If the coordinates of the mid-points of the sides of a triangle are

Question:

If the coordinates of the mid-points of the sides of a triangle are (1, 1) (2, −3) and (3, 4), find the vertices of the triangle.

Solution:

The co-ordinates of the midpoint $\left(x_{m}, y_{m}\right)$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by,

$\left(x_{m}, y_{m}\right)=\left(\left(\frac{x_{1}+x_{2}}{2}\right),\left(\frac{y_{1}+y_{2}}{2}\right)\right)$

Let the three vertices of the triangle be $A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right)$ and $C\left(x_{C}, y_{C}\right)$.

The three midpoints are given. Let these points be $M_{A B}(1,1), M_{B C}(2,-3)$ and $M_{C a}(3,4)$.

Let us now equate these points using the earlier mentioned formula,

$(1,1)=\left(\left(\frac{x_{A}+x_{H}}{2}\right),\left(\frac{y_{A}+y_{H}}{2}\right)\right)$

Equating the individual components we get,

$x_{A}+x_{B}=2$

 

$y_{A}+y_{B}=2$

Using the midpoint of another side we have,

$(2,-3)=\left(\left(\frac{x_{R}+x_{C}}{2}\right),\left(\frac{y_{B}+y_{C}}{2}\right)\right)$

Equating the individual components we get,

$x_{B}+x_{C}=4$

 

$y_{B}+y_{C}=-6$

Using the midpoint of the last side we have,

$(3,4)=\left(\left(\frac{x_{A}+x_{C}}{2}\right),\left(\frac{y_{A}+y_{C}}{2}\right)\right)$

Equating the individual components we get,

$x_{A}+x_{C}=6$

 

$y_{A}+y_{C}=8$

Adding up all the three equations which have variable ‘x’ alone we have,

$x_{t}+x_{\theta}+x_{\theta}+x_{C}+x_{t}+x_{C}=2+4+6$

$2\left(x_{1}+x_{3}+x_{C}\right)=12$

$x_{A}+x_{B}+x_{C}=6$

Substituting $x_{B}+x_{C}=4$ in the above equation we have,

$x_{A}+x_{B}+x_{C}=6$

 

$x_{4}+4=6$

$x_{A}=2$

Therefore,

$x_{1}+x_{c}=6$

$x_{c}=6-2$

$x_{C}=4$

And

$x_{A}+x_{B}=2$

$x_{p}=2-2$

$x_{R}=0$

Adding up all the three equations which have variable ‘y’ alone we have,

$y_{4}+y_{8}+y_{8}+y_{C}+y_{4}+y_{C}=2-6+8$

$2\left(y_{A}+y_{B}+y_{C}\right)=4$

$y_{A}+y_{B}+y_{C}=2$

Substituting $y_{s}+y_{C}=-6$ in the above equation we have,

$y_{A}+y_{B}+y_{C}=2$

 

$y_{A}-6=2$

$y_{A}=8$

Therefore,

y_{A}+y_{c}=8

y_{c}=8-8

$y_{C}=0$

And

$y_{A}+y_{B}=2$

$y_{B}=2-8$

$y_{B}=-6$

Therefore the co-ordinates of the three vertices of the triangle are $A(2,8)$

$B(0,-6)$

 

$C(4,0)$ .

 

 

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now