## NCERT Class 10 Maths Chapter 7: Coordinate Geometry## Exercise 7.1
**(2, 3), (4, 1)****(- 5, 7), (- 1, 3)****(a, b), (- a, - b)**
Distance between the given points: Hence, the distance between the given pair of point is 2√2 units.
Distance between the given points: Hence, the distance between the given pair of point is 4√2 units.
Distance between the given points: Hence, the distance between the given pair of point is 2√(a
Let (0, 0) = (x Distance between the given points: (The two towns A and B in section 7.2 share the same coordinates as given in first part of question, hence, distance between them will be 39 km.)
Let the given coordinates (1, 5), (2, 3) and (- 2, - 11) be A, B and C respectively. If the points are collinear then the sum of the shorter distances will be the same as the longest distance. Distance between A and B: Distance between B and C: Distance between A and C: Sum of shorter distance = √5 + √212 = √217 units Since √217 ? √265, therefore, the given points are not collinear.
Let the given coordinates (5, -2), (6, 4) and (7, -2) be A, B and C respectively. If the points form an isosceles triangle then two of the distances between the a pair of points must be equal. Since, AB = BC. Therefore, the given points will form an isosceles triangle.
If ABCD forms a square then all sides must be equal, i.e, AB = BC = CD = AD and the diagonals AC and BD must be equal as well. Since, AB = BC = CD = AD. Therefore, ABCD forms a quadrilateral with all sides equal, i.e, a rhombus or a square. We will find length of diagonals to find out whether ABCD is a rhombus or a square. Since, the diagonals of ABCD are equal. Therefore, the quadrilateral is a square. Hence, Champa is correct.
**(- 1, - 2), (1, 0), (- 1, 2), (- 3, 0)****(-3, 5), (3, 1), (0, 3), (-1, - 4)****(4, 5), (7, 6), (4, 3), (1, 2)**
Since, AB = BC = CD = AD. Therefore, ABCD forms a quadrilateral with all sides equal, i.e, a rhombus or a square. We will find length of diagonals to find out whether ABCD is a rhombus or a square. Hence, the given points make up a square.
No quadrilateral is formed by the given points.
Since, AB = CD and BC = AD. Therefore, ABCD forms a quadrilateral with opposite sides equal, i.e., a parallelogram or a rectangle. We will find length of diagonals to find out whether ABCD is a parallelogram or a rectangle. The diagonals are unequal. Hence, the given points make up a parallelogram.
The required point lies on the x-axis, which implies that its y-coordinate is 0. Then, let the required point be (x, 0). According to question: Distance between (2, -5) and (x, 0) = Distance between (x, 0) and (-2, 9) Hence, the required point will be (-7, 0).
It is given that: Distance between P and Q = 10 units
According to question, we have: Distance between Q and P = Distance between Q and R Therefore, x is either 4 or -4.
Hence, x = �4 and distances QR and PR are √41 units and √82 units respectively OR √41 units and 9√2 units respectively.
According to the question, we have: Distance between (3, 6) and (x, y) = Distance between (-3, 4) and (x, y) Hence, 3x - y = 5 is the required relation. ## Exercise 7.2
Let the given points be A (-1, 7) and C (4, -3), whose join is divided in the ratio by point B (x, y). The coordinates of B will be: Hence, the required point is (1, 3).
Let the points be A (4, -1) and D (-2, -3) which are being trisected by two point B (x, y) and C (X, Y). Since the line AD is being trisected, therefore, AD is divided by B in 1 : 2 ratio and by C in 2 : 1 ratio. Therefore, Hence, the required point are (2, -5/3) and (0, -7/3).
Total distance of AD = 100 m Niharika ran in the 2 Preet ran in the 8 Distance between the two flags : Hence, distance between the red flag and the green flag is √61 m. The blue flag needs to be posted at the midpoint of the line formed by joining the red and the green flags. Then, the blue flag's coordinates are: Hence, Rashmi needs to post the blue flag at 45/2 m in the 5
Let the given points be A (-3, 10) and C (6, -8) being divided by the point B (-1, 6). According to the section formula: Hence, the required ratio is m : n = 2 : 7.
Since, the given line segment is being divided by the x-axis, therefore the coordinates of the dividing point will be (x, 0). According to the section formula, we can say that: Hence, the line segment AB is being divided in 1 : 1 ratio. This implies that the dividing point on x-axis is the mid-point of AB. Therefore, Hence, the required point is (-3/2, 0).
Let the given points be A (1, 2), B (4, y), C (x, 6) and D (3, 5). Since, ABCD forms a parallelogram, the diagonals AC and BD will bisect each other at (X, Y). Therefore, Hence, x = 6 and y = 3.
Let A be (x The centre of the circle is the mid-point of diameter. Therefore, (2, -3) is mid-point of AB. This implies that: Hence, the point A will be (3, -10).
Therefore, point P divides AB in 3 : 4 ratio. Then, coordinates of P will be: Hence, the coordinates of P are (-2/7, -20/7).
Let P (a, b), Q (x, y) and R (α, β) be the three required points. It is given that P, Q and R divide the line segment AB into four equal parts. This implies that Q will be the mid-point of AB, P will be the mid-point of AQ while R will be the mid-point of BQ.
Therefore, Q is (0, 5).
Therefore, P is (-1, 7/2).
Therefore, R is (1, 13/2). Hence, the required points are (-1, 7/2), (0, 5), and (1, 13/2).
Let the vertices of rhombus be A (3, 0), B (4, 5), C (-1, 4) and D (-2, -1). Area of the given rhombus = ½ × AC × BD = ½ × 4√2 × 6√2 = ½ × 48 = 24 square units. Hence, the area of given rhombus is 24 sq. units. ## Exercise 7.3
**(2, 3), (-1, 0), (2, - 4)****(-5, -1), (3, -5), (5, 2)**
= ½[2(0 - (-4)) - 1(-4 - 3) + 2(3 - 0)] = ½[2(4) - 1(-7) + 2(3)] = ½[8 + 7 + 6] = ½[21] = 21/2 = 10.5 square units
= ½[-5(-5 - 2) + 3(2 - (-1)) + 5(-1 - (-5))] = ½[-5(-7) + 3(3) + 5(4)] = ½[35 + 9 + 20] = ½[64] = 64/2 = 32 square units **(7, -2), (5, 1), (3, k)****(8, 1), (k, - 4), (2, -5).**
Area of triangle formed by the points = ½[x 0 = ½[x 0 = [7(1 - k) + 5(k - (-2)) + 3(-2 - 1)] 0 = [7 - 7k + 5(k + 2) + 3(-3)] 0 = [7 - 7k + 5k + 10 - 9] 0 = [- 2k + 8] -8 = -2k
Area of triangle formed by the points = ½[x 0 = ½[x 0 = [8(-4 - (-5)) + k(-5 - 1) + 2(1 - (-4))] 0 = [8(-4 + 5) + k(-6) + 2(5)] 0 = [8(1) - 6k + 10] 0 = [8 - 6k + 10] 0 = - 6k + 18 -18 = -6k
Let the given points be A (0, -1), B (2, 1) and C (0, 3) with D, E and F being mid-points of AB, BC, and AC respectively. Coordinates of D : D is (1, 0). Coordinates of E : E is (1, 2). Coordinates of F : F is (0, 1). Area of DEF = ½[x = ½[1(2 - 1) + 1(1 - 0) + 0(0 - 2)] = ½[1(1) + 1(1) + 0] = ½[1 + 1 + 0] = ½[2] = 1 square unit Area of ABC = = ½[0(1 - 3) + 2(3 + 1) + 0(-1 - 1)] = ½[0 + 2(4) + 0] = ½[0 + 8 + 0] = ½[8] = 4 square units Ratio of the areas of DEF and ABC = Area of DEF/ Area of ABC = ¼ = 1 : 4 Hence, the area of the triangle formed by joining mid-points of given triangle is 1 square unit and the ratio of the areas of two triangles is 1 : 4.
Let the given quadrilateral's vertices be A (-4, -2), B(-3, -5), C(3, -2), and D(2, 3). We can divide ABCD into two triangles ABC and ADC and find the sum of their areas in order to find the area of ABCD. Area of a triangle = ½[x Area of ABC = ½[-4(-5 - (-2)) - 3(-2 - (-2)) + 3(-2 - (-5))] = ½[-4(-5 + 2) - 3(-2 + 2) + 3(3)] = ½[-4(-3) - 3(0) + 9] = ½[12 + 9] = ½[21] = 21/2 = 10.5 square units Area of ADC = ½[-4(-2 - 3) + 3(3 - (-2)) + 2(-2 - (-2))] = ½[-4(-5) + 3(3 + 2) + 2(-2 + 2)] = ½[20 + 3(5) + 2(0)] = ½[20 + 15] = ½[35] = 35/2 = 17.5 square units Area of ABCD = 10.5 + 17.5 = 28 square units Hence, the area of given quadrilateral is 28 square units.
Let AD be the median of triangle ABC. D will be the mid-point of BC. Therefore, coordinates of D will be D is (4, 0). Area of a triangle = ½[x Area of triangle ADB = ½[4(0 - (-2)) + 4(-2 - (-6)) + 3(-6 - 0)] = ½[4(2) + 4(-2 + 6) + 3(-6)] = ½[8 + 4(4) - 18] = ½[8 + 16 - 18] = ½[6] = 6/2 = 3 square units Area of triangle ADC = ½[4(0 - 2) + 4(2 - (-6)) + 3(-6 - 0)] = ½[4(-2) + 4(2 + 6) + 3(-6)] = ½[-8 + 4(8) - 18] = ½[-8 + 32 - 18] = ½[6] = 6/2 = 3 square units Hence, proved that a median of a triangle divides it into two triangles of equal areas. ## Exercise 7.4 (Optional)
Let the ratio in which AB is divided by 2x + y - 4 = 0 be k : 1. Coordinates of the point of intersection can be obtained as : Since, the point of intersection will also lie on 2x + y - 4 = 0. Therefore, Hence, the ratio in which AB is divided by 2x + y - 4 = 0 is 2 : 9.
Since, it is given that the points are collinear. Therefore, the area of triangle formed by them will be 0. Area of triangle = ½[x 0 = ½[x(2 - 0) + 1(0 - y) + 7(y - 2)] 0 = ½[x(2) + 1(-y) + 7y - 14] 0 = ½[2x - y + 7y - 14] 0 = ½[2x + 6y - 14] 2x + 6y - 14 = 0 2(x + 3y - 7) = 0
Hence, x + 3y - 7 = 0 is the required relation.
Let the centre of the circle be O (x, y) and the three points it passes through be A (6, -6), B (3, -7), and C (3, 3). Since, OA, OB and OC are all radii of the same circle. Therefore, OA = OB = OC From Equation (II), we have: x - 3y = 9 x = 9 + 3y Substitute x = 9 + 3y in Equation (I), 3x + y = 7 3(9 + 3y) + y = 7 27 + 9y + y = 7 10y = -20
x = 9 + 3y = 9 + 3(-2) = 9 - 6
Hence, the centre of the circle lies at (3, -2).
Let the square be ABCD, with A (-1, 2) and C (3, 2). AB = BC as ABCD is a square. AD = DC as ABCD is a square. By applying Pythagoras theorem in ABC, we get AB (x (x (1 + 1) (2) 4 + (y (y (y 2(y (y y y Similarly, we can find that y Since B and D have the same x coordinate, their y coordinates cannot be the same as that would mean the points B and D of the square coincide, which is impossible. Therefore, B and D are (1, 4) and (1, 0) respectively or B and D are (1, 0) and (1, 4) respectively. Hence, the other two vertices of the given square are (1, 4) and (1, 0).
**Taking A as origin, find the coordinates of the vertices of the triangle.****What will be the coordinates of the vertices of ∆ PQR if C is the origin?**
Area of PQR with A as origin = ½[x = ½[4(2 - 5) + 3(5 - 6) + 6(6 - 2)] = ½[4(-3) + 3(-1) + 6(4)] = ½[-12 - 3 + 24] = ½[9] = 9/2 = 4.5 m
Area of PQR with C as origin = ½[x = ½[12(6 - 3) + 13(3 - 2) + 10(2 - 6)] = ½[12(3) + 13(1) + 10(-4)] = ½[36 + 13 - 40] = ½[9] = 9/2 = 4.5 m Observation : Irrespective of the origin, the area of the triangle remains the same.
D and E divide AB and AC is 1 : 3 ratio. Coordinates of D : D is (13/4, 23/4). Coordinates of E : E is (19/4, 5). Area of triangle ADE = ½[x = ½[4(23/4 - 5) + 13/4(5 - 6) + 19/4(6 - 23/4)] = ½[4(23/4 - 20/4) + 13/4(-1) + 19/4(24/4 - 23/4)] = ½[4(3/4) - 13/4 + 19/4(1/4)] = ½[3 - 13/4 + 19/16] = ½[48/16 - 52/16 + 19/16] = ½[15/16] = 15/32 square units Area of triangle ABC = ½[4(5 - 2) + 1(2 - 6) + 7(6 - 5)] = ½[4(3) + 1(-4) + 7(1)] = ½[12 - 4 + 7] = ½[15] = 15/2 square units Ratio of the areas of the two triangles = Area of ADE/ Area of ABC = (15/32)?(15/2) = 1 : 16 Hence, ratio of the area of ADE to ABC is 1 : 16.
**The median from A meets BC at D. Find the coordinates of the point D.****Find the coordinates of the point P on AD such that AP : PD = 2 : 1****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.****What do you observe?**
## [Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]
D is (7/2, 9/2).
P is (11/3, 11/3).
E is (5/2, 3). Now, coordinates of Q: Q is (11/3, 11/3). Similarly, we can find that R is also (11/3, 11/3).
Since, P is the mid-point of AB. Therefore, its coordinates are: P is (-1, 3/2). Similarly, we can find that Q is (2, 4) R is (5, 3/2) and S is (2, -1). Distance between P and Q: Similarly, we can find that This implies that all sides are equal, therefore PQRS is either a rhombus or a square. Therefore, we can conclude that PQRS is a quadrilateral with all sides equal and unequal diagonals. Hence, PQRS is a rhombus. Next TopicClass 10 Maths Chapter 8 |