ML Aggarwal Section Formula Exe-11 Solutions ICSE Class-10 Maths Ch-11. We Provide Step by Step Answer of Exercise-11 **Section Formula** Exe-11 Questions for ICSE Class-10 APC Understanding Mathematics. Visit official Website CISCE for detail information about ICSE Board Class-10.

## ML Aggarwal Section Formula Exe-11 Solutions ICSE Class-10 Maths Ch-11

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-11 | Section Formula |

Writer / Book | Understanding |

Topics | Solutions of Exe-11 |

Academic Session | 2024-2025 |

**ML Aggarwal Section Formula Exe-11 Solutions**

ICSE Class-10 Maths Ch-11

**Question -1. ****Find the co-ordinates 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)

**Answer:**

**(i) Co-ordinates of the mid-point of (2, -3), ( -6, 7)**

{(x_{2} + x_{2})/2, (y_{1 }+ y_{2})/2} or

{(2 – 6)/2, (-3 + 7)/2} or

(-4/2, 4/2) or (-2, 2)

**(ii) Mid-point of (5, – 11) and (4, 3)**

= {(x_{1} + x_{2})/2, (y_{1} + y_{2})/2} or

{(5+4)/2, (-11+3)/2}

Or (9/2, -8/2) or (9/2, – 4)

**(iii) Mid-point of (a + 3, 5b) and (2a – 1, 3b + 4)**

= (x_{1} + x_{2})/2, (y_{1} + y_{2})/2

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

Or (3a + 2)/2 , (8b + 4)/2

Or {(3a + 2)/2, (4b + 2)}

**Question -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.**

**Answer :**

Points are A (-2, 1) and B (1, 4) and

Let P (x, y) divides AB in the ratio of m_{1} : m_{2} i.e. 2 : 1

Co-ordinates of P will be

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

= {(2×1 + 1×(-2)}/(2+1)

= (2 – 2)/3

= 0/3

= 0

y = (m_{1}y_{2 }+ m_{2}y_{1})/(m_{1} + m_{2})

= (2×4 + 1×1)/(2+1)

= (8 + 1)/3

= 9/3

= 3

∴ Co-ordinates of point P are (0, 3).

**Question -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 coordinates of P and Q are (p, – 2) and respectively, find the values of p and q.**

**Answer :**

**(i) Let P (x**_{1}, y_{1}) and Q (x_{2}, y_{2}) be the points which trisect the line segment joining the points A (3, -3) and B (6, 9)

_{1}, y

_{1}) and Q (x

_{2}, y

_{2}) be the points which trisect the line segment joining the points A (3, -3) and B (6, 9)

∵ P(x_{1}, y_{1}) divides AB in the ratio of 1 : 2

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

= (1×6 + 2×3)/(1+2)

= (6 + 6)/3

= 12/3

= 4

y_{1} = (m_{1}y_{2 }+ m_{2}y_{1})/(m_{1} + m_{2})

= {(1×9 + 2×(-3)}/(1+2)

= (9 – 6)/3

= 3/3

= 1

∴ Co-ordinates of Pare (4, 1)

∵ Q (x_{2}, y_{2}) divides the line segment

AB in the ratio of 2 : 1

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

= (2×6 + 1×3)/(2+1)

= (12 + 3)/3

= 15/3

= 5

y_{2} = (m_{1}y_{2} + m_{2}y_{1})/(m_{1} + m_{2})

= {2×9 + 1(-3)}/(2+1)

= (18 – 3)/3

= 15/3

= 5

∴ Co-ordinates of Q are (5, 5)

**(ii) point P and Q trisect the line AB.**

In other words, P divides it in the ratio 1 : 2 and Q divides it in the ratio 2 : 1

∴ p = (mx_{2} + nx_{1})/(m + n)

= (1×1 + 2×3)/(1+2)

= (1 + 6)/3

= 7/3

q = (my_{2 }+ ny_{1})/(m + n)

= {2×2 + 1×(-4)}/(2+1)

= (4 – 4)/2

= 0

∴ p = 7/3, q = 0

**Question -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.**

**Answer :**

#### The point P (x, y) divides the line segment joining the points

A (3, 2) and B (5, 1) in the ratio 1 : 2

∴ x = (mx_{2} + nx_{1})/(m + n)

= (1 × 5 + 2 × 3)/(1+2)

= (5 + 6)/2

= 11/3

y = (my_{2 }+ ny_{1})/(m + n)

= (1 × 1 + 2 × 2)/(1+2)

= (1 + 4)/3

= 5/3

∵ P lies on the line 3x – 18y + k = 0

∴ It will satisfy it.

3(11/3) – 18(5/3) + k = 0

⇒ 11 – 30 + k = 0

⇒ – 19 + k = 0

⇒ k = 19

**Question -5. m ****Find the coordinates of the point which is three-fourth of the way from A (3, 1) to B ( – 2, 5).**

**Answer:**

Let P be the required point, then

AP/AB = 3/4

∴ AP/AB = 3/4

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

⇒ 4AP = 3AP + 3PB

⇒ 4AP – 3AP = 3PB

AP = 3PB

AP/PB = 3/1

∴ m_{1 }= 3, m_{2} = 1

Let co-ordinates of P be (x, y)

∴ x = (m_{1}x_{2 }+ m_{2}x_{1})/(m_{1} + m_{2})

= {3 × (-2) + 1 × (3)}/(3 + 1)

= (-6 + 3)/ 4

= -3/4

y = (m_{1}y_{2} + m_{2}y_{1})/(m_{1} + m_{2})

= (3 × 5 + 1 × 1)/(3 + 1)

= (15 + 1)/4

= 16/4

= 4

∴ Co-ordinates of P will be (-3/4, 4)

**Question -6. ****The line segment joining 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. **

**Answer:**

∵ C is the centre of the circle and AB is the diameter

C is the midpoint of AB.

Let co-ordinates of C (x, y)

∴ x = (-3 + 5)/2, x = (1 – 4)/2

⇒ x = 2/2, y = -3/2

⇒ x = 1, y = -3/2

∴ Co-ordinates of C are (1, -3/2)

**Question -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.**

**Answer :**

Let the mid-point of the line segment joining two points

A(3m, 6) and (-4, 3n) is P( 1, 2m – 1)

∴ 1 = (x_{1} + x_{2})/2

= (3m – 4)/2

⇒ 3m – 4 = 2

⇒ 3m = 2 + 4 = 6

⇒ m = 6/3 = 2

And 2m – 1 = (6 + 3n)/2

⇒ 4m – 2 = 6 + 3n

⇒ 4 × 2 – 2 = 6 + 3n = 8 – 2 = 6 + 3n

⇒ 3n = 8 – 2 – 6 = 0

⇒ n = 0

Hence,

m = 2, n = 0

**Question-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.**

**Answer :**

Let the co-ordinates of Q be (x, y)

co-ordinates of P are (-3, 2) and mid-point of PQ are (1, -2) then

1 = (-3 + x)/2

⇒ – 3 + x = 2

⇒ x = 2 + 3 = 5

And -2 = (2 + y)/2

⇒ 2 + y = -4

⇒ y = – 4 – 2 = – 6

∴ x = 5, y = – 6

Hence, co-ordinates of Q are (5, -6)

**Question -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 coordinates of B. (2013)**

**Answer :**

Let co-ordinate of B are (x, y)

∴ (3 + x)/2 = -2 and (y – 7)/2 = 5

⇒ 3 + x = – 4 and y – 7 = 10

⇒ x = – 4 – 3 and y = 10 + 7

⇒ x = – 7 and y = 17

∴ B is (-7, 17)

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

**Answer :**

Let the co-ordinates of the images of the point A (5, -3) be

A1 (x, y) in the point (-1, 3) then

the point (-1, 3) will be the midpoint of AA1.

∴ – 1 = (5 + x)/2

⇒ 5 + x = – 2

x = – 2 – 5 = – 7

and 3 = (-3 + y)/2

⇒ – 3 + y = 6

⇒ y = 6 + 3 = 9

∴ Co-ordinates of the image A, will be (-7, 9).

**Question -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**

**Answer :**

Let P (x, y) divides the line segment joining

the points -1,5/3, B(a, 5) in the ratio 1 : 3

∴ x = {1×a + 3×(-1)}/(1+3)

= (a – 3)/4

y = {1×a + 3×(-1)}/(1+3)

= (a – 3)/4

= (5 + 5)/4

= 10/4

= 5/2

**(i)** ∵ AB intersects y-axis at P

∴ x = 0

⇒ (a – 3)/4 = 0

⇒ a – 3 = 0

∴ a = 3

**(ii)** ∴ Co-ordinates of P are (0, 5/2)

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

**Answer :**

Let the co-ordinates of B be (x, y)

Co-ordinates of A (2, -2) and point P (-4, 1)

divides AB in the ratio of 3 : 5

∴ -4 = {3×x + 5×(2)}/(3+5)

= (3x +10)/8

And 3x + 10 = – 32

⇒ 3x = – 32 – 10

= – 42

∴ x = – 42/3 = – 14

l = {3×y + 5×(-2)}/(3+5)

⇒ l = (3y – 10)/8

⇒ 3y – 10 = 8

⇒ 3y = 8 + 10

= 18

∴ y = 18/3 = 6

∴ Co-ordinates of B = (- 14, 6)

**Question -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.**

**Answer :**

**(i) Let the ratio be m**_{1} : m_{2} that the point (5, 4) divides

_{1}: m

_{2}that the point (5, 4) divides

the line segment joining the points (2, 1), (7, 6).

5 = (m_{1}×7 + m_{2}×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

**(ii)** **The point (- 4, b) divides the line segment joining the points P (2, -2) and Q (-14, 6) in the ratio m**_{1} : m_{2}.

_{1}: m

_{2}.

∴ -4 = {m_{1}(-14) + m_{2}×2}/(m_{1}+m_{2})

⇒ – 4 m_{1 }– 4m_{2} = – 14 m_{1} + 2 m_{2}

⇒ – 4m_{1} + 14 m_{1} = 2m_{2} + 4m_{2}

⇒ 10m_{1} = 6 m_{2}

⇒ m_{1}/m_{2} = 6/10 = 3/5

⇒ m_{1} : m_{2} = 3 : 5

Again b = {m_{1}×6 + m_{2}×(-2)}/(m_{1}+m_{2})

= (6m_{1} – 2m_{2})/(m_{1}+m_{2})

⇒ b = (6×3 – 2×5)/(3+5)

= (18 – 10)/8

= 8/8

= 1

b = 1

ML Aggarwal Section Formula Exe-11 Solutions

**ICSE Class-10 Maths Ch-11**

Page-223

**Question-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 the point k. Hence, find the ratio in which K divides AB. Also, find the coordinates of the point K. (2006)**

**Answer :**

Let the co-ordinates of K be (x, 0) as it intersects x-axis.

Let point K divides the line segment joining the points

A (2, 3) and B (6, -5) in the ratio m_{1} : m_{2}.

∴ 0 = (m_{1}y_{2} + m_{2}y_{1})/(m_{1}+m_{2})

⇒ 0 = {m_{1}×(-5) + m_{2}×3}/(m_{1}+m_{2})

⇒ -5m_{1} + 3m_{2} = 0

⇒ -5m_{1} = -3m_{2}

⇒ m_{1}/m_{2} = 3/5

⇒ m_{1} : m_{2} = 3 : 5

Now,

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

= (3×6 + 5×2)/(3 + 5)

= (18 + 10)/8

= 28/8

= 7/2

Co-ordinate of K are (7/2, 0)

**Question -15. ****If A ( – 4, 3) and B (8, – 6), **

**(i) find the length of AB.**

**(ii) in what ratio is the line joining AB, divided by the x-axis? (2008)**

**Answer :**

Given A (-4, 3), B (8, -6)

Let O divides AB in the ratio m_{1} : m_{2}

∴ x = (m_{1}x_{2} + m_{2}x_{1})/(m_{1}+m_{2})

⇒ 0 = {m_{1}×8 + m_{2}(-4)}/(m_{1} + m_{2})

⇒ 8m_{1} – 4m_{2} = 0

⇒ 8m_{1} = 4m_{2}

⇒ m_{1}/m_{2} = 4/8 = 1/2

∴ m_{1} : m_{2} = 1 : 2

∴ O, divides AB in the ratio 1 : 2

**Question -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 coordinates of the point of division.**

**Answer :**

Let the point A (3, – 1) and point B (8, 9) and let **the line x – y – 2 = 0 divide the line segment joining the points A (3, – 1) and B (8, 9) in the ratio m _{1} : m_{2} at point P (x, y) then**

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

= (m_{1}×8 + m_{2}×3)/(m_{1}+m_{2})

and y = (m_{1}y_{2} + m_{2}y_{1})/(m_{1}+m_{2})

= {m_{1}×9 + m_{2}(-1)}/(m_{1}+m_{2})

= (9m_{1} – m_{2})/(m_{1}+m_{1})

∵ The point P (x, y) lies on the line x – y – 2 = 0

∴ (8m_{1}+3m_{2})/(m_{1}+m_{2}) – (9m_{1 }– m_{2})/(m_{1}+m_{2}) – 2 = 0

⇒ 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

**∴ Co-ordinates of point P are (5, 3)**

**Question -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 coordinates of the point of intersection.**

**(iii)the length of AB. (2012)**

**Answer :**

**(i) Let the y-axis divide AB in the ratio m : 1.**

therefore

0 = (m×8 – 4×1)/(m+1)

⇒ 8m – 4 = 0

⇒ m = 4/8

⇒ m = 1/2

So, required ratio = 1/2 : 1 or 1 : 2

**(ii)** Also, y = {1×(-3) + 2×6}/(1+2)

= 9/3

= 3

So, coordinates of the point of intersection are (0, 3)

**(iii)**

** **

**Question -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)**

**Answer :**

Let D (x, y) be the median of ΔABC through A to BC.

∴ D will be the midpoint of BC

∴ Co-ordinates of D will be,

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

**Answer :**

Let O in the mid-point of AC the diagonal of ABCD

∴ Co-ordinates of O will be

{(1 + 4)/2, (2 + 0)/2} or (5/2, 1)

∵ OA also find the mid point of second diagonal BD and let co-ordinates of D be (x, y)

∴ 5/2 = (1 + x)/2

⇒ 10 = 2 + 2x

⇒ 2x = 10 – 2 = 8

∴ x = 8/2 = 4

And l = (0 + y)/2

⇒ y = 2

∴ Co-ordinates of D are (4, 2)

**Question- 20. ****If the points A ( – 2, – 1), B (1, 0), C (p, 3) and D (1, q) from a parallelogram ABCD, find the values of p and q.**

**Answer :**

A (-2, -1), B (1, 0), C (p, 3) and D (1, q)

are the vertices of a parallelogram ABCD

∴ Diagonal AC and BD bisect each other at O

O is the midpoint of AC as well as BD

Let co-ordinates of O be (x, y)

When O is mid-point of AC, then

Again when O is the mid-point of BD

Then x = (1 + 1)/2 = 2/2 = 1

And y = (0 + q)/2 = q/2

Now comparing, we get

(p – 2)/2 = 1

⇒ p – 2 = 2

⇒ p = 2 + 2 = 4

∴ p = 4 and q/2 = 1

⇒ q = 2

Hence p = 4, q = 2

**Question -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.**

**Answer :**

Two vertices of a ||gm ABCD are A (3, 2), B (-1, 0)

and point of intersection of its diagonals is P (2, -5)

P is mid-point of AC and BD.

Let co-ordinates of C be (x, y), then

2 = (x + 3)/2

⇒ x + 3 = 4

⇒ x = 4 – 3 = 1

And – 5 = (y + 2)/2

⇒ y + 2 = – 10

⇒ y = – 10 – 2 = – 12

∴ Co-ordinates of C are (1, – 12)

Similarly we shall find the co-ordinates of D also

2 = (x – 1)/2

⇒ x – 1 = 4

⇒ x = 4 + 1 = 5

– 5 = (y + 0)/2

⇒ – 10 = y

∴ Co-ordinates of D are (5, – 10)

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

**Answer:**

Let A (-1, 4) and B (5, 2) be the two points and let D (0, 3)

be its the midpoint of AC and co-ordinates of C be (x, y).

∴ 0 = (x – 1)/2

⇒ x – 1 = 0

⇒ x = 1

3 = (y + 4)/2

⇒ y + 4 = 6

⇒ y = 6 – 4 = 2

∴ Co-ordinates of will be (1, 2)

If we take mid-point D (0, 3) of BC, then

0 = (5 + x)/2

⇒ x + 5 = 0

⇒ x = – 5

and 3 = (2 + y)/2

⇒ 2 + y = 6

⇒ y = 6 – 2 = 4

∴ Co-ordinates of C will be (-5, 4)

Hence co-ordinates of C, third vertex will be (1, 2) or (5, -4)

**Question -23. ****Find the coordinates of the vertices of the triangle the middle points of whose sides are (0,1/2),(1/2,1/2), and (1/2,0)**

**Answer :**

Let ABC be a ∆ in which **(0,1/2),(1/2,1/2), and (1/2,0)**,

the mid-points of sides AB, BC and CA respectively.

Let co-ordinates of A be (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})

(x_{1} + x_{2})/2

⇒ x_{1 }+ x_{2} = 0 **…(i)**

1/2 = (y_{1} + y_{2})/2

⇒ y_{1} + y_{2} = 1 **…(ii)**

Again 1/2 = (x_{2 }+ x_{3})/2

⇒x_{2} + x_{3} = 1 **…(iii)**

And 1/2 = (y_{2} + y_{3})/2

⇒y_{2} + y_{3 }= 1 **…(iv)**

And 1/2 = (x_{3 }+ x_{1})/2

⇒ x_{3} + x_{1} = 1 **…(v)**

0 = (y_{3} + y_{1})/2

⇒ y_{3} + y_{1} = 0 **…(vi)**

Adding (i), (iii) and (v)

2(x_{1} + x_{2} + x_{3}) = 0 + 1 + 1 = 2

∴ x_{1 }+ x_{2} + x_{3} = 1

Now substituting (iii), (v) and (i) respectively, we get

x_{1} = 0, x_{2} = 0, x_{3} = 1

Again Adding (ii), (iv) and (vi)

2 (y_{1} + y_{2} + y_{3}) = 1 + 1 + 0 = 2

∴ y_{1} + y_{2} + y_{3} = 1

Now subtracting (iv), (vi) and (ii) respectively we get,

y_{1} = 0, y_{2} = 1, y_{3} = 0

∴ Co-ordinates of A, B and C will be (0, 0), (0, 1) and (1, 0)

**Question-24. ****Show by section formula that the points (3, – 2), (5, 2) and (8, 8) are collinear.**

**Answer :**

Let the point (5, 2) divides the line joining the points (3, -2) and (8, 8)

in the ratio of m_{1} : m_{2}

∴ 5 = (m_{1} × 8 + m_{2 }× 3)/(m_{1 }+ m_{2})

⇒ 8m_{1} + 3m_{2} = 5m_{1} + 5m_{2}

⇒ 8m_{1} – 5m_{1}

⇒ 5m_{2} – 3m_{2}

⇒ 3m_{1} = 2m_{2}

⇒ m_{1}/m_{2} = 2/3 **….(i)**

Again 2 = (8m_{1} – 2m_{2})/(m_{1}+m_{2})

⇒ 8m_{1} – 2m_{2 }= 2m_{1} + 2m_{2}

⇒ 8m_{1} – 2m_{1} = 2m_{2} + 2m_{2}

⇒ 6m_{1 }= 4m_{2}

⇒ m_{1}/m_{2} = 4/6 = 2/3 **…(ii)**

From (i) and (ii) it is clear that point (5, 2) lies on the line joining the points (3, – 2) and (8, 8).

Hence, proved.

**Question -25. ****Find the value of p for which the points ( – 5, 1), (1, p) and (4, – 2) are collinear.**

**Answer:**

Let points A (-5, 1), B (1, p) and C (4, -2)

are collinear and let point A (-5, 1) divides

BC in the ratio in m_{1} : m_{2}

∴ x = (m_{1}x_{2} + m_{2}x_{1})/(m_{1 }+ m_{2})

⇒ -5 = (m_{1}×4 + m_{1}×1)/(m_{1} + m_{2}) = (4m_{1} + m_{2})/(m_{1} + m_{2})

⇒ – 5m_{1} – 5m_{2 }= 4m_{1} + m_{2}

⇒ – 5m_{1} – 4m_{1} = m_{2} + 5m_{2}

⇒ – 9m_{1} = 6m_{2}

⇒ m_{1}/m_{2 }= 6/-9 = 2/-3 **…(i)**

And {m_{1}×(-2) + m_{2}×p}/(m_{1}+m_{2})

= (-2m_{1} + m_{2}p)/(m_{1}+m_{2})

⇒ m_{1} + m_{2 }= -2m_{1} + m_{2}p

⇒ m_{1} + 2m_{1} = m_{2}p – m_{2}

⇒ 3m_{1} – m_{2} (p – 1)

⇒ m_{1}/m_{2} = (p – 1)/3 **…(ii)**

From (i) and (ii)

(p – 1)/3 = 2/(-3)

⇒ – 3p + 3 = 6

⇒ – 3p = 6 – 3

⇒ – 3p = 3

⇒ p = 3/-3

= – 1

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

**Answer :**

A lies on x-axis and B on the y-axis.

Let co-ordinates of A be (x, 0) and of B be (0, y)

P (4, -3) is the mid-point of AB

∴ 4 = (x + 0)/2

⇒ x = 8

And – 3 = (0 + y)/2

⇒ y = – 6

Co-ordinates of A will be (8, 0) and of B will be (0, – 6)

**ML Aggarwal Section Formula Exe-11 Solutions**

ICSE Class-10 Maths Ch-11

Page-224

**Question -27. ****Find the co-ordinates of the centroid of a triangle whose vertices are A ( – 1, 3), B(1, – 1) and C (5, 1) (2006)**

**Answer :**

Co-ordinates of the centroid of a triangle,

whose vertices are (x1, y1), (x2, y2) and

(x_{3}, y_{3}) are {(x_{1} + x_{2} + x_{3})/3 , (y_{1} + y_{2} + y_{3})/3}

∴ Co-ordinates of the centroid of the given triangle are {(-1 + 1 + 5)/3, (3 – 1 + 1)/2} i.e., (5/3, 1)

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

**Answer :**

Let the co-ordinates of third vertices be (x, y)

and other two vertices are (3, -5) and (-7, 4)

and centroid = (2, -1).

∴ 2 = (3 – 7 + x)/3

⇒ (x – 4)/3 = 2

x – 4 = 6

⇒ x = 6 + 4

⇒ x = 10

And ⇒ – 1 = (- 5 + 4 + y)/3

⇒ – 3 = – 1 + y

⇒ y = – 3 + 1

= 2

∴ Co-ordinates are (10, – 2)

**Question -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 centroid of the triangle ABC is the point (1, – 1).**

**Answer :**

The vertices of ∆ABC are A (-5, 3), B (p, -1), C (6, q)

and the centroid of ∆ABC is O (1, -1)

co-ordinates of the centroid of ∆ABC will be

**[(-5 + p + 6)/3, (3–1+q)/3]**

⇒ {(1+p)/3, (2+q)/3}

But centroid is given (1, -1)

∴ Comparing, we get (1 + p)/3 = 1

⇒ 1 + p = 1

⇒ 1 + p = 3

⇒ p = 3 – 1 = 2

And (2 + q)/3 = – 1

⇒ 2 + q = – 3

⇒ q = – 3 – 2

⇒ q = – 5

Hence,

p = 2, q = – 5

— : End of ML Aggarwal Section Formula Exe-11 Solutions ICSE Class-10 Maths Ch-11 : –

Return to :- ML Aggarwal Solutions for ICSE Class-10

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