Question

If A (1,x), B (5,2)C (5, 6) and D (1,6) are vertices of square ABCD then find x.
SECTION-B

Answer 1

$\textbf{Formula used:}$

$\text{The distance between two points }$

$(x_1,y_1)\:\text{and}\:(x_2,y_2)\;\text{is}$

$\boxed{\bf\;d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}$

$\textbf{Given:The vertices of a square are}$

$\text{A (1,x), B(5,2), C(5, 6) and D(1,6)}$

$\text{We know that,}$

$\textbf{Diagonals of a square are equal}$

$\implies\bf\,AC=BD$

$\implies\sqrt{(1-5)^2+(x-6)^2}=\sqrt{(5-1)^2+(2-6)^2}}$

$\implies\sqrt{16+(x-6)^2}=\sqrt{16+16}}$

$\text{Squaring on bothsides, we get}$

$16+(x-6)^2=32$

$(x-6)^2=16$

$x-6=\pm\,4$

$x=6\pm\,4$

$\implies\boxed{\bf\,x=10,2}$

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