# Class 10 RD Sharma Solutions – Chapter 14 Coordinate Geometry – Exercise 14.3 | Set 3

### Question 41. Three consecutive vertices of a parallelogram are (-2, -1), (1, 0) and (4, 3). Find the fourth vertex.

**Solution:**

Let the coordinates of three vertices are A (-2, -1), B (1, 0), and C (4, 3)

And let the diagonals AC and BD bisect each other at O

As O is the mid-point of AC

Therefore,

Vertices of O will be

Assume coordinates of the forth vertex D be (x, y)

As O is the mid-point of BD

Thus, coordinates of O will be

Therefore(1 + x)/2 = 1

1 + x = 2

x = 1

and

y/2 = 1

y = 2

Hence, the co-ordinates of D will be (1, 2).

### Question 42. The points (3, -4) and (-6, 2) are the extremities of a diagonal of a parallelogram. If the third vertex is (-1, -3). Find the co-ordinates of the fourth vertex.

**Solution:**

Assume the edges of a diagonal AC of a parallelogram ABCD are A (3, -4) and C (-6, 2)

Consider AC and BD bisect each other at O.

Therefore,

Mid-point of AC will be

Assume the fourth vertex of the parallelogram be (x, y)

Therefore,

Mid-point of BD will be

-1 + x = -3

x = -2

and

(-3 + y)/2 = -1

-3 + y = -2

y = -2 + 3 = 1

Hence, the coordinates of D are (-2, 1).

### Question 43. If the co-ordinates of the mid-points of the sides of a triangle are (1, 1), (2, -3), and (3, 4), find the vertices of the triangle.

**Solution:**

Assume A (x

_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) be the vertices of the ∆ABCD, E and F are the mid-points of BC, CA and AB respectively such that their co-ordinates are D (1, 1), E (2, -3) and F (3, 4)

D is mid-point of BC

Therefore,

x

_{2 }+ x_{3 }= 2and

y

_{2 }+ y_{3 }= 2Similarly, E is the mid-point of AC

Therefore,

x

_{3 }+ x_{1 }= 4and

y

_{3 }+ y_{1 }= -6And

F is the mid-point of AB

Therefore,

x

_{2 }+ x_{1 }= 6and

y

_{2 }+ y_{1 }= 8Now,

x

_{1}+ x_{2}= 6 …………..(i)x

_{2}+ x_{3}= 2 ……………(ii)x

_{3}+ x_{1}= 4 …………….(iii)On adding we will get

2(x

_{1}+ x_{2}+ x_{3}) = 12x

_{1}+ x_{2}+ x_{3}= 6 …………(iv)On subtracting (ii), (iii) and (i) from (iv), we get

x

_{1}= 4, x_{2}= 2, x_{3}= 0Similarly

y

_{1}+ y_{2}= 8 ……….(v)y

_{2}+ y_{3}= 2 ……….(vi)y

_{3}+ y_{1}= -6 ………(vii)On adding we will get

2(y

_{1}+ y_{2}+ y_{3}) = 4y

_{1}+ y_{2}+ y_{3}= 2 ………(viii)On subtracting (vi), (vii) and (v) from (viii), we get

y

_{1}= 0y

_{2}= 8y

_{3}= -6Hence, the vertices of ∆ABC are A (4, 0), B(2, 8), C(0, -6)

### Question 44. Determine the ratio in which the straight line x – y – 2 = 0 divides the line segment joining (3, -1) and (8, 9).

**Solution:**

Assume the straight line x – y – 2 = 0 divides the line segment joining the points (3, -1), (8, 9) in the ratio m : n

Co-ordinates of the point will be

and

This point (x, y) lies on the line on the line x – y – 2 = 0

⇒ (8m + 3n) – (9m – n) – 2(m + n) = 0

⇒ 8m + 3n – 9m + n – 2m – 2n = 0

⇒ -3m + 2n = 0

⇒ 2n = 3m

Hence, the ratio = 2 : 3 internally

### Question 45. Three vertices of a parallelogram are (a + b, a – b), (2 a + b, 2a – b), (a – b, a + b). Find the fourth vertex.

**Solution:**

In parallelogram ABCD co-ordinates are of A (a + b, a – b), B (2a + b, 2a – b), C (a – b, a + b)

Assume coordinates of D be (x, y)

Join diagonal AC and BD

Which bisect each other at O

O is the mid-point of AC as well as BD

If O is the mid-point of AC, Then its coordinates will be

and

If O is mid-point of BD, then coordinates will be

x + 2a + b = 2a

x = 2a – 2a – b = -b

and

y + 2a – b = 2a

y = 2a – 2a + b = b

Hence, the coordinates of D will be (-b, b).

### Question 46. If two vertices of a parallelogram are (3, 2), (-1, 0) and the diagonals cut at (2, -5), find the other vertices of the parallelogram.

**Solution:**

Two vertices of a parallelogram ABCD are A (3,2), and B (-1, 0) and its diagonals bisect each other at O (2, -5)

Assume the coordinates of C be (x

_{1}, y_{1}) and of D be (x_{2}, y_{2})If O is the mid-point of AC, then

3 + x

_{1 }= 4x

_{1 }= 4 – 3 = 1and

and

2 + y

_{1}= -10y

_{1 }= -10 – 2 = -12Therefore, the coordinates of C will be (1, -12)

Again if O is mid-point of BD then

-1 + x

_{2 }= 4x

_{2 }= 4 + 1 = 5and

y

_{2 }= -10Hence, the coordinates of D will be (5, -10).

### Question 47. If the coordinates of the mid-points of the sides of a triangle ar6 (3, 4), (4, 6), and (5, 7), find its vertices.

**Solution:**

The coordinates of the mid-points of the sides BC, CA and AB are D (3, 4), E (4, 6) and F (5, 7) of the ∆ABC.

Assume the coordinates of the vertices of the triangle be A (x,

_{1}y_{1}), B (x_{2}, y_{2}), and C (x_{3}, y_{3}).Now the coordinates of D will be

x

_{2 }+ x_{3 }= 6and

y

_{2}+ y_{3}= 4y

_{2}+ y_{3}= 8Similarly, the coordinates of E will be

x

_{3 }+ x_{1 }= 8and

and

y

_{1 }+ y_{2 }= 14Now,

x

_{2}+ x_{3}= 6 …….(i)x

_{3}+ x_{1}= 8 ……..(ii)x

_{1}+ x_{2}= 10 ……..(iii)On adding we will get

2(x

_{1}+ x_{2}+ x_{3}) = 24⇒ x

_{1}+ x_{2}+ x_{3}= 24/2 = 12 ………..(iv)On subtracting each from (iv),

We will get

x

_{1}= 6, x_{2}= 4 and x_{3}= 2Similarly,

y

_{2}+ y_{3}= 8 ……..(v)y

_{3}+ y_{1}= 12 ………(vi)y

_{1}+ y_{2}= 14 ……..(vii)On Adding, we will get

2(y

_{1}+ y_{2}+ y_{3}) = 34⇒ y

_{1}+ y_{2}+ y_{3}= 34/2 = 17 ……..(viii)On subtracting each from (viii),

We will get

y

_{1}= 9y

_{2}= 5y

_{3}= 3Hence, the coordinates will be of A (6, 9), B (4, 5) and C (2, 3)

### Question 48. The line segment joining the points P (3, 3) and Q (6, -6) is trisected at the points A and B such that A is nearer to P. If A also lies on the line given by 2x+ y + k =0, find the value of k.

**Solution:**

Two points A and B trisect the line segment joining the points P (3, 3) and Q (6, -6) and A is nearer to P and A lies also on the line 2x + y + k = 0

Now,

A divides the line segment PQ in the ratio of 1 : 2

i.e.,

PA = AQ = 1 : 2

Assume coordinates of A be (x, y), then

and

Therefore,

Coordinates of A are (4, 0)

As A lies on the line 2x + y + k = 0

Hence,

It will satisfy it

2 × 4 + 0 + k = 0

8 + k = 0

k = -8

### Question 49. If three consecutive vertices of a parallelogram are (1, -2), (3, 6), and (5, 10), find its fourth vertex.

**Solution:**

A (1, -2), B (3, 6) and C (5, 10) are the three consecutive vertices of the parallelogram ABCD

Assume (x, y) be its fourth vertex

AC and BD are its diagonals which bisect each other at O

As O is the mid-point of AC

Therefore,

Coordinates of O will be

On comparing,

3 + x = 3

3 + x = 6

x = 3

and

(6 + y)/2 = 4

6 + y = 8

y = 2

Hence, the coordinates of fourth vertex D are (3, 2).

### Question 50. If the points A (a, -11), B (5, b), C (2, 15), and D (1, 1) are the vertices of a parallelogram ABCD, find the values of a and b.

**Solution:**

A (a, -11), B (5, b), C (2, 15) and D (1, 1) are the vertices of a parallelogram ABCD

Diagonals AC and BD bisect each other at O

As O is the mid-point of AC

Therefore,

Coordinates of O will be

Similarly, O is the mid-point of BD also

Therefore,

Coordinates of O will be

(2 + a)/2 = 3

b + 1 = 4

b = 3

Hence,

a = 4

b = 3

### Question 51. If the co-ordinates of the mid-points of the sides of a triangle be (3, -2), (-3, 1), and (4, -3), then find the coordinates of its vertices.

**Solution:**

In a ∆ABC,

D, E and F are the mid-points of the sides BC, CA and AB respectively and co-ordinates of D, E and F are (3, -2), (-3, 1) and (4, -3) respectively.

Assume the coordinates of A are (x

_{1}, y_{1}), of B are (x_{2}, y_{2}) and of C are (x_{3}, y_{3})Now,

As D is the mid-point BC

Therefore,

and

As E is the mid-point of CA

Therefore,

and

and F is the mid-point of AB

Therefore,

and

Now,

x

_{2}+ x_{3}= 6 ……..(i)x

_{3}+ x_{1}= -6 ……..(ii)x

_{1}+ x_{2}= 8 ………(iii)On adding we get

2(x

_{1}+ x_{2}+ x_{3}) = 8⇒ x

_{1}+ x_{2}+ x_{3}= 4 …….(iv)On subtracting from (iv) we get

x

_{1}= -2,x

_{2}= 10,x

_{3}= -4and

y

_{2}+ y_{3}= -4 …….(v)y

_{3}+ y_{1}= 2 …….(vi)y

_{1}+ y_{2}= -6 ………(vii)On Adding, We will get

2(y

_{1}+ y_{2}+ y_{3}) = -8⇒ y

_{1}+ y_{2}+ y_{3}= -4 ……..(viii)On subtracting from (viii) we get

y

_{1}= 0,y

_{2}= -6,y

_{3}= 2Hence, the coordinates of A are (-2, 0) of B are (10, -6) and of C (-4, 2).

### Question 52. The line segment joining the points (3, -4) and (1, 2) is trisected at the points P and Q. If the co-ordinates of P and Q are (p, -2) and (53 , q) respectively, find the values of p and q.

**Solution:**

Assume line AB whose ends points are A (3, -4) and B (1, 2)

Coordinates of P and Q which trisect AB are P (p, -2) and Q

Now,

As P divides AB in the ratio 1 : 2

Therefore,

Coordinates of P will be

Similarly, Q divides AB in the ratio 2 : 1

Hence,

p = 7/3, q = 0

### Question 53. The line joining the points(2, 1), (5, -8) is trisected at the points P and Q. If point P lies on the line 2x – y + k = 0, find the value of k.

**Solution:**

Points A (2, 1), and B (5, -8) are the ends points of the line segment AB

Points P and Q trisect it and P lies on the line 2x – y + k = 0

As P divides AB in the ratio of 1 : 2

Therefore,

Coordinates of P will be

Therefore,

Coordinates of P are (3, -2)

As it lies on 2x – y + k = 0

Therefore,

It will satisfy it

2 × 3 – (-2) + k = 0

⇒ 6 + 2 + k = 0

⇒ 8 + k = 0

⇒ k = -8

### Question 54. A (4, 2), B (6, 5), and C (1, 4) are the vertices of ∆ABC,

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

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

**(iii) Find the coordinates of the 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? **

**Solution:**

In ∆ABC, co-ordinates of A (4, 2) of (6, 5) and of (1, 4) and AD is BE and CF are the medians

such that D, E and F are the mid-points of the sides BC, CA and AB respectively

P is a point on AD such that AP : PD = 2 : 1

(i)Now coordinates of D will be

(ii)As P divides AD in the ratio of 2 : 3Therefore,

Coordinates of P will be

Thus,

Coordinates of P are

(iii)As E and F are the mid-point if CA and AB respectivelyCoordinates of E will be

and of F will be

As Q and R divides BE and CF in such a way that BQ : QE = 2 : 1 and CR : RF = 2 : 1

Therefore,

Coordinates of Q will be

and

i.e., Q

_{1}is (11/3, 11/3)and similarly the coordinates of R will be

and

R is (11/3, 11/3)

(iv)We can see that coordinates of P, Q and R are samei.e., P, Q and R coincides each other.

Medians of the sides of a triangle pass through the same point which is called the centroid of the triangle.

### Question 55. If the points A (6, 1), B (8, 2), C (9, 4), and D (k, p) are the vertices of a parallelogram taken in order, then find the values of k and p.

**Solution:**

The diagonals of a parallelogram bisect each other

O is the mid-point of AC and also of BD

O is the mid-point of AC

Therefore,

Coordinates of O will be

As O is also the mid-point of BD

Therefore,

and

⇒ 8 + k = 15

⇒ k = 15 – 8 = 7

and

⇒ 2 + p = 5

⇒ p = 5 – 2 = 3

Hence,

k = 7, p = 3

### Question 56. A point P divides the line segment joining the points A (3, -5) and B (-4, 8) such that . If P lies on the line x + y = 0, then find the value of k.

**Solution:**

Point P divides the line segment by joining the points A (3, -5) and B (-4, 8)

Such that

⇒ AP : PB = k : 1

Assume coordinates of P be (x, y), then

and

As x + y = 0

Therefore,

4k – 2 = 0

4k = 2

k = 2/4

k = 1/2

### Question 57. The mid-point P of the line segment joining the points A (-10, 4) and B (-2, 0) lies on the line segment joining the points C (-9, -4) and D (-4, y). Find the ratio in which P divides CD. Also, find the value of y.

**Solution:**

P is the mid-point of line segment joining the points A (-10, 4) and B (-2, 0)

Coordinates of P will be

P lies on CD also,

Let P divides C (9, 4) and D (-4, y) in the ratio m

_{1}: m_{2}Therefore,

⇒ -6m

_{1}– 6m_{2}= -4m_{1}– 9m_{2}⇒ -6m

_{1}+ 4m_{1}= -9m_{2}+ 6m_{2}⇒ -2m

_{1}= -3m_{2}⇒

Therefore,

m

_{1}: m_{2}= 3 : 2and

⇒ 10 = 3y – 8x

⇒ 3y = 10 + 8 = 18

⇒ y = 18/3 = 6

⇒ y = 6

### Question 58. If the point C (-1, 2) divides internally the line segment joining the points A (2, 5) and B (x, y) in the ratio 3 : 4, find the value of x^{2} + y^{2}.

**Solution:**

As we know that if a point (x, y) divides the line segment joining

the points (x

_{1}, y_{1}) and (x,_{2}y_{2}) in the ration m : n, thenand

Now here, C (-1, 2) divides the line segment joining A (2, 5) and B (x, y) in the ratio of 3 : 4

Now,

and

⇒ 3x + 8 = -7

and

⇒ 20 – 3y = 14

⇒ x = -5

and

⇒ y = 2

Now,

x

^{2}+ y^{2}= (-5)^{2}+ (2)^{2}x

^{2}+ y^{2}= 25 + 4 = 29

### Question 59. ABCD is a parallelogram with vertices A (x_{1}, y_{1}), B (x_{2}, y_{2}), and C (x_{3}, y_{3}). Find the coordinates of the fourth vertex D in terms of x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, and y_{3 }

**Solution:**

Assume the coordinates of D be (x, y). We know that diagonals of a parallelogram bisect each other.

Hence,

Mid-point of AC = Mid-point of BD

i.e., x

_{1}+ x_{3}= x_{2}+ x and y_{1}+ y_{3}= y_{2}+ yi.e, x

_{1}+ x_{3}– x_{2}= x and y_{1}+ y_{3}+ y_{2}= yHence, the coordinates of D are (x

_{1}+ x_{3}– x_{2}, y_{1}+ y_{3}+ y_{2})

### Question 60. The points A (x_{1}, y_{1}), B (x_{2}, y_{2}), and C (x_{3}, y_{3}) are the vertices of ∆ABC.

**(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 points of coordinates Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.**

**(iv) What are the coordinates of the centroid of the triangle ABC? **

**Solution:**

Given: The points A (x

_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) are the vertices of ∆ABC.

(i)We know that, the median bisect the line segment into two equal parts i.e., here D is the mid-point of BC.Therefore,

Coordinates of mid-point of BC =

D =

(ii)Assume the coordinates of a point P be (x, y)Given that, the point P (x, y), divide the line joining A (x

_{1,}y_{1}) andD in the ration of 2 : 1,

Then the coordinates of P

Using internal section formula, we get

Therefore,

Required coordinates of points P =

(iii)Assume the coordinates of a point Q be (p, q)Given: The point Q (p, q) divide the line joining B(x

_{2,}y_{2}) and EIn the ratio of 2 : 1

Then the coordinates of Q

[Since, BE is the median of side CA. So, BE divides AC into two equal parts]

Therefore,

Mid-point of AC = Coordinates of E

⇒ E =

So, the required coordinate of point Q =

Now, let us considered that the coordinates of a point E be (α, β).

It is given that, the point R (α, β), divide the line joining C (x

_{3}, y_{3}) andF in the ration 2 : 1,

So, the coordinates of R be

Here, CF is the median of side AB, hence CF divides AB into two equal parts

Hence,

Mid-point of AB = Coordinates of CF

⇒ F

So, required coordinate of point R

(iv)Coordinate of the centroid of the ∆ABC = (sum of abscissa of all vertices/3, sum of all vertices/2)

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