What are the coordinates of point B in the xy-plane above?
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17 Nov 2021, 01:05
\(AB = BC\), which means \(\triangle ABC\) is a isosceles triangle.
\(\overline{AC} = \sqrt{(-8 - 20)^2 + (0 - 0)^2} = \sqrt{28^2} = \textbf{28}\)
\(BD = AC = 28\), y-ordinate of point \(B\) will be equal to \(\overline{BD}\).
Therefore, \(y_B = 28\)
Since \(\triangle ABC\) is an isosceles triangle which means \(D\) will be mid-point of \(AC\).
Point D = \((\frac{-8+20}{2} , \frac{0+0}{2}) = (\frac{12}{2} , 0) = \textbf{(6,0)}\)
We can easily tell that, \(x_B = 6\)
Therefore, Point B will be \((x_B, y_B) = \textbf{(6,28)}\)
Hence, Answer is B