Let D(3,-2),E(-3,1) and F(4,-3) be the midpoint of the sides BC,CA and AB respectively of / ABC Then find the - Maths - Coordinate Geometry

Question

Let D(3,-2),E(-3,1) and F(4,-3) be the midpoint of the sides
BC,CA and AB respectively of / ABC Then find the coordinates
of the vertices A,B and C

Ajanta Trivedi · Accepted Answer

given: D, E and F are the mid-points of the sides BC, CA and AB respectively.

let the coordinates of the vertices of the triangles be $A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})$

now from the mid-point formula, the coordinates of D, E and F are given as:

$D (\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2}), E (\frac{x_{3} + x_{1}}{2}, \frac{y_{3} + y_{1}}{2}), F (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

therefore:

$\frac{x_{2} + x_{3}}{2} = 3 \Rightarrow x_{2} + x_{3} = 6 ............ (1) \frac{y_{2} + y_{3}}{2} = - 2 \Rightarrow y_{2} + y_{3} = - 4........ (2) \frac{x_{3} + x_{1}}{2} = - 3 \Rightarrow x_{3} + x_{1} = - 6 ......... (3) \frac{y_{3} + y_{1}}{2} = 1 \Rightarrow y_{3} + y_{1} = 2 ............. (4) \frac{x_{1} + x_{2}}{2} = 4 \Rightarrow x_{1} + x_{2} = 8 .......... (5) \frac{y_{1} + y_{2}}{2} = - 3 \Rightarrow y_{1} + y_{2} = - 6......... (6)$

from (1) and (3), $x_{2} - x_{1} = 12$ , solving this with (5),

$x_{2} - x_{1} = 12 x_{1} + x_{2} = 8 2 x_{2} = 20 \Rightarrow x_{2} = 10 x_{1} = 8 - 10 = - 2$

substitute this value in (3) $x_{3} - 2 = - 6 \Rightarrow x_{3} = - 6 + 2 = - 4$

now subtracting (4) from (2) $y_{2} - y_{1} = - 6$ solving this with (6)

$y_{2} - y_{1} = - 6 y_{1} + y_{2} = - 6 2 y_{2} = - 12 \Rightarrow y_{2} = - 6 y_{1} = - 6 + 6 = 0$

substitute this value in (4), $y_{3} + 0 = 2 \Rightarrow y_{3} = 2$

therefore the coordinates of vertices are A(-2,0) , B(10,-6) and C(-4,2).

Let D(3,-2),E(-3,1) and F(4,-3) be the midpoint of the sides BC,CA and AB respectively of / ABC.Then find the coordinates of the vertices A,B and C.