ML Aggarwal ICSE Solutions Class 10 Math Eleventh Chapter Section Formula

ML Aggarwal ICSE Solutions Class 10 Math 11th Chapter Section Formula

Class 10 Chapter 11 Section Formula

Chapter 11 Section Formula

(1) Find the Coordinates of the mid – point of the line segments joining the following pairs of points:

(i) (2, -3), (-6, 7)

(ii) (5, – 11), (4,3)

(iii) (a + 3, 5b), (2a – 1, 3b + 4)

Solution:

(i) Given two points are (2, -3) and (-6, 7)

Mid Point formula is given by,

(x, y) = (x₁ + x₂/2, y₁ + y₂/2)

(x, y) = (2 – 6)/2, (-3 + 7)/2 = (-2, 2)

Thus, the Co – ordinates of the mid Point of line segment joining the points (2, -3) & (-6, 7) is found to be (-2, 2)

(ii) Given two points are (5, – 11) and (4,3)

Mid – Point formula is given by

(x, y) = (x₁ + x₂/2, y₁ + y₂/2)

(x, y) = (5 + 4/2, -11 +3/2) = (9/2, -8/2) = (4.5, – 4)

Thus, the Co – ordinates of the mid – point of line segment joining the points (5, – 11) and (4,3) is found to be (4.5, – 4).

(iii) Given two points are (a + 3, 5b) and (2a – 1, 3b + 4)

Mid – point formula is given by,

(x, y) = (x₁ + x₂/2, y₁ + y₂/2)

= (a + 3 + 2a – 1/2, 5b+ 3b + 4/2) = (3a + 2/2, 8b + 4/2)

(x, y) = (3a + 2/2, 4b + 2)

Thus, the Co – ordinates of mid – point of line – segment joining the points (a + 3, 5b) and (2a – 1, 3b + 4) is found to be (3a + 2/2, 4b + 2)

 

(2) P divides the distance between A (-2, 1) and B (1, 4) in the ratio of 2: 1. Calculate the Co – ordinates of the Point P.

Solution:

Given that, point P divides the distance between A (-2, 1) and B (1,4) in the ratio of 2 :1.

Let us Consider the point P (x, y) divides AB in the ratio of m₁ : m₂ i.e. 2 : 1.

X = (m₁ x₂ + m₂ x₁/m₁ + m₂)

X = (2 × 1 + 1 × (-2)/ 2 + 1)

= (2 – 2/3) = 0/3 = 0  x = 0

And y = (m₁ y₁ + m₂ y₁/ m₁ + m₂) = (2 × 4 + 1 × 1/2 + 1) = (8 + 1/3) = 9/3 = 3 y = 3

Hence, the Co – ordinates of point P are found to be P = (0,3).

 

(3) (i) Find the Co – ordinates of the points of trisection of the line Segment joining the point (3, -3) and (6, 9).

(ii) 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 (5/3, 9) respectively, find the values P and q.

Solution:

(i) Given that,

The Points of line segment the points (3, -3) and (6, 9).

Let us consider the points P (x₁, y₁) and Q (x₂, y₂) are the points which trisect the line segment joining Points A (3, -3) and B (6, 9).

Here, point P (x₁, y₁) divides AB in the ratio 1:2.

Thus, the Co – ordinates of point P (x₁, y₁) are found to be P (4,1).

Let us consider, Q (x₂, y₂) divides the line segment AB in the ratio 2 :1.

Thus, the Co – ordinates of point Q (x₂, y₂) are found to be Q ( 5, 5).

 

(ii) Given that, the line segment joining the points (3, -4) and (1,2) is trisected at the points P and Q.

And the Co – ordinates of point P and Q are (P, -2) and (5/3, q) respectively.

Here, point P (p, -2) divides lines AB in the ratio 1:2 and point Q divides line AB in the ratio 2:1.

Hence, the co – ordinates of point Q are found to be Q (5/3, 0) and P (7/3, -2).

 

(4) The line segment joining the points A (3, 2) and B (5, 1) is divided at the point p in the ratio 1: 2 and it lies on the line 3x – 18y + k = 0. Find the value of K.

Solution:

Let us consider the point P (x, y) divides the line segment joining the points A (3, 2) and B (5, 1) in the ratio 1: 2.

Given that point P is on the line 3x – 18y + k = 0 ….. (1)

Put x = 11/3 and y = 5/3 in equine (1)

= 3 (11/3) – 18 (5/3) + k = 0

= 11 – 30 + k = 0

= – 19 + k = 0

K = 19

 

(5) Find the co – ordinates of the point which is three – fourth of the way from A (3, 1) to B (-2, 5).

Solution:

Given that, the points having co – ordinates A (3, 1) and B (-2, 5) are the end points of the line segment.

Let us consider, P (x, y) be the point which is three – fourth of the way from A to B.

= AP/AB = 3/4

= AP/(AP + PB) = 3/4

4 AP = 3 AP + 3 PB

4 AP – 3 AP = 3PB

= AP = 3 PB

Hence, AP/PB = 3/1 = m₁ = 3, m₂ = 1

Now, to find the co – ordinates of point P (x, y):

Thus, the co – ordinates of point P will be (-3/4, 4).

 

(6) The line segment joining points A (-3, 1) and B (5, – 4) is a diameter of a circle whose centre is C. Find the co – ordinates of the point c.

Solution:

Given that, the line segment joining points A (-3, 1) and B (5, -4) is a diameter of a circle whose centre is C.

i.e. C is the midpoint of AB.

Let us consider the point C (x, y).

X = (-3 + 5)/2 = (1 – 4)/2 = 2/2 = 1

Y = (1 – 4)/2 = -3/2

Thus, the co – ordinates of Point c are found to be (1, -3/2).

 

(7) The mid – point of the line segment joining the points (3m, 6) and (-4, 3n) is (1, 2m -1). Find the values of m and n.

Solution:

Let us consider, the point P (1, 2m – 1) is the mid – point of the line segment joining the points (3m, 6) and (-4, 3n).

1 = (x₁ + x₂)/2

1 = (3m – 4)/2

2 = 3m – 4

m = 6/3

m = 2

and 2m – 1 = (6 + 3n) /2

4m – 2 = 6 + 3n

4 × 2 – 2 = 6 + 3n

8 – 2 = 6 + 3n

3n = 8 – 2 – 6

n = 0

Thus, Here the values of m and n are found to be m = 2 and n = 0

 

(8) The co – ordinates of the mid – point of the line segment PQ are (1, -2). The co – ordinates of P are (-3, 2). Find the co – ordinates of Q.

Solution:

Let us consider the co – ordinates of point Q be (x, y). Given that, the co – ordinates of the mid – point of the line segment PQ are (1, -2) and P (-3, 2).

1 = (-3 + x)/2

-3 + x = 2

X = 2 + 3 = 5

and -2 = (2 + y) /2

-4 = 2 + y

Y = -4 – 2

Y = -6

Thus, the co – ordinates of point Q are found to be Q (5, -6).

 

(9) AB is a diameter of a circle with centre C (-2, 5). If point A is (3, -7).

Find (i) the length of radius AC.

(ii) the co – ordinates of B.

Solution:

Here, given that

AB is a diameter of a circle with centre c (-2, 5).

and point A having co – ordinates A (3, – 7).

(i) Thus, By distance formula

AC = √(3 + 2)² + (-7 – 5)² = √5² + 12² = √25 + 144

= √169 = 13

Thus, the length of radius AC is found to be 13 units.

(ii) Let us consider, co – ordinates of points B are (x, y).

∴ (3 + x)/2 = -2

3 + x = -4

X = -7

and (y – 7)/2 = 5

y = 10 + 7

y = 17

Thus, the co – ordinates of point B are found to be B (-7, 17).

 

(10) Find the reflection (image) of the point (5, – 3) in the point (-1, 3).

Solution:

Let us consider, the co – ordinates of the image of the point A (5, -3) be A¹ (x, y) in the point (-1, 3).

= Point (-1, 3) is the midpoint of AA.

= -1 = (5 + x) /2

5 + x = -2

X = -7

and 3 = (-3 + Y) /2

-3 + y = 6

Y = 9

Hence, the co – ordinates of the image A will be A (-7, 9).

 

(11) The line segment joining A (-1, 5/3) the Points B (a, 5) is divided in the ratio 1:3 at P, the point where the line segment AB intersects y – axis Calculate

(i) The value of a

(ii) The co – ordinates of P.

Solution:

Given that, the line segment joining A (-1, 5/3) the points B (a, 6) is divided in the ratio 1:3 at P.

(i) Thus,

(ii) Given that, the line segment AB intersects Y axis at P.

∴ x = 0

= (a – 3)/4 = 0

a – 3 = 0

a = 3

And hence the co – ordinates of point P are found to be P (0, 5/2).

 

(12) The point P (-4, 1) divides the line segment joining the points A (-2, 2) and B in the ratio 3: 5. Find the point B.

Solution:

Given that,

The point P (-4, 1) divides the line segment joining the points A (2, -2) and B in the ratio 3: 5.

Let us consider, B be any point whose co – ordinates are found to be B (x, y).

= -4 = [3x + 5 (2)/ 3 + 5] = (3x + 10)/8

Also, 3x + 10 = -32

3x = -32 – 10

3x = -42

∴ x = -14

and 1 = [3y + 5 (-2)/ 3 + 5]

1 = (3y – 10) / 8

8 = 3y – 10

3y = 8 + 10

Y = 18/3

Y = 6

Thus, the co – ordinates of point B are found to be B (-14, 6).

 

(13) (i) In what ratio does the point (5, 4) divide the line segment joining the points (2, 1) and (7, 6)?

(ii) In what ratio does the point (-4, b) divide the line segment joining the points P (2, -2) Q (-14, 6)? Hence find the value of b.

Solution:

(i) Given that, the point (5, 4) divide the line segment joining the points (2, 1) and (7, 6).

Let us consider the points be A (2, 1) & B (7, 6) and P (5, 4).

Let us consider point P divides AB in the ratio m₁ : m₂.

Then, 5 = (m₁ x 7 + m₂ x 2/m₁+m₂)

5m₁ + 5m₂ = 7m₁ + 2m₂

5m₂ – 2m₂ = 7m₁ – 5m₁

3m₂ = 2m₁

m₁/m₂ = 3/2

m₁ : m₂ = 3 : 2

Thus, point P (5, 4) divides the line segment A (2,1) & B (7, 6) joining points in the ratio 3 : 2.

(ii) Given that, point (-4, b) divide the line segment joining the points P (2, – 2), Q (-14, 6).

Let us consider point A (-4, b) divide the line segment joining points P & Q in the ratio m₁ : m₂.

Then, – 4 = (m₁ x (-14) + m₂ x 2/m₁+m₂)

-4m₁ – 4m₂ = -14m₁ + 2m₂

-4m₁ + 14 m₁ = 2m₂ + 4m₂

10m₁ = 6m₂

m₁/ m₂ = 6/10 = 3/5

m₁/ m₂ = 3: 5

Also, b =

Thus, point A divide the line segment joining points P and Q in the ratio 3: 5.

And the value of b is found to be b = 1.

 

(14) The line segment joining A (2, 3) and B (6, -5) is intercepted by the X – axis at the point K. Write the ordinate of point K. Hence, find the ratio in which K divides AB Also, find the co – ordinates of the point K.

Solution:

Given that,

The line segment joining A (2, 3) and B (6, -5) is intercepted by the x – axis at the point K.

Let us consider the co – ordinates of point K be K (x, o) Which intersects x – axis.

Let us consider point K divides AB in the ratio m₁ : m₂.

Thus, co – ordinates of point K are found to be K (7/2, 0).

 

(16) In what ratio does the line x – y – 2 = 0 divide the line Segment joining the points (3, -1) and (8, 9)? Also, find the co – ordinates of the point of division?

Solution:

Given that,

The line x – y – 2 = 0 divide the line segment the points (3, -1) and (8, 9).

Let us consider it divide the segment in the ratio m₁ : m₂ at point P (x, y).

8m₁ + 3m₂ – 9m₁ + m₂ – 2m₁ – 2m₂ = 0

-3m₁ + 2m₂ = 0

3m₁ = 2m₂

∴ m₁: m₂ = 2/3

Thus, the co – ordinates of point P are found to be P (5, 3)

 

(17) Given a line segment AB joining the points A (-4, 6) and B (8, -3).

Find (i) The ratio in which AB is divided by the y. axis.

(ii) Find the co – ordinates of the point of intersection.

(iii) The length of AB

Solution:

Given that, a line segment AB joining the points A (-4, 6) & B (8, -3).

(i) Let us consider AB is divided by the y – axis in the ratio m : 1.

O = (m × 8 – 4 × 1/ m + 1)

8m – 4 = 0

8m = 4

m = 1/2

Thus, AB is divided by the y – axis in the ratio 1/2 : 1 or 1 : 2

(ii) Here, y = (1 × (-3) + 2 × 6)/ (1 + 2) = 9/3 = 3

Thus, co – ordinates of the point of intersection are found to be (0, 3).

(iii) Here, By distance formula,

AB = √(8 – 4)² + (-3 – 6)²

= √144 + 81

AB = √225 = 15 units

Thus, the length of line segment AB is found to be 15 units.

 

(18) Calculate the length of the median through the vertex A of the triangle ABC with vertices A (7, -3), B (5, 3) and C (3, – 1).

Solution:

Given that, In triangle ABC,

The vertices are having co – ordinates A (7, -3), B (5, 3) and C (3, -1)

Let us consider the Point D (x, y) be the midpoint of BC.

Hence, co – ordinates of point D are calculated by,

X = (5 + 3)/ 2 = 8/2 = 4

and y = (3 – 1)/2 = 2/2 = 1

Hence, co – ordinates of point D are found to be D (4, 1).

Thus, length of DA = √(7 – 4)² + (-3 -1)²

= √(3)² + (-4)²

= √9 + 16

= √25

l (DA) = 5 units

Thus, the length of median is found to be 5 units.

 

(19) Three consecutive vertices of a parallelogram ABCD are A (1, 2), B (1, 0) and C (4, 0). Find the fourth vertex D.

Solution:

Given that, the three consecutive vertices of a Parallelogram ABCD are A (1, 2), B (1, 0) and C (4, 0).

Let us consider Point ‘P’ is the mid- point of diagonal AC.

Thus, co – ordinates of Point P = (1 + 4/2, 2 + 0/2) = (5/2, 1)

Let us consider the co – ordinates of point D (x, y).

Thus, 5/2 = (1 + x)/2

10 = 2 + 2x

2x = 10 – 2 = 8

X = 4

And 1 = (0 + y)/2

Y = 2

Thus, co – ordinates of point D are found to be D = (4, 2).

(20) If the Points A (-2, -1), B (1, 0), C (P, 3) and D (1, q) form a Parallelogram ABCD, Find the values of p and q.

Solution:

Given that, the points A (-2, -1), B (1, 0), C (P, 3) and D (1, q) forms a parallelogram ABCD.

• Let us consider, the diagonals of parallelogram ABCD which are AC and BD bisects each other at point P.
• Thus, By the properties of a parallelogram P is the mid – point of AC as well as BD.
• Let us consider point P has co – ordinates P(x, y).

Then, x = (P – 2)/2

and y = (3 – 1)/2 = 2/2 = 1

x = (P – 2)/ 2,

Y = 1

[Since, Point P is the mid – point of diagonal AC.]

• Now, x = (1 + 1)/ 2 = 2/2 = 1

(P – 2)/2 = 1

P – 2 = 2

P = 2 + 2 = 4

P = 4

and q/2 =1

q = 2

Thus, the values of P and q are found to be P = 4 and q = 2.

 

(21) If two vertices of a Parallelogram are (3, 2), (-1, 0) and its diagonals meet at (2,-5). Find the other two vertices of the Parallelogram.

Solution:

Given that, the two vertices of a Parallelogram are (3, 2), (-1, 0) and its diagonals meet at (2, -5).

Let us consider, A (3, 2), B (-1, 0) & P (2, – 5).

Then, P is the mid – point of diagonals Ac & BD.

Let us take, the co – ordinates P of point C be (x, y).

2 = (x + 3)/2

X + 3 = 4

X = 1

And, -5 = (y + 2)/2

Y = -10 – 2

Y = -12

Hence, co ordinates of point C are found to be C (1, -12).

Let us consider, co – ordinates of point ‘D’ be D (x, y).

2 = (x – 1)/2

X – 1 = 4

X = 5

and -5 = (y + 0)/2

y = -10

Thus, the co – ordinates of point D are found to be D (5, -10).

 

(22) Find the third vertex of a triangle if its two vertices are (-1, 4) and (5, 2) and mid – point of one side is (0, 3).

Solution:

Given that,

The third vertex of a triangle if its two vertices are A (-1, 4) and B (5, 2).

And also, the mid – point of one side is P (0, 3).

Thus, By mid – point formula,

0 = (x – 1)/2

X = 1

and 3 = (y + 4)/ 2

y + 4 = 6

y = 2

Hence, co – ordinates of third vertex be C (1, 2).

But, if we considered point P (0,3) is the mid – point of side BC then, By mid – point formula

O = (5 + x)/2

X + 5 = 0

X = – 5

and 3 = (2 + y)/2

2 + y = 6

Y = 4

Hence, co – ordinates of third vertex C be C (-5, 4).

Finally, the co – ordinates of point C found to be (1, 2) or (-5, 4).

 

(24) Show that by Section Formula the points (3, -2), (5, 2), and (8, 8) are collinear.

Solution:

Given points are (3, -2), (5, 2) and (8, 8).

To show the given points are collinear:

Let us consider (5, 2) divides the line joining points (3, -2) & (8, 8) in the ratio m: n.

Then, 5 = (m × 8 + n × 3/ m + n)

8m + 3n = 5m + 5n

8m – 5m = 5n – 3n

3m = 2n

m/n = 2/3 —– (1)

From (1) & (2), it is proved that point (5, 2) lies on the line joining the points (3, -2) & (8, 8).

Hence, the Points (3, -2), (5, 2) & (8, 8) are collinear. Hence proved.

 

(25) Find the value of P for which the points (-5, 1), (1, P) & (4, -2) are collinear.

Solution:

Given that, the Points (-5, 1), (1, P) & (4, -2) are collinear.

Let us take, A (-5, 1), B (1, P) & C (4, – 2).

And point A (-5, 1) divides BC in the ratio m: n.

From (1) & (2) = (P – 1)/3 = -2/3

= -3P + 3 = 6

-3P = 3

P = -3/3

P = 1 is the required value of P.

 

(26) The mid – point of the line segment AB shown in the adjoining diagram is (4, -3). Write down the co – ordinates of A and B.

Solution:

Given that, The mid – point of the line Segment AB shown in the adjoining diagram is (4, – 3).

A lies on the X – axis & B on the Y – axis as shown in fig.

Then, co – ordinates of A and B are found to be:

A (x, 0) and B (0, y).

As P (+4, -3) is the mid – point of AB, then by mid – point formula,

4 = (x + 0)/2

X = 8

and -3 = (0 + y)/ 2

y = -6

Thus, the co – ordinates of points A and B are found to be:

A (8, 0) and B (0, -6).

 

(27) Find the co – ordinates of the centroid of a triangle whose vertices are A (-1, 3), B (1, – 1) and C (5, 1).

Solution:

Given that, the three vertices of a triangle are A (-1, 3), B (1, -1) and C (5, 1).

Then, by Centroid Formula,

C (x, y) ≡ (-1 + 1 + 5/3, 3 – 1 + 1/3) ≡ (5/3, 3/3) ≡ (5/3, 1)

Thus, the co – ordinates of a centroid of a given triangle is found to be C (5/3, 1).

 

(28) Two vertices of a triangle are (3, -5) and (-7, 4). Find the third vertex given that the centroid is (2, -1).

Solution:

Given that, two vertices of a triangle are A (3, -5) & B (-7, 4).

And co – ordinates of a centroid of a triangle are C (2, 1).

Let us consider, the co – ordinates of third vertex of a triangle are D (x, y).

Then, By centroid formula,

2 = (3 – 7 + x/3)

(x – 4)/3 = 2

X – 4 = 6

X = 10

and -1 = (- 5 + 4 + y/3)

-3 = -1 + y

Y = -3 + 1

Y = -2

Thus, the co – ordinates of third vertex of a given triangle are found to be D (10, -2).

 

(29) The vertices of a triangle are A (-5, 3), B (P, -1) and C (6, q). Find the values of p and q, if the centriod of the triangle ABC is the point (1, -1).

Solution:

Given that, the vertices of a triangle are A (-5, 3), B (P, – 1), and C (6, q).

And centroid of a triangle is found to be D (1, -1).

Then, By Centroid formula,

(x, y) ≡ (-5 + P + 6/3, 3 – 1 + q/3) ≡ (1 + P/3, 2 + q/3)

= 1 + P/3 = 1

1 + P = 3

P = 2

and (2 + q/3) = -1

2 + q = – 3

q = -5

Thus, the values of p and q are found to be

P = 2 and q = – 5