If the coordinates of the midpoints of the sides of a triangle are p(1, 1), Q (2, -3) and R (3, 4). Find
vertices of triangle also find sides of triangle.

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Solution

Dear Student,
$L e t A (x 1, y 1) B (x 2, y 2) a n d C (x 3, y 3) b e t h e 3 v e r t i c e s o f t h e t r i a n g l e L e t P b e m i d p o i n t o f A B, Q b e m i d p o i n t o f B C a n d R b e m i d p o i n t o f A C T h e f o l l o w i n g s e t o f e q u a t i o n s w i l l b e : - (U \sin g m i d p o i n t f o r m u l a (\frac{x 1 + x 2}{2}, \frac{y 1 + y 2}{2})) x 1 + x 2 = 2 y 1 + y 2 = 2 x 2 + x 3 = 4 y 2 + y 3 = - 6 x 1 + x 3 = 6 y 1 + y 3 = 8 O n s o l v i n g w e g e t, A = (2, 8), B = (0, - 6) a n d C = (4, 0)$
Using distance formula, the sides of triangle are:-
$A B = \sqrt{{(2 - 0)}^{2} + {(8 + 6)}^{2}} = \sqrt{4 + 196} = \sqrt{200} = 10 \sqrt{2} B C = \sqrt{{(0 - 4)}^{2} + {(- 6 - 0)}^{2}} = \sqrt{16 + 36} = \sqrt{52} = 2 \sqrt{13} A C = \sqrt{{(2 - 4)}^{2} + {(8 - 0)}^{2}} = \sqrt{4 + 64} = \sqrt{68} = 2 \sqrt{17}$
Regards.

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