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Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

If A(1,2) , B(4,3) and C(6,6) are three vertices of parallelogram ABCD.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The co-ordinate of D.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

Let the co-ordinate of D be (r,m)

We know that formula of the midpoint (M) :

$\boxed{\bf{Midpoint\:(M)=\frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2} }}}}$

Show diagram :

$\setlength{\unitlength}{1.3cm}\begin{picture}(8,2)\thicklines\put(8.6,3){\large{A\:(1,2)}}\put(9,1.3){\sf{}}\put(9.9,1.3){\sf{}}\put(7.7,0.9){\large{B\:\:(4,3)}}\put(9.2,0.7){\sf{\large{}}}\put(11.1,0.9){\large{C(6,6)}}\put(9.9,2.1){\large{O}}\put(8,1){\line(1,0){3}}\put(11,1){\line(1,2){1}}\put(9,3){\line(3,0){3}}\put(11,1){\line(-1,1){2}}\put(8,1){\line(2,1){4}}\put(8,1){\line(1,2){1}}\put(12.1,3){\large{D(r,m)}}\end{picture}$

As we know that diagonal of parallelogram bisect to each other.

For AC :

$\bullet\sf{x_{1}=1\:\:\:x_{2}=6}\\\bullet\sf{y_{1}=2\:\:\:y_{2}=6}$

For BD :

$\bullet\sf{x_{1}=4\:\:\:x_{2}=r}\\\bullet\sf{y_{1}=3\:\:\:y_{2}=m}$

A/q

$\longrightarrow\bf{Mid-point\:of\:AC=Mid-point\:of\:BD}$

$\longrightarrow\sf{\bigg(\dfrac{1+6}{2} ,\dfrac{2+6}{2} \bigg)=\bigg(\dfrac{4+r}{2} ,\dfrac{3+m}{2} \bigg)}\\\\\\\longrightarrow\sf{\bigg(\dfrac{7}{2} ,\cancel{\dfrac{8}{2}} \bigg)=\bigg(\dfrac{r+4}{2} ,\dfrac{m+3}{2} \bigg)}\\\\\\\longrightarrow\sf{\bigg(\dfrac{7}{2} ,4\bigg)=\bigg(\dfrac{r+4}{2} ,\dfrac{m+3}{2} \bigg)}\\\\\\\longrightarrow\sf{\dfrac{r+4}{\cancel{2}} =\dfrac{7}{\cancel{2}} \:\:Or\:\:\dfrac{m+3}{2} =4}\\\\\\\longrightarrow\sf{r+4=7\:\:\:Or\:\:\:m+3=8}$

$\longrightarrow\sf{r=7-4\:\:\:Or\:\:\:m=8-3}\\\\\\\longrightarrow\sf{\green{r=3\:\:\:Or\:\:\:m=5}}$

Thus;

The co-ordinate of D is (3,5) .

Answer 2

Let D(a,b) be the 4th vertex.

Midpoint of AC = (1+6/2,2+6/2) = (7/2,4)        

Midpoint of BD = (a+4/2,b+3/2)

a+4/2 = 7/2 , b+3/2 = 4

a + 4 = 7, b + 3 = 8

a = 3, b = 5

Hope this helps!