**1. Determine the ratio in which the line 2x + y − 4 = 0 divides the line segment joining the points A(2, − 2) and B(3, 7) **

Solution:

Let the given line divide the line segment joining the points A(2, -2) and B(3,7) in a ratio k:1.

Coordinates of the point of division =

This point also lies on 2x + y – 4 = 0

Therefore, the ratio in hich the line 2x + y – 4 = 0 divides the line segment joining the points A(2, -2) and B(3, 7) is 2:9

**2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear. **

Solution:

If the given points are collinear, then the area of triangle formed by these points will be 0.

This is the required relation between x and y.

**3. Find the centre of a circle passing through the points (6, − 6), (3, − 7) and (3, 3). **

Solution:

Let O (x, y) be the centre of the circle. And let the points (6, −6), (3, −7), and (3, 3) be representing the points A, B, and C on the circumference of the circle.

On adding equation (1) and (2), we obtain

10y = −20 y = −2

From equation (1), we obtain

3x − 2 = 7 3x = 9 x = 3

Therefore, the centre of the circle is (3, −2).

**4. The two opposite vertices of a square are (− 1, 2) and (3, 2). Find the coordinates of the other two vertices. **

Solution:

Let ABCD be a square having (−1, 2) and (3, 2) as vertices A and C respectively. Let (x, y), (x1, y1) be the coordinate of vertex B and D respectively.

We know that, the sides of a square are equal to each other.

∴ AB = BC

We know that in a square, all interior angles are of 90°. In ∆ABC, AB² + BC² = AC²

⇒ 4 + y² + 4 − 4y + 4 + y² − 4y + 4 =16

⇒ 2y² + 16 − 8 y =16

⇒ 2y² − 8 y = 0

⇒ y (y − 4) = 0

⇒ y = 0 or 4

We know that in a square, the diagonals are of equal length and bisect each other at 90°. Let O be the mid-point of AC. Therefore, it will also be the mid-point of BD.

⇒ y + y_{1} = 4

If y = 0, y_{1} = 4

If y = 4, y_{1} = 0

Therefore, the required coordinates are (1, 0) and (1, 4).

**5. The class X students of MRV Public School in Krishna Park have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the following figure. The students are to sow seeds of flowering plants on the remaining area of the plot. **

**(i) Taking A as origin, find the coordinates of the vertices of the triangle. **

**(ii)What will be the coordinates of the vertices of ∆ PQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe?**

Solution:

(i) Taking A as origin, we will take AD as x-axis and AB as y-axis. It can be observed that the coordinates of point P, Q, and R are (4, 6), (3, 2), and (6, 5) respectively.

(ii) Taking C as origin, CB as x-axis, and CD as y-axis, the coordinates of vertices P, Q, and R are (12, 2), (13, 6), and (10, 3) respectively.

It can be observed that the area of the triangle is same in both the cases.

**6. **The vertices of a ∆ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that 𝐴𝐷/𝐴𝐵 = 𝐴𝐸/𝐴𝐶 = 1/4 . Calculate the area of the ∆ADE and compare it with the area of ∆ABC. (Recall Converse of basic proportionality theorem and Theorem 6.6 related to ratio of areas of two similar triangles

Solution:

Given that:

𝐴𝐷/𝐴𝐵 = 𝐴𝐸/𝐴𝐶 = 1/4

Therefore, D and E are two points on side AB and AC respectively such that they divide side AB and AC in a ratio of 1:3.

Clearly, the ratio between the areas of ∆ADE and ∆ABC is 1:16.

**7. Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ∆ABC. **

**(i) The median from A meets BC at D. Find the coordinates of point D. **

**(ii) Find the coordinates of the point P on AD such that AP: PD = 2:1 **

**(iii)Find the coordinates of point Q and R on medians BE and CF respectively such that BQ: QE = 2:1 and CR: RF = 2:1. **

**(iv)What do you observe? **

**(v) If A(x _{1}, y_{1}), B(x_{2}, y_{2}), and C(x_{3}, y_{3}) are the vertices of ∆ABC, find the coordinates of the centroid of the triangle.**

Solution:

(i) Median AD of the triangle will divide the side BC in two equal parts. Therefore, D is the mid-point of side BC.

(ii) Point P divides the side AD in a ratio 2:1.

(iii)Median BE of the triangle will divide the side AC in two equal parts. Therefore, E is the mid-point of side AC.

Point Q divides the side BE in a ratio 2:1.

Median CF of the triangle will divide the side AB in two equal parts. Therefore, F is the mid-point of side AB

Point R divides the side CF in a ratio 2:1.

(iv)It can be observed that the coordinates of point P, Q, R are the same. Therefore, all these are representing the same point on the plane i.e., the centroid of the triangle.

(v) Consider a triangle, ∆ABC, having its vertices as A(x_{1}, y_{1}), B(x_{2}, y_{2}), and C(x_{3}, y_{3}). Median AD of the triangle will divide the side BC in two equal parts. Therefore, D is the mid-point of side BC.

Let the centroid of this triangle be O. Point O divides the side AD in a ratio 2:1.

**8. ABCD is a rectangle formed by the points A (−1, −1), B (− 1, 4), C (5, 4) and D (5, −1). P, Q, R and S are the mid-points of AB, BC, CD, and DA respectively. Is the quadrilateral PQRS is a square? a rectangle? or a rhombus? Justify your answer. **

Solution:

P is the mid-point of side AB.

therefore, the coordinates of P are

Similarly, the coordinates of Q, R and S are (2, 4) , (5, 3/2) and (2, -1) respectively.

It can be observed that all sides of the given quadrilateral are of the same measure. However, the diagonals are of different lengths. Therefore, PQRS is a rhombus.

## 9 responses to “Coordinate Geometry – Exercise 7.4 – Class 10”

[…] Coordinate Geometry – Exercise 7.4 […]

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Omg! I wish I would’ve come here and followed your blog earlier. Just cleared my 10th boards😉😁😂😀 But never mind, do you post something on Humanities subjects too?

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Haha! No dear.. I’m mathematics lecturer…

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Oh hehe!!!! Wow. A maths lecturer? So glad to have been followed by you.😀

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😀

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Your username tells all about it. Such a fool I am. Sorry!

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Haha that’s okay 🙂

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You blog made me “follow you”… nice! Keep posting

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Thanks a lot!😃😃

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