NCERT Grade 10-Coordinate Geometry-Answers – MySchoolPage

NCERT Grade 10-Coordinate Geometry-Answers

NCERT Solutions for Class 10 Maths

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1.    (i) Applying Distance Formula to find distance between points (2, 3) and (4,1), we get

(ii) Applying Distance Formula to find distance between points (-5, 7) and (-1, 3), we get

(iii) Applying Distance Formula to find distance between points (a, b) and (-a, -b), we get

2.    Applying Distance Formula to find distance between points (0, 0) and (36, 15), we get  Town B is located at 36 km east and 15 km north of town A. So, the location of town A and B can be shown as:

Clearly, the coordinates of point A are (0, 0) and coordinates of point B are (36, 15). To find the distance between them, we use Distance formula:

km

3.    Let A = (1, 5), B = (2, 3) and C = (-2, -11)

Using Distance Formula to find distance AB, BC and CA.

Since AB + AC ≠ BC, BC + AC ≠ AB and AC ≠ BC.

Therefore, the points A, B and C are not collinear.

4.   Let A = (5, -2), B = (6, 4) and C = (7, -2)

Using Distance Formula to find distances AB, BC and CA.

Since AB = BC.

Therefore, A, B and C are vertices of an isosceles triangle.

5.    We have A = (3, 4), B = (6, 7), C = (9, 4) and D = (6, 1)

Using Distance Formula to find distances AB, BC, CD and DA, we get

Therefore, All the sides of ABCD are equal here.        … (1)

Now, we will check the length of its diagonals.

Diagonals of ABCD are also equal.                  … (2)

From (1) and (2), we can definitely say that ABCD is a square. Therefore, Champa is correct.

6.    (i) Let A = (-1, -2), B = (1, 0), C= (-1, 2) and D = (-3, 0)

Using Distance Formula to find distances AB, BC, CD and DA, we get

Therefore, all four sides of quadrilateral are equal.        … (1)

Now, we will check the length of diagonals.

Therefore, diagonals of quadrilateral ABCD are also equal.        … (2)

From (1) and (2), we can say that ABCD is a square.

(ii) Let A = (-3, 5), B= (3, 1), C= (0, 3) and D= (-1, -4)

Using Distance Formula to find distances AB, BC, CD and DA, we get

We cannot find any relation between the lengths of different sides.

Therefore, we cannot give any name to the quadrilateral ABCD.

(iii) Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2)

Using Distance Formula to find distances AB, BC, CD and DA, we get

Here opposite sides of quadrilateral ABCD are equal.         … (1)

We can now find out the lengths of diagonals.

Here diagonals of ABCD are not equal.          … (2)

From (1) and (2), we can say that ABCD is not a rectangle therefore it is a parallelogram.

7.    Let the point be (x, 0) on x-axis which is equidistant from (2, -5) and (-2, 9).

Using Distance Formula and according to given conditions we have:

Squaring both sides, we get

x2 + 4 – 4x + 25 = x2 + 4 + 4x + 81 -4x + 29 = 4x + 85

8x = -56

x = -7

Therefore, point on the x-axis which is equidistant from (2, -5) and (-2, 9) is (-7, 0)

8.   Using Distance formula, we have

Squaring both sides, we get

100 = 73 + y2 + 6y               ⇒        y2 + 6y – 27 = 0

Solving this Quadratic equation by factorization, we can write

⇒ y2 + 9y – 3y – 27 = 0              ⇒                    y (y + 9) -3 (y + 9) = 0

⇒ (y + 9 ) (y – 3) = 0                  ⇒                     y = 3, -9

9.   It is given that Q is equidistant from P and R. Using Distance Formula, we get

Squaring both sides, we get

25 + 16 = x2 + 25                        ⇒                     x2 = 16

x = 4, -4

Thus, Q is (4, 6) or (-4, 6).

Using Distance Formula to find QR, we get

Using Distance Formula to find PR, we get

10.   It is given that (x, y) is equidistant from (3, 6) and (-3, 4).

Using Distance formula, we can write

Squaring both sides, we get

⇒ x2 + 9 – 6x + y2 + 36 – 12y = x2 + 9 + 6x + y2 + 16 – 8y

⇒ -6x – 12y + 45 = 6x – 8y + 25

⇒ 12x + 4y = 20

⇒ 3x + y = 5

11.  Let x1 = -1, x2= 4, y1 = 7 and y2 = -3, m1 = 2 and m2 = 3

Using Section Formula to find coordinates of point which divides join of (-1, 7) and (4, -3) in the ratio 2:3, we get

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

12.   

We want to find coordinates of the points of trisection of the line segment joining (4, – 1) and (-2, -3).

We are given AC = CD = DB

We want to find coordinates of point C and D.

Let coordinates of point C be (x1, y1) and let coordinates of point D be (x2, y2).

Clearly, point C divides line segment AB in 1:2 and point D divides line segment AB in 2:1.

Using Section Formula to find coordinates of point C which divides join of (4, -1) and (-2, -3) in the ratio 1:2, we get

Using Section Formula to find coordinates of point D which divides join of (4, -1) and (-2, -3) in the ratio 2:1, we get

Therefore, coordinates of point C are  and coordinates of point D are

13.  Niharika runs 14th of the distance AD on the 2nd line and posts a green flag.

There are 100 flower pots. It means, she stops at 25th flower pot.

Therefore, the coordinates of point where she stops are (2 m, 25 m).

Preet runs 15th of the distance AD on the eighth line and posts a red flag. There are 100 flower pots. It means, she stops at 20th flower pot.

Therefore, the coordinates of point where she stops are (8, 20).

Using Distance Formula to find distance between points (2 m, 25 m) and (8 m, 20 m), we get

Rashmi posts a blue flag exactly halfway the line segment joining the two flags.

Using section formula to find the coordinates of this point, we get

Therefore, coordinates of point, where Rashmi posts her flag are 

It means she posts her flag in 5th line after covering = 22.5 m of distance.

14.  Let (-1, 6) divides line segment joining the points (-3, 10) and (6, -8) in k:1.

Using Section formula, we get

Therefore, the ratio is  :1 which is equivalent to 2:7.

Therefore, (-1, 6) divides line segment joining the points (-3, 10) and (6, -8) in 2:7.

15.  Let the coordinates of point of division be (x, 0) and suppose it divides line segment joining A (1, -5) and B (-4, 5) in k:1.

According to Section formula, we get

Therefore, point  on x-axis divides line segment joining A (1, -5) and B (-4, 5) in 1:1.

16.  Let A= (1, 2), B= (4, y), C= (x, 6) and D = (3, 5)

We know that diagonals of parallelogram bisect each other. It means that coordinates of midpoint of diagonal AC would be same as coordinates of midpoint of diagonal BD.          …(1)

Using Section formula, the coordinates of midpoint of AC are:

Using Section formula, the coordinates of midpoint of BD are:

According to condition (1), we have

Again, according to condition (1), we also have

Therefore, x = 6 and y = 3

17.   We want to find coordinates of point A. AB is the diameter and coordinates of center are (2, -3) and, coordinates of point B are (1, 4).

Let coordinates of point A are (x, y). Using section formula, we get

Using section formula, we get

Therefore, Coordinates of point A are (3, -10).

18.   A = (-2, -2) and B = (2, -4)

It is given that AP =  AB

PB = AB – AP = AB

So, we have AP : PB = 3 :4

Let coordinates of P be (x, y)

Using Section formula to find coordinates of P, we get

Therefore, Coordinates of point P are 

19.  A = (-2, 2) and B = (2, 8)

Let P, Q and R are the points which divide line segment AB into 4 equal parts.

Let coordinates of point P = (x1, y1), Q = (x2, y2) and R = (x3, y3)

We know AP = PQ = QR = RS.

It means, point P divides line segment AB in 1:3.

Using Section formula to find coordinates of point P, we get

Since, AP = PQ = QR = RS.

It means, point Q is the mid-point of AB.

Using Section formula to find coordinates of point Q, we get

20. Let A = (3, 0), B = (4, 5), C = (-1, 4) and D = (-2, -1)

Using Distance Formula to find length of diagonal AC, we get

Distance Formula to find length of diagonal BD, we get

∵ Area of rhombus = 1/2 (product of its diagonals)

21.  (i) (2, 3), (-1, 0), (2, -4)

Area of Triangle = 

= 1/2 [2 {0 – (-4)} – 1 (-4 – 3) + 2 (3 – 0)]

= 1/2 [2 (0 + 4) – 1 (-7) + 2 (3)] = 1/2 (8 + 7 + 6) =  sq. units

(ii) (-5, -1), (3, -5), (5, 2)

Area of Triangle = 

= 1/2 [-5 (-5 – 2) + 3 {2 – (-1)} + 5 {-1 – (-5)}]

1/2 [-5 (-7) + 3 (3) + 5 (4)] = 1/2 (35 + 9 + 20) = 1/2 (64) = 32 sq. units

22.  (i) (7, -2), (5, 1), (3, k)

Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.

Area of Triangle = = 0

⇒ 1/2 [7 (1 – k) + 5 {k – (-2)} + 3 (-2 – 1)] = 1/2 (7 – 7k + 5k + 10 – 9) = 0

⇒ 1/2 (7 – 7k + 5k + 1) = 0

⇒ 1/2 (8 – 2k) = 0 8 – 2k = 0

⇒ 2k = 8

⇒ k = 4

(ii) (8, 1), (k, -4), (2, -5)

Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.

Area of Triangle = = 0

⇒ 1/2 [8 {-4 – (-5)} + k (-5 – 1) + 2 {1 – (-4)}] = 1/2 (8 – 6k + 10) = 0

⇒ 1/2 (18 – 6k) = 0

⇒ 18 – 6k = 0

⇒ 18 = 6k

⇒ k = 3

23.  Let A = (0, -1) = (x1, y1), B = (2, 1) = (x2, y2) and C = (0, 3) = (x3, y3)

⇒ Area of ΔABC

= 1/2[0 (1 – 3) + 2 {3 – (-1)} + 0 (-1 – 1)] =  1/2 x 8

= 4 sq. units

P, Q and R are the mid-points of sides AB, AC and BC respectively.

Applying Section Formula to find the vertices of P, Q and R, we get

Applying same formula, Area of ΔPQR = 1/2 [1 (1 – 2) + 0 (2 – 0) + 1 (0 – 1)] = 1/2 |-2|

= 1 sq. units (numerically)

24.  Area of Quadrilateral ABCD = Area of Triangle ABD + Area of Triangle BCD ,.. (1)

Using formula to find area of triangle: Area of Δ ABD

=

= 1/2[-4 (-5 – 3) – 3 {3 – (-2)} + 2 {-2 – (-5)}]

= 1/2 (32 – 15 + 6) = 1/2 (23) = 11.5 sq units   … (2)

Again using formula to find area of triangle:

Area of Δ BCD = 

= 1/2 [-3 (-2 – 3) + 3 {3 – (-5)} + 2 {-5 – (-2)}]

= 1/2 (15 + 24 – 6) = 1/2 (33) = 16.5 sq units

Putting (2) and (3) in (1), we get

Area of Quadrilateral ABCD = 11.5 + 16.5 = 28 sq units

25.  We have, ΔABC whose vertices are given. We need to show that ar(ΔABD) = ar(ΔACD). Let coordinates of point D are (x, y)

Using section formula to find coordinates of D, we get

Therefore, coordinates of point D are (4, 0)

Using formula to find area of triangle:

Area of ΔABD = 

= 1/2 [4 (-2 – 0) + 3 {0 – (-6)} + 4 {-6 – (-2)}]

= 1/2 (-8 + 18 -16) = 1/2 (-6) = -3 sq units

Area cannot be in negative.

Therefore, we just consider its numerical value.

Therefore, area of ΔABD = 3 sq units      … (1)

Again using formula to find area of triangle:

Area of AACD = 

= 1/2 [4 (2 – 0) + 5 {0 – (-6)} + 4 {-6 -2 )}]

= 1/2 (8 + 30 – 32) = 1/2 (6) = 3 sq units          … (2)

From (1) and (2), we get ar(ΔABD) = ar(ΔACD) Hence Proved

26.  Let 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 

But C lies on 2x + y – 4 = 0, therefore 

⇒ 6k + 4 + 7k – 2 – 4k – 4 = 0

⇒ 9k – 2 = 0

Hence, the required ratio if 2 : 9 internally.

27.  The points A (x, y), B (1, 2) and C (7, 0) will be collinear if

Area of triangle = 0

28.  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.

Taking               AP = BP

AP2 = BP2

⇒ (x- 6)2 + (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)

Again, taking BP = CP

⇒ BP2 = CP2

⇒ (x-3)2 +(y+ 7)2 = (x-3)2 + (y -3)2

⇒ x2 – 6x + 9 + y2 + 14y +4 9 = x2 – 6x + 9 + y2 – 6y + 9

⇒ -6x + 6x + 14y + 6y + 58 – 18 = 0

⇒ 20y + 40 = 0

⇒ y = -2

Putting the value of y in eq. (i),

3x + y – 7 = 0            ⇒                3x = 9               ⇒             x = 3

Hence, the centre of the circle is (3,-2).

29.  Let ABCD be a square and B (x, y) be the unknown vertex. AB = BC

⇒ AB2 = BC2

⇒ (x+1)2 +(y — 2)2 = (x-3)2 +(y— 2)2

⇒ 2x + 1 = -6x + 9

⇒ 8x = 8

⇒ x = 1 …………………….. (i)

In Δ ABC,              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 + 2y2 — 4x —8y + 2 = 0

⇒ x2 +y2 —2x-4y+1= …………………… (ii)

Putting the value of x in eq. (ii),

1 + y– 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).

30.  (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.

31.  Since,   

∴ DE || BC            [By Thales theorem]

∴ Δ ADE – Δ ABC

Area (Δ ADE) : Area (Δ ABC) = 1 : 16

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

(i) Since AD is the median of Δ ABC.

∴ D is the mid-point of BC.

∴ Its coordinates are 

(ii) Since P divides AD in the ratio 2 : 1

∴ Its coordinates are 

(iii) Since BE is the median of Δ ABC

∴ E is the mid-point of AD.

∴ Its coordinates are 

Since Q divides BE in the ratio 2 : 1.

∴ Its coordinates are 

Since CF is the median of Δ ABC.

∴ F is the mid-point of AB.

∴ Its coordinates are 

Since R divides CF in the ratio 2 : 1.

∴ Its coordinates are 

(iv) We observe that the points P, Q and R coincide, i.e., the medians AD, BE and CF are concurrent at the point  This point is known as the centroid of the triangle.

(v) According to the question, D, E, and F are the mid-points of BC, CA and AB respectively.

∴ Coordinates of D are 

Coordinates of a point dividing AD in the ratio 2 : 1 are

The coordinates of E are 

∴ The coordinates of a point dividing BE in the ratio 2 : 1 are

Similarly the coordinates of a point dividing CF in the ratio 2 : 1 are

Thus, the point  is common to AD, BE and CF and divides them in the ratio 2 : 1.

∴ The median of a triangle are concurrent and the coordinates of the centroid are  

33.  Use distance formula, 

 

Since all the sides are equal but the diagonals are not equal.

∴ PQRS is a rhombus.

 

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