0, 8), (-1, 1)$

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(A) Let the vertices of the square be $A(-4, 5)$ and $B(3, 4)$. The diagonal $L$ is $7x - y + 8 = 0$. Since $A$ and $B$ are vertices of the square,the midpoint of the diagonal $AB$ is $M = (\frac{-4+3}{2}, \frac{5+4}{2}) = (-0.5, 4.5)$. Check if $M$ lies on $L$: $7(-0.5) - 4.5 + 8 = -3.5 - 4.5 + 8 = 0$. Thus,$AB$ is the other diagonal. The slope of $AB$ is $m_1 = \frac{4-5}{3-(-4)} = \frac{-1}{7}$. The slope of diagonal $L$ is $m_2 = 7$. Since $m_1 	imes m_2 = -1$,the diagonals are perpendicular,which is consistent with a square. The vertices on $L$ are equidistant from $M$ and at a distance equal to half the length of diagonal $AB$. Length $AB = \sqrt{(3 - (-4))^2 + (4 - 5)^2} = \sqrt{7^2 + (-1)^2} = \sqrt{50} = 5\sqrt{2}$. Half-length $d = \frac{5\sqrt{2}}{2} = \frac{5}{\sqrt{2}}$. The line $L$ has direction vector $(1, 7)$. The unit vector is $u = \pm(\frac{1}{\sqrt{50}}, \frac{7}{\sqrt{50}}) = \pm(\frac{1}{5\sqrt{2}}, \frac{7}{5\sqrt{2}})$. The vertices on $L$ are $M \pm d \cdot u = (-0.5, 4.5) \pm \frac{5}{\sqrt{2}} \cdot (\frac{1}{5\sqrt{2}}, \frac{7}{5\sqrt{2}}) = (-0.5, 4.5) \pm (0.5, 3.5)$. Vertex $1: (-0.5 + 0.5, 4.5 + 3.5) = (0, 8)$. Vertex $2: (-0.5 - 0.5, 4.5 - 3.5) = (-1, 1)$. Thus,the vertices are $(0, 8)$ and $(-1, 1)$.

$L \equiv 7x - y + 8 = 0$ is one of the diagonals of a square for which $(-4, 5)$ and $(3, 4)$ are two vertices. Find the coordinates of the two vertices lying on the diagonal $L = 0$.

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