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

## 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 \(A(x_1, y_1)\) and \(B(x_2, y_2)\), internally, in the ratio \(m_1 : m_2\) is given by the Section Formula.

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

**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

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

If \(O\) is the mid-point of \(BD\), then the coordinates of \(O\) are

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

Since both the coordinates are of the same point \(O\),

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

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

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