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

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  • Last Updated : 02 Nov, 2022
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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

=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{-2+4}{2},\ \frac{-1+3}{2}\right)\\ =\left(\frac{2}{2},\ \frac{2}{2}\right)\\ =(1,\ 1)

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

As O is the mid-point of BD

Thus, coordinates of O will be 

\left(\frac{1+x}{2},\ \frac{0+y}{2}\right)\\ =\left(\frac{1+x}{2},\ \frac{y}{2}\right)

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

=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{3-6}{2},\ \frac{-4+2}{2}\right)\\ =\left(\frac{-3}{2},\ \frac{-2}{2}\right)\\ =(\frac{-3}{2},\ -1)

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

Therefore,

Mid-point of BD will be 

=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{-1+x}{2},\ \frac{-3+y}{2}\right)\\ =\frac{-1+x}{2}=\frac{-3}{2}

-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 (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of the ∆ABC

D, 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,

\frac{x_2+x_3}{2}=1

x2 + x3 = 2

and

\frac{y_2+y_3}{2}=1

y2 + y3 = 2

Similarly, E is the mid-point of AC

Therefore,

\frac{x_3+x_1}{2}=2

x3 + x1 = 4

and

\frac{y_3+y_1}{2}=-3

y3 + y1 = -6

And

F is the mid-point of AB

Therefore,

\frac{x_2+x_1}{2}=3

x2 + x1 = 6

and

\frac{y_2+y_1}{2}=4

y2 + y1 = 8

Now,

x1 + x2 = 6  …………..(i)

x2 + x3 = 2  ……………(ii)

x3 + x1 = 4  …………….(iii)

On adding we will get

2(x1 + x2 + x3) = 12

x1 + x2 + x3 = 6  …………(iv)

On subtracting (ii), (iii) and (i) from (iv), we get

 x1 = 4, x2 = 2, x3 = 0

Similarly

y1 + y2 = 8  ……….(v)

y2 + y3 = 2  ……….(vi)

y3 + y1 = -6  ………(vii)

On adding we will get 

2(y1 + y2 + y3) = 4

y1 + y2 + y3 = 2  ………(viii)

On subtracting (vi), (vii) and (v) from (viii), we get

y1 = 0

y2 = 8

y3 = -6

Hence, 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

x=\frac{mx_2+nx_1}{m+n}\\ =\frac{m\times8+n\times3}{m+n}\\ =\frac{8m+3n}{m+n}

and

y=\frac{my_2+ny_1}{m+n}\\ =\frac{m\times9+n(-1)}{m+n}\\ =\frac{9m-n}{m+n}

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

\frac{m}{n}=\frac{2}{3}

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

=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{a+b+a-b}{2},\ \frac{a-b+a+b}{2}\right)\\ =\left(\frac{2a}{a},\ \frac{2a}{a}\right)\\ =(a,\ a)

and 

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

\left(\frac{x+2a+b}{2},\ \frac{y+2a-b}{2}\right)\\ \therefore\frac{x+2a+b}{2}=a

x + 2a + b = 2a

x = 2a – 2a – b = -b

and

\frac{y+2a-b}{2}=a

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 (x1, y1) and of D be (x2, y2)

If O is the mid-point of AC, then

x=\frac{x_1+x_2}{2}

3 + x1 = 4

x1 = 4 – 3 = 1

and 

y=\frac{y_1+y_2}{2}

2=\frac{3+x_1}{2}

and

-5=\frac{2+y_1}{2}

2 + y1 = -10

y1 = -10 – 2 = -12

Therefore, the coordinates of C will be (1, -12)

Again if O is mid-point of BD then

2=\frac{-1+x_2}{2}

-1 + x2 = 4

x2 = 4 + 1 = 5

and

-5=\frac{0+y_2}{2}

y2 = -10

Hence, 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 y1), B (x2, y2), and C (x3, y3).

Now the coordinates of D will be

\left(\frac{x_2+x_3}{2},\ \frac{y_2+y_3}{2}\right)\\ \therefore\frac{x_2+x_3}{2}=3

x2 + x3 = 6

and

y2 + y3 = 4

y2 + y3 = 8

Similarly, the coordinates of E will be

\frac{x_3+x_1}{2}=4

 x3 + x1 = 8

and

\frac{y_3+y_1}{2}=6\\ \Rightarrow y_3+y_1=12\\ \frac{x_1+x_2}{2}=5\\ \Rightarrow x_1+x_2=10

and

\frac{y_1+y_2}{2}=7

y1 + y2 = 14

Now,

x2 + x3 = 6  …….(i)

x3 + x1 = 8  ……..(ii)

x1 + x2 = 10  ……..(iii)

On adding we will get

2(x1 + x2 + x3) = 24

⇒ x1 + x2 + x3 = 24/2 = 12  ………..(iv)

On subtracting each from (iv), 

We will get

x1 = 6, x2 = 4 and x3 = 2

Similarly,

y2 + y3 = 8   ……..(v)

y3 + y1 = 12  ………(vi)

y1 + y2 = 14   ……..(vii)

On Adding, we will get

2(y1 + y2 + y3) = 34

⇒ y1 + y2 + y3 = 34/2  = 17  ……..(viii)

On subtracting each from (viii),

We will get

y1 = 9

y2 = 5

y3 = 3

Hence, 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

x=\frac{m_1x_2+m_2x_1}{m_1+m_2}\\ =\frac{1\times6+2\times3}{1+2}\\ =\frac{6+6}{3}\\ =\frac{12}{3}\\ =4

and

y=\frac{m_1y_2+m_2y_1}{m_1+m_2}\\ =\frac{1\times(-6)+2\times3}{1+2}\\ =\frac{-6+6}{3}\\ =\frac{0}{3}\\ =0

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

\left(\frac{3+x}{2},\ \frac{6+y}{2}\right)

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

\left(\frac{2+a}{2},\ \frac{-11+15}{2}\right)\\ \left(\frac{2+a}{2},\ \frac{4}{2}\right)\\ \left(\frac{2+a}{2},\ 2\right)

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 (x1, y1), of B are (x2, y2) and of C are (x3, y3)

Now,

As D is the mid-point BC

Therefore,

\frac{x_2+x_3}{2}=3\\ \Rightarrow x_3+x_3=6

and 

\frac{y_2+y_3}{2}=-2\\ \Rightarrow y_2+y_3=-4

As E is the mid-point of CA

Therefore,

-3=\frac{x_3+x_1}{2}\\ \Rightarrow x_3+x_1=-6

and

1=\frac{y_3+y_1}{2}\\ \Rightarrow y_3+y_1=2

and F is the mid-point of AB

Therefore,

\frac{x_1+x_2}{2}=4\\ \Rightarrow x_1+x_2=8

and

\frac{y_1+y_2}{2}=-3\\ \Rightarrow y_1+y_2=-6

Now,

x2 + x3 = 6  ……..(i)

x3 + x1 = -6  ……..(ii)

x1 + x2 = 8   ………(iii)

On adding we get

2(x1 + x2 + x3) = 8

⇒ x1 + x2 + x3 = 4   …….(iv)

On subtracting from (iv) we get

x1 = -2, 

x2 = 10,

x3 = -4

and

y2 + y3 = -4   …….(v)

y3 + y1 = 2   …….(vi)

y1 + y2 = -6  ………(vii)

On Adding, We will get

2(y1 + y2 + y3) = -8

⇒ y1 + y2 + y3 = -4   ……..(viii)

On subtracting from (viii) we get

y1 = 0,

y2 = -6,

y3 = 2

Hence, 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 (\frac{5}{3},\ q)

Now,

As P divides AB in the ratio 1 : 2

Therefore,

Coordinates of P will be

p=\frac{mx_2+nx_1}{m+n}\\ =\frac{1\times1+2\times3}{1+2}\\ =\frac{1+6}{3}\\ =\frac{7}{3}

Similarly, Q divides AB in the ratio 2 : 1

q=\frac{my_2+ny_1}{m+n}\\ =\frac{2\times2+1\times(-4)}{2+1}\\ =\frac{4-4}{3}\\ =0

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

\left(\frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n}\right)\\ =\left(\frac{1\times5+2\times2}{1+2},\ \frac{1\times(-8)+2\times1}{1+2}\right)\\ =\left(\frac{5+4}{3}=\frac{9}{3}=3,\ \frac{-8+2}{3}=\frac{-6}{3}=-2\right)

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

\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ \left(\frac{6+1}{2},\ \frac{5+4}{2}\right)\\ \left(\frac{7}{2},\ \frac{9}{2}\right)

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

Therefore,

Coordinates of P will be

x=\frac{mx_2+nx_1}{m+n}\\ =\frac{2\times\frac{7}{2}+1\times4}{2+1}\\ =\frac{9+2}{3}\\ =\frac{11}{3}

Thus,

Coordinates of P are \left(\frac{11}{3},\ \frac{11}{3}\right)

(iii) As E and F are the mid-point if CA and AB respectively

Coordinates of E will be 

\left(\frac{1+4}{2},\ \frac{4+2}{2}\right)\\ \left(\frac{5}{2},\ \frac{6}{2}\right)\\ \left(\frac{5}{2},\ 3\right)

and of F will be

\left(\frac{4+6}{2},\ \frac{2+5}{2}\right)\\ \left(\frac{10}{2},\ \frac{7}{2}\right)\\ \left(5,\ \frac{7}{2}\right)

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

x=\frac{2\times\frac{5}{2}+1\times6}{2+1}\\ =\frac{5+6}{3}\\ =\frac{11}{3}

and 

\frac{2\times3+1\times5}{2+1}\\ \frac{6+5}{3}\\ \frac{11}{3}

i.e., Q1 is (11/3, 11/3)

and similarly the coordinates of R will be

x=\frac{2\times5+1\times1}{2+1}\\ =\frac{10+1}{3}\\ =\frac{11}{3}

and

y=\frac{2\times\frac{7}{2}+1\times4}{2+1}\\ =\frac{7+4}{3}\\ =\frac{11}{3}

R is (11/3, 11/3)

(iv) We can see that coordinates of P, Q and R are same 

i.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

\left(\frac{6+9}{2},\ \frac{1+4}{2}\right)\\ \left(\frac{15}{2},\ \frac{5}{2}\right)

As O is also the mid-point of BD

Therefore,

\frac{15}{2}=\frac{8+k}{2}

and

\frac{5}{2}=\frac{2+p}{2}

⇒ 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 \frac{AP}{PB}=\frac{k}{1}   . 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 \frac{AP}{PB}=\frac{k}{1}    

⇒ AP : PB = k : 1

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

x=\frac{m_1x_2+m_2x_1}{m_1+m_2}\\ =\frac{k\times(-4)+1\times3}{k+1}\\ =\frac{-4k+3}{k+1}

and

y=\frac{m_1y_2+m_2y_1}{m_1+m_2}\\ =\frac{k\times(8)+1\times(-5)}{k+1}\\ =\frac{8k-5}{k+1}

As x + y = 0

Therefore,

\frac{-4k+3}{k+1}+\frac{8k-5}{k+1}=0\\ \Rightarrow \frac{-4k+3+8k-5}{k+1}=0

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

=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{-10-2}{2},\ \frac{4+0}{2}\right)\\ =\left(\frac{-12}{2},\ \frac{4}{2}\right)\\ =(-6,\ 2)

P lies on CD also,

Let P divides C (9, 4) and D (-4, y) in the ratio m1 : m2

Therefore,

-6=\frac{m_1(-4)+m_2(-9)}{m_1+m_2}

⇒ -6m1 – 6m2 = -4m1 – 9m2

⇒ -6m1 + 4m1 = -9m2 + 6m2

⇒ -2m1 = -3m2

⇒ \frac{m_1}{m_2}=\frac{-3}{-2}=\frac{3}{2}

Therefore,

m1 : m2 = 3 : 2

and

2=\frac{3\times y+2\times(-4)}{3+2}\\ ⇒ 2=\frac{3y-8}{5}

⇒ 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 x2 + y2

Solution:

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

the points (x1, y1) and (x,2 y2) in the ration m : n, then

x=\frac{mx_2+nx_1}{m+n}

and

y=\frac{my_2+ny_1}{m+n}

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

Now, 

-1=\frac{3x+8}{3+4}

and

2=\frac{20-3y}{3+4}

⇒ 3x + 8 = -7

and

⇒ 20 – 3y = 14

⇒ x = -5

and

⇒ y = 2

Now,

x2 + y2 = (-5)2 + (2)2

x2 + y2 = 25 + 4 = 29

Question 59. ABCD is a parallelogram with vertices A (x1, y1), B (x2, y2), and C (x3, y3). Find the coordinates of the fourth vertex D in terms of x1, x2, x3, y1, y2, and y

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 

\left(\frac{x_1+x_3}{2},\ \frac{y_1+y_3}{2}\right)=\left(\frac{x_2+x}{2},\ \frac{y_2+y}{2}\right)

i.e., x1 + x3 = x2 + x and y1 + y3 = y2 + y

i.e, x1 + x3 – x2 = x and y1 + y3 + y2 = y

Hence, the coordinates of D are (x1 + x3 – x2, y1 + y3 + y2)

Question 60. The points A (x1, y1), B (x2, y2), and C (x3, y3) 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 (x1, y1), B (x2, y2) and C (x3, y3) 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 = \left(\frac{x_2+x_3}{2},\ \frac{y_2+y_3}{2}\right)

D = \left(\frac{x_2+x_3}{2},\ \frac{y_2+y_3}{2}\right)

(ii) Assume the coordinates of a point P be (x, y)

Given that, the point P (x, y), divide the line joining A (x1, y1) and 

\left(\frac{x_2+x_3}{2},\ \frac{y_2+y_3}{2}\right)    in the ration of 2 : 1,

Then the coordinates of P

=\left[\frac{2\left(\frac{x_2+x_3}{2}\right)+1.x_1}{2+1},\ \frac{2\left(\frac{y_2+y_3}{2}\right)+1.y_1}{2+1}\right]

Using internal section formula, we get

=\left(\frac{x_2+x_3+x_1}{3},\ \frac{y_2+y_3+y_1}{3}\right)

Therefore,

Required coordinates of points P = \left(\frac{x_2+x_3+x_1}{3},\ \frac{y_2+y_3+y_1}{3}\right)

(iii) Assume the coordinates of a point Q be (p, q)

Given: The point Q (p, q) divide the line joining B(x2, y2) and E =\left(\frac{x_1+x_3}{2},\ \frac{y_1+y_3}{2}\right)

In the ratio of 2 : 1

Then the coordinates of Q

=\left[\frac{2\left(\frac{x_1+x_3}{2}\right)+1.x_2}{2+1},\ \frac{2\left(\frac{y_1+y_3}{2}\right)+1.y_2}{2+1}\right]\\ \left(\frac{x_2+x_3+x_1}{3},\ \frac{y_2+y_3+y_1}{3}\right)

[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 = =\left(\frac{x_1+x_3}{2},\ \frac{y_1+y_3}{2}\right)

So, the required coordinate of point Q = \left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)

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 (x3, y3) and 

\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)    in the ration 2 : 1, 

So, the coordinates of R be

=\left[\frac{2\left(\frac{x_1+x_2}{2}\right)+1.x_3}{2+1},\ \frac{2\left(\frac{y_1+y_2}{2}\right)+1.y_3}{2+1}\right]\\ \left(\frac{x_2+x_3+x_1}{3},\ \frac{y_2+y_3+y_1}{3}\right)

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 =\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)

So, required coordinate of point R =\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)

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

=\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)


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