# Ex.7.2 Q6 Coordinate Geometry Solution - NCERT Maths Class 10

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## Question

If $$(1, 2)$$, $$(4, y)$$, $$(x, 6)$$ and $$(3, 5)$$ are the vertices of a parallelogram taken in order, find $$x$$ and $$y$$.

## Text Solution

Reasoning:

The coordinates of the point $$P(x, y)$$ which divides the line segment joining the points $$A(x1, y1)$$ and $$B(x2, y2)$$, internally, in the ratio $$\rm m1 : m2$$ is given by the Section Formula.

\begin{align}{\text{P}}({\text{x}},{\text{y}}) = \left[ {\frac{{{\text{m}}{{\text{x}}_2} + {\text{n}}{{\text{x}}_1}}}{{{\text{m}} + {\text{n}}}},\frac{{{\text{m}}{{\text{y}}_2} + {\text{n}}{{\text{y}}_1}}}{{{\text{m}} + {\text{n}}}}} \right] & & \end{align}

What is the known?

The $$x$$ and $$y$$ co-ordinates of the vertices of the parallelogram.

What is the unknown?

The missing $$x$$ and $$y$$ co-ordinate.

Steps:

From the Figure, Given,

• Let $$A \;(1, 2)$$, $$B \;(4, y)$$, $$C\;(x, 6)$$, and $$D \;(3, 5)$$ are the vertices of a parallelogram $$ABCD$$.
• Since the diagonals of a parallelogram bisect each other, Intersection point $$O$$ of diagonal $$AC$$ and $$BD$$ also divides these diagonals

Therefore, $$O$$ is the mid-point of $$AC$$ and $$BD$$.

If $$O$$ is the mid-point of $$AC$$, then the coordinates of $$O$$ are

\begin{align}\left( {\frac{{1 + {\text{x}}}}{2},\;\frac{{2 + 6}}{2}} \right) \Rightarrow \left( {\frac{{{\text{x}} + 1}}{2},\;4} \right)\end{align}

If $$O$$ is the mid-point of $$BD$$, then the coordinates of $$O$$ are

\begin{align}\left( {\frac{{4 + 3}}{2},\;\frac{{5 + {\text{y}}}}{2}} \right) \Rightarrow \left( {\frac{7}{2},\;\frac{{5 + {\text{y}}}}{2}} \right)\end{align}

Since both the coordinates are of the same point $$O$$,

\begin{align} &\therefore \frac{{{\text{x}} + 1}}{2} = \frac{7}{2}{\text{ and }}\;4 = \frac{{5 + {\text{y}}}}{2} \end{align}

\begin{align} &\Rightarrow {\text{x}} + 1 = 7\;\;{\text{and}}\;\;5 + {\text{y}} = 8 \, \end{align} (By cross multiplying & transposing)

\begin{align} &\Rightarrow {\text{x}} = 6{\text{ and y}} = 3\end{align}

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