If the midpoints of sides of the triangle PQR are (1,2) (0,1) (1,0), then find the cordinates of thevertices of - Maths - Coordinate Geometry

Question

If the midpoints of sides of the triangle PQR are (1,2) (0,1)
(1,0), then find the cordinates of thevertices of the triangle

Priyanka Kedia · Accepted Answer

Let $A (x_{1}, y_{1}), B (x_{2}, y_{2}) a n d C (x_{3}, y_{3})$ be the vertices of the triangle ABC and $P (1, 2), Q (0, 1) a n d R (1, 0)$ be the mid points of sides AB, BC and CA respectively.

Since P is the mid point of side AB. Therefore,

$\frac{x_{1} + x_{2}}{2} = 1 \Rightarrow x_{1} + x_{2} = 2 ...... (1) \frac{y_{1} + y_{2}}{2} = 2 \Rightarrow y_{1} + y_{2} = 4 ...... (2)$

Similarly, Q and R is the mid point of side BC and CA respectively. So,

$\frac{x_{2} + x_{3}}{2} = 0 \Rightarrow x_{2} + x_{3} = 0 ...... (3) \frac{y_{2} + y_{3}}{2} = 1 \Rightarrow y_{2} + y_{3} = 2 ...... (4)$

And,

$\frac{x_{3} + x_{1}}{2} = 1 \Rightarrow x_{3} + x_{1} = 2 ...... (5) \frac{y_{3} + y_{1}}{2} = 0 \Rightarrow y_{3} + y_{1} = 0 ...... (6)$

Adding equation (1), (3), (5) and (2), (4), (6), we get,

$(x_{1} + x_{2}) + (x_{2} + x_{3}) + (x_{3} + x_{1}) = 2 + 0 + 2 \Rightarrow 2 (x_{1} + x_{2} + x_{3}) = 4 \Rightarrow x_{1} + x_{2} + x_{3} = 2 ...... (7)$

And,

$(y_{1} + y_{2}) + (y_{2} + y_{3}) + (y_{3} + y_{1}) = 4 + 2 + 0 \Rightarrow 2 (y_{1} + y_{2} + y_{3}) = 6 \Rightarrow y_{1} + y_{2} + y_{3} = 3 ...... (8)$

Subtracting equation (1), (3) and (5) from equation (7) respectively, we get,

$x_{3} = 0, x_{1} = 2 a n d x_{2} = 0$

Subtracting equation (2), (4) and (6) from equation (8) respectively, we get,

$y_{3} = - 1, y_{1} = 1 a n d y_{2} = 3$

Hence, the vertices of triangle ABC are A(2,1) , B(0,3) and C(0,-1).

If the midpoints of sides of the triangle PQR are (1,2) (0,1) (1,0), then find the cordinates of thevertices of the triangle.