NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Exercise 7.1

Q.1) Find the distance between the following pairs of points:
(i) (2, 3), (4, 1) (ii) (−5, 7), (−1, 3) (iii) (a, b), (− a, − b)
Sol.1) (i) Distance between the points is given by

The distance between two points $({�}_1,{�}_1)$ and $({�}_2,{�}_2)$ in a Cartesian coordinate system is given by the distance formula:

$�=\sqrt{({�}_2-{�}_1{)}^2+({�}_2-{�}_1{)}^2}$

Let's apply this formula to find the distances for the given pairs of points:

(i) $(2,3),(4,1)$:

$�=\sqrt{(4-2{)}^2+(1-3{)}^2}=\sqrt{2^2+(-2{)}^2}=\sqrt{4+4}=\sqrt{8}$

(ii) $(-5,7),(-1,3)$:

$�=\sqrt{((-1)-(-5){)}^2+(3-7{)}^2}=\sqrt{4^2+(-4{)}^2}=\sqrt{16+16}=\sqrt{32}$

(iii) $(�,�),(-�,-�)$:

$�=\sqrt{((-�)-�{)}^2+((-�)-�{)}^2}=\sqrt{(-2�{)}^2+(-2�{)}^2}=\sqrt{4{�}^2+4{�}^2}$

So, the distances are: (i) $\sqrt{8}$ (ii) $\sqrt{32}$ (iii) $\sqrt{4{�}^2+4{�}^2}$

Q.2) Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
Sol.2) Distance between points (0, 0) and (36, 15)

Yes, Assume town A at origin point (0, 0).
Therefore, town B will be at point (36, 15) with respect to town A.
And hence, as calculated above, the distance between town A and B will be 39 km.

Q.3) Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Sol.3) Let the points (1, 5), (2, 3), and (- 2,-11) be representing the vertices A, B, and C of the given triangle respectively.
Let A = (1, 5), B = (2, 3) and C = (- 2,-11)

Therefore, the points (1, 5), (2, 3), and ( - 2, - 11) are not collinear.

Q.4) Check whether (5, - 2), (6, 4) and (7, - 2) are the vertices of an isosceles triangle.
Sol.4) Let the points (5, - 2), (6, 4), and (7, - 2) are representing the vertices A, B, and C of the given triangle respectively.

Therefore, 𝐴𝐵 = 𝐵𝐶
As two sides are equal in length, therefore, ABC is an isosceles triangle

Q.5) In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don't you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Sol.5) Clearly from the figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6, 1). By using distance formula, we get

The distance formula is given by:

$�=\sqrt{({�}_2-{�}_1{)}^2+({�}_2-{�}_1{)}^2}$

Let's calculate the distances:

1. Distance between A and B: $��=\sqrt{(6-3{)}^2+(7-4{)}^2}=\sqrt{3^2+3^2}=\sqrt{9+9}=\sqrt{18}$

2. Distance between B and C: $��=\sqrt{(9-6{)}^2+(4-7{)}^2}=\sqrt{3^2+(-3{)}^2}=\sqrt{9+9}=\sqrt{18}$

3. Distance between C and D: $��=\sqrt{(6-9{)}^2+(1-4{)}^2}=\sqrt{(-3{)}^2+(-3{)}^2}=\sqrt{9+9}=\sqrt{18}$

4. Distance between D and A: $��=\sqrt{(3-6{)}^2+(4-1{)}^2}=\sqrt{(-3{)}^2+3^2}=\sqrt{9+9}=\sqrt{18}$

It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are of the same length.
Therefore, ABCD is a square and hence, Champa was correct

Q.6) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (- 1, - 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, - 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Sol.6) Let the points ( - 1, - 2), (1, 0), ( - 1, 2), and ( - 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
By using distance formula, we get

It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square.
(ii) Let the points (- 3, 5), (3, 1), (0, 3), and ( - 1, - 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively. It can be observed that all sides of this quadrilateral are of different lengths. Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc.
By using distance formula, we get

It can be observed that opposite sides of this quadrilateral are of the same length. 
However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.

Q.7) Find the point on the x-axis which is equidistant from (2, - 5) and (- 2, 9).
Sol.7) We have to find a point on x-axis. Therefore, its y-coordinate will be 0.
Let the point on 𝑥 − 𝑎𝑥𝑖𝑠 be (𝑥, 0)

Q.8) Find the values of y for which the distance between the points 𝑃 (2, − 3) and 𝑄 (10, 𝑦) is 10 units.
Sol.8) It is given that the distance between (2, - 3) and (10, y) is 10.

Q.9) If 𝑄 (0, 1) is equidistant from 𝑃 (5, − 3) and 𝑅 (𝑥, 6), find the values of 𝑥. Also find the distance QR and PR.
Sol.9) 𝑃𝑄 = 𝑄𝑅

Q.10) Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
Sol.10) Point (𝑥, 𝑦) is equidistant from (3, 6) and ( - 3, 4)

(𝑥 − 3)2 + (𝑦 + 6)2 = (𝑥 + 3)2 + (𝑦 − 4)2
(𝑥 − 3)2 + (𝑦 + 6)2 = (𝑥 + 3)2 + (𝑦 − 4)2
𝑥2 + 9 − 6𝑥 + 𝑦2 + 36 − 12𝑦 = 𝑥2 + 9 + 6𝑥 + 𝑦2 + 16 − 8𝑦
36 − 16 = 6𝑥 + 6𝑥 + 12𝑦 − 8𝑦
20 = 12𝑥 + 4𝑦
3𝑥 + 𝑦 = 5
3𝑥 + 𝑦 − 5 = 0

Exercise 7.2

Q.1) Find the coordinates of the point which divides the join of (- 1, 7) and (4, - 3) in the ratio 2: 3.
Sol.1) Let 𝑃(𝑥, 𝑦) be the required point. Using the section formula,

Q.2) Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2,-3).
Sol.2) Let 𝑃 (𝑥1 , 𝑦1 ) and 𝑄 (𝑥2 , 𝑦2) are the points of trisection of the line segment joining the given points i.e., 𝐴𝑃 = 𝑃𝑄 = 𝑄𝐵
Therefore, point 𝑃 divides AB internally in the ratio 1: 2.

Q.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 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th 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?

Sol.3) It can be observed that Niharika posted the green flag at 1/4𝑡ℎ of the distance 𝐴𝐷 i.e.,
(1 × 100/4) 𝑚 = 25𝑚 from the starting point of 2nd line. Therefore, the coordinates of this point G is (2, 25).
Similarly, Preet posted red flag at 1/5 of the distance 𝐴𝐷 i.e., (1 × 100/5) 𝑚 = 20𝑚 from the starting point of 8th line. Therefore, the coordinates of this point 𝑅 are (8, 20).
Distance between these flags by using distance formula

Hence, 𝐴(𝑥, 𝑦) = (5,22.5)
Therefore, Rashmi should post her blue flag at 22.5𝑚 on 5th line.

Q.4) Find the ratio in which the line segment joining the points (-3, 10) and (6, - 8) is divided by (-1, 6).
Sol.4) Let the ratio in which the line segment joining ( -3, 10) and (6, -8) is divided by point
( -1, 6) be 𝑘 ∶ 1.
Therefore, −1 = 6𝑘−3/𝑘+1
−𝑘 − 1 = 6𝑘 − 3
7𝑘 = 2
𝑘 = 2/7
Therefore, the required ratio is 2: 7.

Q.5) Find the ratio in which the line segment joining A (1, - 5) and B (- 4, 5) is divided by the xaxis. 
Also find the coordinates of the point of division
Sol.5) Let the ratio in which the line segment joining A (1, - 5) and B ( - 4, 5) is divided by x-axis be 𝑘: 1.

Q.6) If (1, 2), (4, 𝑦), (𝑥, 6) and (3, 5)are the vertices of a parallelogram taken in order, find 𝑥 and 𝑦.
Sol.6) Let A,B,C and D be the points (1,2) (4, 𝑦), (𝑥, 6) and (3,5) respectively

Since the diagonals of a parallelogram bisect each other, the mid-point of AC and BD are same.
∴ 𝑥+1/2 = 7/2 and 4 = 5+𝑦/2
⇒ 𝑥 + 1 = 7 and 5 + 𝑦 = 8
⇒ 𝑥 = 6 and 𝑦 = 3

Q.7) Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, -3) and B is (1, 4).
Sol.7) Let the coordinates of point A be (𝑥, 𝑦).
Mid-point of AB is (2, - 3), which is the centre of the circle.

⇒ 𝑥 + 1 = 4 and 𝑦 + 4 = −6
⇒ 𝑥 = 3 and 𝑦 = −10
Therefore, the coordinates of A are (3,-10).

Q.8) If A and B are (–2, –2) and (2, –4), respectively, find the coordinates of P such that 𝐴𝑃 = 3/7 𝐴𝐵 and P lies on the line segment 𝐴𝐵.
Sol.8)

Q.9) Find the coordinates of the points which divide the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.
Sol.9)

Q.10) 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 = ½ (product of its diagonals)]
Sol.10) Let (3, 0), (4, 5), ( - 1, 4) and ( - 2, - 1) are the vertices A, B, C, D of a rhombus ABCD.

Exercise 7.3

Q.1) Find the area of the triangle whose vertices are:

(i) (2, 3), (-1, 0), (2, -4) (ii) (-5, -1), (3, -5), (5, 2)
Sol.1) (i) Area of a triangle is given by Area of triangle = 1/2
{𝑥1(𝑦2 − 𝑦3) + 𝑥2(𝑦3 − 𝑦1) + 𝑥3 (𝑦1 − 𝑦2)}
Area of the given triangle = 1/2 [2 { 0 − (−4)} + (−1){(−4) − (3)} + 2(3 − 0)] = 1/2
{8 + 7 + 6} = 21/2 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠.
(ii) Area of the given triangle = 1/2 [−5 { (−5) − (4)} + 3(2 − (−1)) + 5{−1 −
(−5)}] = 1/2 {35 + 9 + 20} = 32 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠

Q.2) In each of the following find the value of ′𝑘′, for which the points are collinear.
(i) (7, -2), (5, 1), (3, -k) (ii) (8, 1), (k, -4), (2, -5)
Sol.2) (i) For collinear points, area of triangle formed by them is zero.
Therefore, for points (7, −2) (5, 1), and (3, 𝑘), area = 0
1/2 [7 { 1 − 𝑘} + 5(𝑘 − (−2)) + 3{(−2) + 1}] = 0
7 − 7𝑘 + 5𝑘 + 10 − 9 = 0 − 2𝑘 + 8 = 0
𝑘 = 4

(ii) For collinear points, area of triangle formed by them is zero.
Therefore, for points (8, 1), (𝑘, − 4), and (2, − 5), area = 0
1/2 [8 { −4 − (−5)} + 𝑘{(−5) − (1)} + 2{1 − (−4)}] = 0
8 − 6𝑘 + 10 = 0
6𝑘 = 18
𝑘 = 3

Q.3) Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Sol.3) Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides of this triangle.
Coordinates of D, E, and F are given by
Area of a triangle = 1/2 {𝑥1(𝑦2 − 𝑦3) + 𝑥2(𝑦3 − 𝑦1) + 𝑥3 (𝑦1 − 𝑦2)}

Area of 𝛥𝐷𝐸𝐹 = 1/2 {1(2 − 1) + 1(1 − 0) + 0(0 − 2)} = 1/2 (1 + 1) = 1 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠
Area of 𝛥𝐴𝐵𝐶 = 1/2 [0(1 − 3) + 2{3 − 1)} + 0(−1 − 1)] = 1/2 {8} 
= 4 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠
Therefore, the required ratio is 1: 4.

Q.4) Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, - 2) and (2, 3)
Sol.4)

Let the vertices of the quadrilateral be A( - 4, - 2), B ( - 3, - 5), C (3, - 2), and D (2, 3).
Join AC to form two triangles 𝛥𝐴𝐵𝐶 and 𝛥𝐴𝐶𝐷.
Area of a triangle = 1/2 {𝑥1(𝑦2 − 𝑦3) + 𝑥2(𝑦3 − 𝑦1) + 𝑥3(𝑦1 − 𝑦2)}
Area of 𝛥𝐴𝐵𝐶 = 1/2 [(−4){(−5) − (−2)} + (−3){(−2) − (−2)} + 3 {(−2) − (−5)}]
= 1/2 (12 + 0 + 9) = 21/2 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠
Area of 𝛥𝐴𝐶𝐷 = 1/2 [(−4){(−2)– (3)} + 3{(3)– (−2)} + 2 {(−2)– (−2)}]
= 1/2 (20 + 15 + 0) = 35/2 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠
Area of ☐𝐴𝐵𝐶𝐷 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝛥𝐴𝐵𝐶 + 𝐴𝑟𝑒𝑎 𝑜𝑓 𝛥𝐴𝐶𝐷
= (21/2 + 35/2) 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠 = 28 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠

Q.5) You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, - 6), B (3, - 2) and C (5, 2).
Sol.5)

Let the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2).
Let D be the mid-point of side BC of 𝛥𝐴𝐵𝐶.
Therefore, AD is the median in 𝛥𝐴𝐵𝐶.
Coordinates of point 𝐷 = (3 + 5/2, −2 + 2/2) = (4,0)
Area of a triangle = 1/2
{𝑥1(𝑦2 − 𝑦3) + 𝑥2(𝑦3 − 𝑦1) + 𝑥3(𝑦1 − 𝑦2)}
Area of 𝛥𝐴𝐵𝐷 = 1/2 [(4){(−2)– (0)} + 3{(0)– (−6)} + (4){(−6)– (−2)}]
= 1/2 (−8 + 18 − 16) = −3 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠
However, area cannot be negative.
Therefore, area of 𝛥𝐴𝐵𝐷 is 3 square units.
Area of 𝛥𝐴𝐵𝐷 = 1/2
[(4){0 – (2)} + 4{(2)– (−6)} + (5){(−6)– (0)}]
= 1/2 (−8 + 32 − 30) = −3 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠
However, area cannot be negative.
Therefore, area of 𝛥𝐴𝐵𝐷 is 3 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠.
The area of both sides is same.
Thus, median AD has divided 𝛥𝐴𝐵𝐶 in two triangles of equal areas

Exercise 7.4

Q.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).

Sol.1) Suppose the line 2𝑥 + 𝑦 – 4 = 0 divides the line segment joining A(2, –2) and B(3, 7) in the ratio 𝑘 ∶ 1 at point C.
Then, the coordinates of C are (3𝑘+2/𝑘+1, 7𝑘−2/𝑘+1)
But C lies on 2x + y – 4 = 0 therefore,
2 (3𝑘+2/𝑘+1) + (7𝑘−2/𝑘+1) − 4 = 0
⇒ 6𝑘 + 4 + 7𝑘 − 2 − 4𝑘 − 4 = 0
⇒ 9k − 2 = 0
⇒ 9𝑘 = 2 ⇒ k = 2/9
So, the required ratio is 2 : 9 internally.

Q.2) Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Sol.2) The points 𝐴(𝑥, 𝑦), 𝐵(1, 2) and C(7, 0) will be are collinear.
𝑥(2 – 0) + 1(0 – 𝑦) + 7(𝑦 – 2) = 0
2𝑥 – 𝑦 + 7𝑦 – 14 = 0
2𝑥 + 6𝑦 – 14 = 0
𝑥 + 3𝑦 – 7 = 0
which is the relation between 𝑥 and 𝑦.

Q.3) Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3).
Sol.3) Let 𝑃(𝑥, 𝑦) be the centre of the circle passing through the points 𝐴(6, – 6), 𝐵(3, – 7) and 𝐶(3, 3).
Then, 𝐴𝑃 = 𝐵𝑃 = 𝐶𝑃.
Now, 𝐴𝑃 = 𝐵𝑃 = 0
⇒ 𝐴𝑃2 = 𝐵𝑃2
⇒ (𝑥 − 6) + (𝑦 + 6)2 = (𝑥 − 3)2 + (𝑦 + 7)2
⇒ 𝑥2 − 12𝑥 + 36 + 𝑦2 + 12𝑦 + 36 = 𝑥2 − 6𝑥 + 9 + 𝑦2 + 14𝑦 + 49
⇒ −12𝑥 + 6𝑥 + 12𝑦 − 14𝑦 + 72 − 58 = 0
⇒ −6𝑥 − 2𝑦 + 14 = 0
⇒ 3𝑥 + 𝑦 − 7 = 0 …. (i)
and, 𝐵𝑃 = 𝐶𝑃 = 0 ...(2)
⇒ 𝐵𝑃2 = 𝐶𝑃2
⇒ (𝑥 − 3)2 + (𝑦 + 7)2 = (𝑥 − 3)2 + (𝑦 − 3)2
⇒ 𝑥2 − 6𝑥 + 9 + 𝑦2 + 14𝑦 + 49 = 𝑥2 − 6𝑥 + 9 + 𝑦2 − 6𝑦 + 9
⇒ −6𝑥 + 6𝑥 + 14𝑦 + 6𝑦 + 58 − 18 = 0
⇒ 20𝑦 + 40 = 0
⇒ 𝑦 = −40/20
= −2 …… (ii)
Putting 𝑦 = – 2 in (i), we get
3𝑥 – 2 – 7 = 0
⇒ 3𝑥 = 9
⇒ 𝑥 = 3
Thus, the centre of the circle is (3, – 2)

Q.4) The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.
Sol.4) Let 𝐴𝐵𝐶𝐷 be a square and let 𝐴(– 1, 2) and 𝐶(3, 2) be the given angular points.
Let 𝐵(𝑥, 𝑦) be the unknown vertex.
Then, 𝐴𝐵 = 𝐵𝐶
⇒ 𝐴𝐵2 = 𝐵𝐶2
⇒ (𝑥 + 1)2 + (𝑦 − 2)2 = (𝑥 − 3)22 + (𝑦 − 2)2
⇒ 𝑥2 + 2𝑥 + 1 + 𝑦2 − 4𝑦 + 4 = 𝑥2 − 6𝑥 + 9 + 𝑦2 − 4𝑦 + 4
⇒ 2𝑥 + 1 = −6𝑥 + 9
⇒ 8𝑥 = 8
⇒𝑥 = 1 ...(1)
In ABC, we have
𝐴𝐵2 + 𝐵𝐶2 = 𝐴𝐶2
⇒ (𝑥 + 1)2 + (𝑦 − 2)2 + (𝑥 − 3)2 + (𝑦 − 2)2 = (3 + 1)2 + (2 − 2)2
⇒ 2𝑥2 + 2𝑦2 + 2𝑥 − 4𝑦 − 6𝑥 − 4𝑦 + 1 + 4 + 9 + 4 = 16
⇒ 2𝑥2 + 2 𝑦2 − 4𝑥 − 8𝑦 + 2 = 0
⇒ 𝑥2 + 𝑦2 − 2𝑥 − 4𝑦 + 1 = 0 ...(2)
Substituting the value of 𝑥 from (1) into (2), we get
1 + 𝑦2 − 2 − 4𝑦 + 1 = 0
⇒ 𝑦2 − 4𝑦 = 0
⇒ 𝑦 (𝑦 − 4) = 0
⇒ 𝑦 = 0 or 4
Hence, the required vertices of the square are (1, 0) and (1, 4).

Q.5) The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the 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 Δ𝑃𝑄𝑅 if C is the origin ? Also calculate the area of the triangle in these cases. What do you observe ?
Sol.5) (i) Taking A as the origin, AD and AB as the coordinate axes. Clearly, the points P,Q and R
are (4, 6), (3, 2) and (6, 5) respectively
(ii) Taking C as the origin, CB and CD as the coordinate axes. Clearly, the points P, Q and R are given by (12, 2), (13, 6) and (10, 3) respectively,
We know that the area of the triangle whose vertices are (𝑥1, 𝑦1), (𝑥2, 𝑦2) and (𝑥3, 𝑦3) is given by
= 1/2 [𝑥1(𝑦2 − 𝑦3) + 𝑥2(𝑦3 − 𝑦1) + 𝑥3(𝑦1 − 𝑦2)]
Therefore, area of Δ𝑃𝑄𝑅 in the 1st case
1/2 [4 (2 − 5) + 3 (5 − 6) + 6 (6 − 2)]
= 1/2 (4 × −3 + 3 × −1 + 6 × 4)
= 1/2 = (−12 − 3 + 24) = 9/2 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
and, area of Δ𝑃𝑄𝑅 in the 2nd case
= 1/2 [12 (6 − 3) + 13 (3 − 2) + 10 (2 − 6)] 
= 1/2 (12 × 3 + 13 × 1 + 10 × −4)
= 1/2 (36 + 13 − 40) = 9/2 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
Thus, we observe that the areas are the same in both the cases.

Q.6) The vertices of a Δ𝐴𝐵𝐶 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 Δ𝐴𝐷𝐸 and compare it with the area of Δ𝐴𝐵𝐶.
Sol.6) We know that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Therefore, 𝐷𝐸 || 𝐵𝐶
Clearly, Δ𝐴𝐷𝐸~Δ𝐴𝐵𝐶 (AAA similarity)
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of .
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that 𝐴𝑃 ∶ 𝑃𝐷 = 2 ∶ 1.
(iii) Find the coordinates of points 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 ?
(Note : The point which is common to all the three medians is called centroid and this point divides each median in the ratio 2 ∶ 1). (v) If 𝐴(𝑥1, 𝑦1), 𝐵(𝑥2, 𝑦2) and 𝐶(𝑥3, 𝑦3) are the vertices of Δ𝐴𝐵𝐶, find the coordinates of the centroid of the triangle.

(i) The median from A meets BC at D. To find the coordinates of point D, we can use the midpoint formula. The coordinates of the midpoint M of BC are given by $({�}_{�},{�}_{�})=(\frac{{�}_{�}+{�}_{�}}{2},\frac{{�}_{�}+{�}_{�}}{2})$. So, the coordinates of D are the same as M.

$�(\frac{{�}_{�}+{�}_{�}}{2},\frac{{�}_{�}+{�}_{�}}{2})$

$�(\frac{6+1}{2},\frac{5+4}{2})$

$�(3.5,4.5)$

(ii) To find the coordinates of the point P on AD such that $��:��=2:1$, we can use the section formula. The coordinates of P are given by:

$�(\frac{2{�}_{�}+1{�}_{�}}{2+1},\frac{2{�}_{�}+1{�}_{�}}{2+1})$

$�(\frac{2(3.5)+1(4)}{3},\frac{2(4.5)+1(2)}{3})$

$�(\frac{11.5}{3},\frac{11}{3})$

(iii) To find the coordinates of points Q and R on medians BE and CF respectively such that $��:��=2:1$ and $��:��=2:1$, we can use the section formula again.

For Q:

$�(\frac{2{�}_{�}+1{�}_{�}}{2+1},\frac{2{�}_{�}+1{�}_{�}}{2+1})$

For R:

$�(\frac{2{�}_{�}+1{�}_{�}}{2+1},\frac{2{�}_{�}+1{�}_{�}}{2+1})$

(iv) What do you observe?

In a triangle, the medians are concurrent at a point called the centroid, and the centroid divides each median in the ratio 2:1. So, the ratios $��:��$ and $��:��$ are indeed 2:1.

(v) If $�({�}_1,{�}_1)$, $�({�}_2,{�}_2)$, and $�({�}_3,{�}_3)$ are the vertices of $\mathrm{\Delta }���$, the coordinates of the centroid G are given by:

$�(\frac{{�}_1+{�}_2+{�}_3}{3},\frac{{�}_1+{�}_2+{�}_3}{3})$

So, for the given triangle with vertices A(4, 2), B(6, 5), and C(1, 4), the coordinates of the centroid G are:

$�(\frac{4+6+1}{3},\frac{2+5+4}{3})$

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.

To determine the nature of the quadrilateral PQRS formed by the midpoints of the sides of rectangle ABCD, we can analyze its properties:

Let's find the coordinates of the midpoints P, Q, R, and S:

Midpoint formula: $({�}_{\text{mid}},{�}_{\text{mid}})=(\frac{{�}_1+{�}_2}{2},\frac{{�}_1+{�}_2}{2})$

1. Midpoint of AB (P): $�=(\frac{-1+(-1)}{2},\frac{-1+4}{2})=(-1,1.5)$

2. Midpoint of BC (Q): $�=(\frac{-1+5}{2},\frac{4+4}{2})=(2,4)$

3. Midpoint of CD (R): $�=(\frac{5+5}{2},\frac{4+(-1)}{2})=(5,1.5)$

4. Midpoint of DA (S): $�=(\frac{-1+5}{2},\frac{-1+(-1)}{2})=(2,-1)$

Now, let's analyze the properties:

1. Equal Side Lengths:

   • Distance between P and Q: $��=\sqrt{(2-(-1){)}^2+(4-1.5{)}^2}=\sqrt{3^2+2.5^2}=\sqrt{9+6.25}=\sqrt{15.25}$
   • Distance between Q and R: $��=\sqrt{(5-2{)}^2+(1.5-4{)}^2}=\sqrt{3^2+(-2.5{)}^2}=\sqrt{9+6.25}=\sqrt{15.25}$
   • Distance between R and S: $��=\sqrt{(2-5{)}^2+(-1-1.5{)}^2}=\sqrt{(-3{)}^2+(-2.5{)}^2}=\sqrt{9+6.25}=\sqrt{15.25}$
   • Distance between S and P: $��=\sqrt{(-1-2{)}^2+(-1-(-1){)}^2}=\sqrt{(-3{)}^2+2^2}=\sqrt{9+4}=\sqrt{13}$

   Since PQ = QR = RS = SP, all sides are equal.

2. Right Angles:

   • The diagonals of a rectangle bisect each other at right angles.

3. Diagonals are Equal:

   • Diagonals PS and QR are both equal to $\sqrt{13}$.

Given that all sides are equal, angles are right angles, and diagonals are equal, we can conclude that PQRS is a square.

quadrilateral PQRS is a square.

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