if\( (3,-4),(-6,5)\) are the exterimities of the diagonal of the parallelogram and \( (-2,-1)\) is it's third vertex then find fourth vertex.
Answer
Using property that diagonals of a parallelogram bisects each other, we can say that the midpoint of the diagonals of a parallelogram is same.
Let the fourth vertex be \((x,y)\).
Now using mid-point formula .
\(\begin{array}{l}\frac{(3,-4)+(-6,5)\ }{2}=\frac{(-2,-1)+\left(x,y\right)}{2}\\\Rightarrow (3,-4)+(-6,5)\ =(-2,-1)+\left(x,y\right)\\\Rightarrow \left(3+\left(-6\right),-4+5\right)=\left(-2+x,-1+y\right)\\\Rightarrow \left(-3,1\right)=\left(-2+x,-1+y\right)\\\Rightarrow -3=-2+x,1=-1+y\\\Rightarrow -3+2=x,1+1=+y\\\Rightarrow x=-1,y=2\end{array}\)
Answer: The fourth vertex is \((-1,2)\)
Let the fourth vertex be \((x,y)\).
Now using mid-point formula .
\(\begin{array}{l}\frac{(3,-4)+(-6,5)\ }{2}=\frac{(-2,-1)+\left(x,y\right)}{2}\\\Rightarrow (3,-4)+(-6,5)\ =(-2,-1)+\left(x,y\right)\\\Rightarrow \left(3+\left(-6\right),-4+5\right)=\left(-2+x,-1+y\right)\\\Rightarrow \left(-3,1\right)=\left(-2+x,-1+y\right)\\\Rightarrow -3=-2+x,1=-1+y\\\Rightarrow -3+2=x,1+1=+y\\\Rightarrow x=-1,y=2\end{array}\)
Answer: The fourth vertex is \((-1,2)\)
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