CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If $$A(1, 2), B(4, 3), C(6, 6)$$ are the vertices of a parallelogram $$ABCD,$$ then find the coordinate of the forth vertex.


Solution

Let $$D\left(x,y\right)$$ be the coordinate of the fourth vertex  of the parallelogram $$ABCD$$
We know that diagonals bisect each other in a parallelogram.
$$\Rightarrow\,$$Mid-point of $$AC=$$Mid-point of $$BD$$
Using mid-point formula,we have $$\left(x,y\right)=\left(\dfrac{{x}_{1}+{x}_{2}}{2},\dfrac{{y}_{1}+{y}_{2}}{2}\right)$$
$$\Rightarrow\,\left(\dfrac{1+6}{2},\dfrac{2+6}{2}\right)=\left(\dfrac{x+4}{2},\dfrac{y+3}{2}\right)$$
$$\Rightarrow\,\dfrac{x+4}{2}=\dfrac{7}{2},\,\dfrac{y+3}{2}=\dfrac{8}{2}$$
$$\Rightarrow\,x+4=7,\,y+3=8$$
$$\Rightarrow\,x=7-4=3,\,y=8-3=5$$
$$\therefore\,$$ the fourth vertex is $$\left(3,5\right)$$

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image