Coordinate Geometry | Exercise 7.2

Question 6

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

Solution

Understand the Question

Let the given vertices taken in order be $A(1, 2)$ , $B(4, y)$ , $C(x, 6)$ , and $D(3, 5)$ .
In a parallelogram, the diagonals bisect each other, meaning the midpoint of diagonal $AC$ is identical to the midpoint of diagonal $BD$ .
The midpoint formula for points $(x_1, y_1)$ and $(x_2, y_2)$ is: $M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$
By equating the corresponding $x$ -coordinates and $y$ -coordinates of both midpoints, we can solve directly for $x$ and $y$ .

Step 1 · Find the Midpoints of Diagonals $AC$ and $BD$

Let the vertices of parallelogram $ABCD$ in order be $A(1, 2)$ , $B(4, y)$ , $C(x, 6)$ , and $D(3, 5)$ . Diagram 1

Midpoint of diagonal $AC$ :

\begin{aligned} M_{\text{AC}} &= \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right) \\[0.6em] &= \left(\dfrac{1 + x}{2}, \dfrac{2 + 6}{2}\right) \\[0.6em] &= \left(\dfrac{1 + x}{2}, \dfrac{8}{2}\right) \\[0.6em] &= \left(\dfrac{1 + x}{2}, 4\right) \end{aligned}

Midpoint of diagonal $BD$ :

\begin{aligned} M_{\text{BD}} &= \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right) \\[0.6em] &= \left(\dfrac{4 + 3}{2}, \dfrac{y + 5}{2}\right) \\[0.6em] &= \left(\dfrac{7}{2}, \dfrac{y + 5}{2}\right) \end{aligned}

Step 2 · Equate Midpoints and Solve for $x$ and $y$

Since diagonals of a parallelogram bisect each other, $M_{\text{AC}} = M_{\text{BD}}$ .

Equating $x$ -coordinates:

\begin{aligned} \dfrac{1 + x}{2} &= \dfrac{7}{2} \\[0.6em] 1 + x &= 7 \\ x &= 7 - 1 \\ x &= 6 \end{aligned}

Equating $y$ -coordinates:

\begin{aligned} \dfrac{y + 5}{2} &= 4 \\[0.6em] y + 5 &= 4 \times 2 \\ y + 5 &= 8 \\ y &= 8 - 5 \\ y &= 3 \end{aligned}

Answer

$x = 6, \quad y = 3$

Common Mistakes

Wrong Diagonal Pairs: The vertices are taken in order ( $A \to B \to C \to D$ ). The diagonals must be $AC$ and $BD$ . Pairing adjacent vertices (like $AB$ or $CD$ ) as diagonals leads to incorrect equations.
Using the Distance Formula: Equating lengths of opposite sides ( $AB = CD$ and $BC = AD$ ) creates complex non-linear simultaneous equations. The midpoint property is far simpler and directly linear.

More questions in Exercise 7.2

Find the coordinates of the point which divides the join of $(-1, 7)$ and $(4, -3)$ in the ratio $2 : 3$ .

Find the coordinates of the points of trisection of the line segment joining $(4, -1)$ and $(-2, -3)$ .

To conduct Sports Day activities, in your rectangular shaped school ground $ABCD$ , lines have been drawn with chalk powder at a distance of $1\text{ m}$ each. $100$ flower pots have been placed at a distance of $1\text{ m}$ from each other along $AD$ , as shown in Fig. 7.12. Niharika runs $\dfrac{1}{4}\text{th}$ the distance $AD$ on the 2nd line and posts a green flag. Preet runs $\dfrac{1}{5}\text{th}$ the distance $AD$ on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Find the ratio in which the line segment joining the points $(-3, 10)$ and $(6, -8)$ is divided by $(-1, 6)$ .

Find the ratio in which the line segment joining $A(1, -5)$ and $B(-4, 5)$ is divided by the $x$ -axis. Also find the coordinates of the point of division.

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

Find the coordinates of a point $\text{A}$ , where $\text{AB}$ is the diameter of a circle whose centre is $(2, -3)$ and $\text{B}$ is $(1, 4)$ .

If A and B are $(-2, -2)$ and $(2, -4)$ , respectively, find the coordinates of P such that $\text{AP} = \dfrac{3}{7} \text{AB}$ and P lies on the line segment AB.

Find the coordinates of the points which divide the line segment joining $A(-2, 2)$ and $B(2, 8)$ into four equal parts.

Q10

Find the area of a rhombus if its vertices are $(3, 0)$ , $(4, 5)$ , $(-1, 4)$ and $(-2, -1)$ taken in order.

[Hint : Area of a rhombus = $\dfrac{1}{2}$ (product of its diagonals)]