**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) = (x_{1} + x_{2}/2, y_{1} + y_{2}/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) = (x_{1} + x_{2}/2, y_{1} + y_{2}/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) = (x_{1} + x_{2}/2, y_{1} + y_{2}/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 m_{1} : m_{2} i.e. 2 : 1.

X = (m_{1} x_{2} + m_{2} x_{1}/m_{1} + m_{2})

X = (2 × 1 + 1 × (-2)/ 2 + 1)

= (2 – 2/3) = 0/3 = 0 x = 0

And y = (m_{1} y_{1} + m_{2} y_{1}/ m_{1} + m_{2}) = (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 (x_{1}, y_{1}) and Q (x_{2}, y_{2}) are the points which trisect the line segment joining Points A (3, -3) and B (6, 9).

Here, point P (x_{1}, y_{1}) divides AB in the ratio 1:2.

Thus, the Co – ordinates of point P (x_{1}, y_{1}) are found to be P (4,1).

Let us consider, Q (x_{2}, y_{2}) divides the line segment AB in the ratio 2 :1.

Thus, the Co – ordinates of point Q (x_{2}, y_{2}) 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 = m_{1} = 3, m_{2 }= 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 = (x_{1} + x_{2})/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} = √5^{2} + 12^{2} = √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 A^{1} (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 m_{1} : m_{2}.

Then, 5 = (m_{1 }x 7 + m_{2} x 2/m_{1}+m_{2})

5m_{1} + 5m_{2} = 7m_{1} + 2m_{2}

5m_{2} – 2m_{2} = 7m_{1} – 5m_{1}

3m_{2} = 2m_{1}

m_{1}/m_{2} = 3/2

m_{1} : m_{2} = 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 m_{1} : m_{2}.

Then, – 4 = (m_{1 }x (-14) + m_{2} x 2/m_{1}+m_{2})

-4m_{1} – 4m_{2 }= -14m_{1} + 2m_{2}

-4m_{1} + 14 m_{1 }= 2m_{2} + 4m_{2}

10m_{1} = 6m_{2}

m_{1}/ m_{2} = 6/10 = 3/5

m_{1}/ m_{2} = 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 m_{1} : m_{2}.

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 m_{1} : m_{2} at point P (x, y).

8m_{1} + 3m_{2} – 9m_{1} + m_{2} – 2m_{1} – 2m_{2 }= 0

-3m_{1} + 2m_{2} = 0

3m_{1} = 2m_{2}

∴ m_{1}: m_{2} = 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