A(0,2) and C(6,4) are opposite vertices of sq | vedclass

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(B) Let the vertices of the square be $A(0,2)$,$B(x,y)$,$C(6,4)$,and $D(x',y')$.
Since $ABCD$ is a square,the diagonal $AC$ has a slope $m_{A

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(B) Let the vertices of the square be $A(0,2)$,$B(x,y)$,$C(6,4)$,and $D(x',y')$.
Since $ABCD$ is a square,the diagonal $AC$ has a slope $m_{AC} = \frac{4-2}{6-0} = \frac{2}{6} = \frac{1}{3}$.
Let the slopes of the sides $AB$ and $AD$ be $m_1$ and $m_2$ respectively.
Since the sides of a square are inclined at $45^\circ$ to the diagonal,the angle between the diagonal $AC$ and the sides $AB$ or $AD$ is $45^\circ$.
Using the formula $	an 	heta = |\frac{m - m_{AC}}{1 + m \cdot m_{AC}}|$,we have $	an 45^\circ = |\frac{m - 1/3}{1 + m/3}| = 1$.
This gives $\frac{m - 1/3}{1 + m/3} = 1$ or $\frac{m - 1/3}{1 + m/3} = -1$.
Case $1$: $m - 1/3 = 1 + m/3 \implies \frac{2m}{3} = \frac{4}{3} \implies m = 2$.
Case $2$: $m - 1/3 = -1 - m/3 \implies \frac{4m}{3} = -2/3 \implies m = -1/2$.
Thus,the slopes of the sides passing through $A$ are $2$ and $-1/2$.
The sum of the slopes is $2 + (-1/2) = 3/2 = 1.5$.

$A(0,2)$ and $C(6,4)$ are opposite vertices of square $ABCD$. The sum of the slopes of the sides passing through vertex $A$ is:

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