Question

If in a parallelogram ABDC, the coordinates of A, B, C are respectively $\left( 1,2 \right)$, $\left( 3,4 \right)$ and $\left( 2,5 \right)$. Then the equation of the diagonal AD is
$\begin{align}
  & \left( A \right)5x-3y+1=0 \\
 & \left( B \right)5x+3y-11=0 \\
 & \left( C \right)3x-5y+7=0 \\
 & \left( D \right)3x+5y-13=0 \\
\end{align}$

Answer 1

Hint: We solve this problem by finding the midpoint of diagonal BC using the formula \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\]. Then we assume that the coordinates of D as $\left( a,b \right)$. Then we equate the obtained midpoints and solve them to find the values of a and b. then we find the equation of the diagonal AD using the formula \[~\left( y-{{y}_{1}} \right)=\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}\left( x-{{x}_{1}} \right)\].

Complete step by step answer:
Let us go through a property of the parallelogram. The two diagonals of a parallelogram bisect each other, that is, the midpoints of the diagonals of the parallelogram are the same.
We are given that in a parallelogram ABDC, the coordinates of A, B, C are respectively $\left( 1,2 \right)$, $\left( 3,4 \right)$ and $\left( 2,5 \right)$.
Let us assume that the coordinates of D is $\left( a,b \right)$.

From the above discussed property, we can say that mid points of AD and BC are equal.
Now, let us find the midpoint of diagonal BC.
Let us consider the formula for midpoint of two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$.
\[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\]
Then the midpoint of diagonal BC is
$\Rightarrow \left( \dfrac{3+2}{2},\dfrac{4+5}{2} \right)=\left( \dfrac{5}{2},\dfrac{9}{2} \right)$
Now let us find the midpoint of diagonal AD.
$\Rightarrow \left( \dfrac{1+a}{2},\dfrac{2+b}{2} \right)$
As midpoints of AD and BC are equal, let us equate the obtained coordinates.
$\begin{align}
  & \Rightarrow \left( \dfrac{1+a}{2},\dfrac{2+b}{2} \right)=\left( \dfrac{5}{2},\dfrac{9}{2} \right) \\
 & \Rightarrow \left( 1+a,2+b \right)=\left( 5,9 \right) \\
 & \Rightarrow \left( a,b \right)=\left( 5-1,9-2 \right) \\
 & \Rightarrow \left( a,b \right)=\left( 4,7 \right) \\
\end{align}$
So, coordinates of D are (4,7).
Now let us consider the formula for equation of line joining two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$.
\[~\left( y-{{y}_{1}} \right)=\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}\left( x-{{x}_{1}} \right)\]
So, the equation of the line AD is
\[\begin{align}
  & \Rightarrow ~\left( y-2 \right)=\dfrac{2-7}{1-4}\left( x-1 \right) \\
 & \Rightarrow ~\left( y-2 \right)=\dfrac{-5}{-3}\left( x-1 \right) \\
 & \Rightarrow ~\left( y-2 \right)=\dfrac{5}{3}\left( x-1 \right) \\
 & \Rightarrow ~3\left( y-2 \right)=5\left( x-1 \right) \\
 & \Rightarrow ~3y-6=5x-5 \\
 & \Rightarrow ~5x-3y+1=0 \\
\end{align}\]
So, the equation of the line AD is 5x-3y+1=0.

So, the correct answer is “Option A”.

Note: We can also find the coordinates of point D using the formula,
For a parallelogram with coordinates $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ occurring successively the coordinates of the other vertex is \[~\left( {{x}_{1}}+{{x}_{3}}-{{x}_{2}},{{y}_{1}}+{{y}_{3}}-{{y}_{2}} \right)\].
So, we are given that in a parallelogram ABDC, the co-ordinates of A, B, C are respectively $\left( 1,2 \right)$, $\left( 3,4 \right)$ and $\left( 2,5 \right)$. Writing them in successive points they are $\left( 2,5 \right)$, $\left( 1,2 \right)$ and $\left( 3,4 \right)$. Then the coordinates of D by applying the above formula are,
\[\Rightarrow ~\left( 2+3-1,5+4-2 \right)=\left( 4,7 \right)\]
So, we get the co-ordinates of D as (4,7).
Then we can apply the formula for equation of line as in the solution above and find the equation of diagonal AD.