Answer:
[tex]4x-y+12=0[/tex]
Step-by-step explanation:
The diagonals of a square are perpendicular bisectors of each other.
Given that (2, 3) and (-6, 5) represent the endpoints of one diagonal, we must find the slope and midpoint of this diagonal to determine the equation of the other diagonal. This is because the midpoint of one diagonal serves as the point of intersection for both diagonals.
To find the slope of the first diagonal, substitute the endpoints (2, 3) and (-6, 5) into the slope formula:
[tex]\textsf{Slope $(m_1$)}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{5-3}{-6-2}=\dfrac{2}{-8}=-\dfrac{1}{4}[/tex]
To find the midpoint of the first diagonal, substitute the endpoints (2, 3) and (-6, 5) into the midpoint formula:
[tex]\textsf{Midpoint}=\left(\dfrac{x_2+x_1}{2},\dfrac{y_2+y_1}{2}\right)\\\\\\\textsf{Midpoint}=\left(\dfrac{-6+2}{2},\dfrac{5+3}{2}\right)\\\\\\\textsf{Midpoint}=\left(\dfrac{-4}{2},\dfrac{8}{2}\right)\\\\\\\textsf{Midpoint}=\left(-2,4\right)[/tex]
Since the diagonals are perpendicular, the negative reciprocal of the slope of the first diagonal gives the slope of the other diagonal.
Therefore, the slope of the other diagonal is:
[tex]m_2=\dfrac{-1}{m_1}=\dfrac{-1}{-\frac{1}{4}}=4[/tex]
Now that we have the slope (m₂ = 4) and the midpoint (-2, 4), we can substitute these values into the point-slope form of a linear equation:
[tex]\begin{aligned}y-y_1&=m(x-x_1)\\\\y-4&=4(x-(-2))\\\\y-4&=4(x+2)\\\\y-4&=4x+8\\\\y-4-y+4&=4x+8-y+4\\\\0&=4x-y+12\\\\4x-y+12&=0\end{aligned}[/tex]
Therefore, the equation of the other diagonal is:
[tex]\Large\boxed{\boxed{4x-y+12=0}}[/tex]
