Question

show that A(6,4) B(5,-2) and C(7,-2) are vertices of an isosceles triangle, also find the length of the median through A​

Answer 1

Step-by-step explanation:

ABC IS AN ISOSCELES TRIANGLE

LENGTH OF MEDIAN THROUGH A IS 6 UNITS.

$ab = \sqrt{ {(6 - 5)}^{2} + {(4 + 2)}^{2} } \\ \\ = \sqrt{1 + 36} \\ \\ = \sqrt{37} \\ \\ bc = \sqrt{ {(7 - 5)}^{2} + {(0)}^{2} } \\ \\ = 2 \\ \\ ca = \sqrt{ {(7 - 6)}^{2} + {(4 + 2)}^{2} } \\ \\ = \sqrt{37} \\ \\ because \: \: ab = ca = \sqrt{ 37} \: units \\ \\ therefore \: abc \: is \: an \: isosceles \: triangle. \\ \\ mid \: point \: of \: bc = ( \frac{7 + 5}{2} \: \: \: \: \: \: \: \frac{ - 2 - 2}{2} ) \\ \\ = (6 \: \: \: \: \: \: \: - 2) \\ \\ length \: of \: median \: through \: a \: is \: \: \: \sqrt{ {0}^{2} + {6}^{2} } = 6 \: units.$