# ML Aggarwal Solutions for Chapter 11 Section Formula Class 10 Maths ICSE

### Exercise 11

**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)}

**2. The co-ordinates of two points A and B are (- 3, 3) and (12, -7) respectively. P is a point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of P.**

**Answer**

Points are A (-3, 3), B (12, - 7)

Let P (x_{1}, y_{1}) be the points which divides AB in the ratio of m_{1} : m_{2} i.e. 2 : 3 then co-ordinates of P will be

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

= {2×12 + 3×(-3)}/(2×3)

= (24 – 9)/5

= 15/5

= 3

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

= {2×(-7) + 3×3}/(2+3)

= (-14 + 9)/5

= -5/5

= -1

∴ Co-ordinates of P are (3, -1)

**3. 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).

**4. (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 (5/3, q) 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 intersect 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)

Again

∵ 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)** Points 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

**5. (i) 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. **

**(ii) A point P divides the line segment joining the points A (3, -5) and B (- 4, 8) such that AP/PB = k/1. If P lies on the line x + y = 0, then find the value of k. **

**Answer**

**(i) **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

**(ii)** A point P divides the line segment joining the points A (3, -5), B (-4, 8) such that AP/BP = k/1

∴ Ratio = AP : PB = k : 1

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

x = (mx_{2} + nx_{1})/(m + n)

= {k×(-4) + 1×3}/(k+1)

x = (- 4k + 3)/(k+1)

and y = (8k – 5)/(k+1) **{∵ (my _{2 }+ my_{1})/(m + n)}**

= (8k – 5)/(k+1)

∵ This point lies on the line x + y = 0

∴ {(- 4k + 3)/(k + 1)} + {(8k – 5)/(k + 1)} = 0

⇒ - 4k + 3 + 8k – 5 = 0

⇒ 4k – 2 = 0

⇒ 4k = 2

⇒ k = 2/4 = 1/2

**6. 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)

**7. Point P (3 -5) is reflected to P’ in the x-axis. Also P on reflection in the y-axis is mapped as P’’. **

**(i) Find the co-ordinates**** of P’ and P’’.**

**(ii) Compute the distance P’ P’’. **

**(iii) Find the middle point of the line segment P’ P’’. **

**(iv) On which co-ordinate axis does the middle point of the line segment P P’’ lie ? **

**Answer**

**(i)** Co-ordinates of P’, the image of P (3, -5) when reflected in x-axis will be (3, 5) and co-ordinates of P’’, the image of P (3, -5) when reflected in y-axis will be (- 3, -5) when reflected in y-axis will be (-3, - 5)

**(ii)**

**(iii)** Let co-ordinates of middle point M be (x, y)

∴ x = (x_{1} + x_{2})/2

= (3 – 3)/2

= 0/2

= 0

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

= (-5 + 5)/2

= 0/2

= 0

∴ middle point is (0, 0)

**(iv)** Middle point of P P’’ be N (x_{1}, y_{1})

∴ x_{1} = (3 – 3)/2 = 0/2 = 0

x_{2} = (- 5 – 5)/2

= -10/2

= -5

∴ Co-ordinates of middle point of PP" are (0, -5)

As x = 0, this point lies on y-axis.

**8. Use graph paper for this question. Take 1 cm = 1 unit on both axes. Plot the points A (3, 0) and B (0, 4). **

**(i) Write down the co-ordinates of A1, the reflection of A in the y-axis.**

**(ii) Write down the co-ordinates of B1, the reflection of B in the x-axis. **

**(iii) Assign the special name to the quadrilateral ABA1B1. **

**(iv) If C is the mid point is AB. Write down the co-ordinates of the point C1, the reflection of C in the origin. **

**(iv) Assign the special name to quadrilateral ABC1B1. **

**Answer**

Two points A (3, 0) and B (0, 4) have been plotted on the graph.

(i) ∵ A1 is the reflection of A (3, 0) in the y-axis. Its co-ordinates will be (-3, 0).

(ii) ∵ B1 is the reflection of B(0, 4) in the x-axis co-ordinates of B, will be (0, - 4).

(iii) The so formed figure ABA1B1 is a rhombus.

(iv) C is the mid point of AB co-ordinates of C’’ will be (AP/AB) = 3/4 ,

∵ C, is the reflection of C in the origin

Co-ordinates of C, will be (-3/2, -2)

(v) The name of quadrilateral ABC1B1 is a trapezium because AB is parallel to B1C1.

**9. 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)

**10.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

**11. 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)

**12. AB is a diameter of a circle centre C (-2, 5). If point A is (3, -7). Find : **

**(i) the length of radius AC. **

**(ii) the coodinates of B. **

**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)

**13. 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 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).

**14. 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)

**15. 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)

**16. (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 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}.

∴ -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

**17. 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. **

**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)

**18. 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. **

**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

**19. (i) Calculate the ratio in which the line segment joining (3, 4) and (- 2, 1) is divided by the x-axis. **

**(ii) 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**

**(i)** Let the point P divided the line segment joining the points A (3, 4) and B (-2, 3) in the ratio of m_{1} : m_{2} and

Let the co-ordinates of P be (0, y) as it intersects the y-axis

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

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

⇒ 0 = - 2m_{1 }+ 3m_{2 }

⇒ 2m_{1} = 3m_{2 }

⇒ m_{1}/m_{2} = 3/2

⇒ m_{1 }: m_{2} = 3 : 2

**(ii)** Let the points be A (3, -1) and B (8, 9) and let line x – y – 2 = 0 divides the line segment joining the points A and B 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

**(i)** ∵ Ratio = m_{1} : m_{2 }= 2 : 3

∴ x = (2×8 + 3×3)/(2+3)

= (16 + 9)/5

= 25/5

= 5

And y = {2×9 + 3×(-1)}/(2+3)

= (18 – 3)/5

= 15/5

= 3

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

**20. 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. **

**Answer**

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

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) **

**21. (i) Write down the co-ordinates of the point P that divides the line joining A (- 4, 1) and B (17, 10) in the ratio 1 : 2. **

**(ii) Calculate the distance OP where O is the origin. **

**(iii) In what ratio does the y-axis divide the line AB ? **

**Answer**

**(i)** Let co-ordinate of P be (x, y) which divides the line segment joining the points

A (- 4, 1) and B (17, 10) in the ratio of 1 : 2.

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

= (1×17 + 2×(-4)}/(1+2)

= (17–8)/3

= 9/3

= 3

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

= (1×10 + 2×1)/(1+2)

= (10 + 2)/3

= 12/3

= 4

∴ Co-ordinates of P are (3, 4)

**(ii)** Distance of OP where O is the origin i.e., co-ordinates are(0, 0)

**(iii)** Let y-axis divides AB in the ratio of m1 : m2 at P and let co-ordinates of P be (0, y)

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

⇒ 0 = {m_{1}×17 + m_{2}×(-4)}/(m_{1}+m_{2})

⇒ 17m_{1} – 4m_{2} = 0

⇒ 17m_{1} = 4m_{2 }

⇒ m_{1}/m_{2 }= 4/17

⇒ m_{1} : m_{2} = 4 : 17

**22. 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 ∆ABC through A to BC.

∴ D will be the midpoint of BC

∴ Co-ordinates of D will be,

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

and y = (3 – 1)/2 = 2/2 = 1

Co-ordinates of D are (4, 1)

**23. 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)

**24. 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

**25. If two vertices of a parallelogram are (2, 3) (-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)

**26. Prove that the points A (- 5, 4), B (- 1, - 2) and C (5, 2) are the vertices of an isosceles right angled triangle. Find the co-ordinates of D so that ABCD is a square. **

**Answer: **

Points A (-5, 4), B (- 1, -2) and C (5, 2) are given.

If these are vertices of an isosceles ABC then

AB = BC

∵ AB = BC

∴ ∆ABC is an isosceles triangle.

^{2}= AB

^{2 }+ BC

^{2}

^{}

**27. 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 it’s 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)

**28. 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 D (0, 1/2), E = (1/2, 1/2) and F = (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)

**29. 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.

**30. 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 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

**31. A (10, 5), B (5, -3) and C (2, 1) are the vertices of triangle ABC. L is the mid point of AB, M is the mid-point of AC. Write down the co-ordinates of L and M. Show that LM = 1/2 BC. **

**Answer**

Co-ordinates of L will be

{(10 + 6)/2, (5 – 3)/2} or (16/2, 2/2) or (8, 1)

Co-ordinates of M will be ={(10 + 2)/2, (5 + 1)/2} or (12/2, 6/2) or (6, 3)

= 4√2 unitsFrom (i) and (ii)

LM = 1/2 BC

**32. A (2, 5), B (- 1, 2) and C (5, 8) are the vertices of a triangle ABC. P and Q are points on AB and AC respectively such that AP : PB = AQ : QC = 1 : 2. **

**(i) Find the co-ordinates of P and Q**

**(ii) Show that PQ = 1/3 BC. **

**Answer**

A (2, 5), B (-1, 2) and C (5, 8) are the vertices of a ∆ABC,

P and Q are points on AB

And AC respectively such that AP/PB = AQ/QC = 1/2

Let co-ordinates of P be (x_{1}, y_{1}) and of Q be (x_{2}, y_{2})

∵ P divides AB in the ratio 1 : 2

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

= {1×(-1) + 2×2}/(1 + 2)

= (- 1 + 4)/3

= 3/3

= 1

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

= (1×2 + 2×5)/(1+2)

= (2 + 10)/3

= 12/3

= 4

∴ Co-ordinates of P will be (1, 4)

Similarly Q divides AC in the ratio 1 : 2

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

= (1×5 + 2×2)/(1 + 2)

= (5 + 4)/3

= 9/3

= 3

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

= (1×8 + 2×5)/(1+2)

= (8+10)/3

= 18/3

= 6

∴ Co-ordinates of Q will be (3, 6)

(ii)

= (6√2)/3

= BC/3

= 1/3 BC

**33. 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)

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

**Answer: **

Co-ordinates of the centroid of a triangle, whose vertices are (x_{1}, y_{1}), (x_{2}, y_{2}) 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)

**35. 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)

**36. 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

**Multiple Choice Questions**

**Choose the correct answer from the given four options (1 to 12): **

**1. The points A (9, 0), B (9, 6), C(- 9, 6) and D (- 9, 0) are the vertices of a **

**(a) rectangle **

**(b) square **

**(c) rhombus**

**(d) trapezium **

**Answer**

(a) rectangle

A(9, 0), B(9, 6), C(-9, 6), D(-9, 0)

AB^{2} = (x_{2} – x_{1})^{2 }+ (y_{2} – y_{1})^{2 }

= (9 – 9)^{2 }+ (6 – 0)^{2 }

= 0^{2} + 6^{2}

= 0^{2} + 36

= 36

CD^{2} = (-9 + 9)^{2} + (6 – 0)^{2}

= 0^{2} + 6^{2 }

= 0 + 36

= 36

BC^{2} = (9 + 9)^{2} + (6 – 6)^{2}

= 18^{2} + 0^{2}

= 324 + 0

= 324

AD^{2} = (9 + 9)^{2} + (0)^{2}

= 18^{2} + 0^{2}

= 324 + 0

= 324

AB = CD and BC = AD

But these are opposite sides of a rectangle

ABCD is a rectangle.

**2. The mid-point of the line segment joining the points A (- 2, 8) and B (- 6, - 4) is **

**(a) (-4, -6) **

**(b) (2, 6) **

**(c) (-4, 2) **

**(d) (4, 2) **

**Answer**

(b) (2, 6)

Mid-point of the line segment joining the points A (- 2, 8), B (- 6, - 4)

= {(-2+6)/2, (8+4)/2}

= (4/2, 12/2)

= (2, 6)

**3. If P (a/3, 4) segment joining the points Q (-6, 5) and R (-2, 3), the value of a is **

**(a) –4 **

**(b) –6 **

**(c) 12 **

**(d) –12 **

**Answer**

(d) -12

P (a/3, 4) is mid-point of the line segment joining the points Q (-6, 5) and R (-2, 3)

∴ a/3 = (-6-2)/2

= -8/2 = -4

a = -4 × 3

⇒ a = -12

**4. If the end points of a diameter of a circle are A (- 2, 3) and B (4, - 5), then the coordinates of its centre are **

**(a) (2, -2)**

**(b) (1, -1)**

**(c) (-1, 1)**

**(d) (-2, 2) **

**Answer**

(b) (1, -1)

End points of a diameter of a circle are (- 2, 3) and B (4, - 5) then co-ordinates of the centre of the circle = {(- 2 + 4)/2, (3 – 5)/2} or (2/2, -2/2)

= (1, -1)

**5. If one end of a diameter of a circle is (2, 3) and the centre is (- 2, 5), then the other end is **

**(a) (-6, 7) **

**(b) (6, -7) **

**(c) (0, 8) **

**(d) (0, 4) **

**Answer**

(a) (-6, 7)

One end of a diameter of a circle is (2, 3) and center is (-2, 5)

Let (x, y) be the other end of the diameter (2 + x)/2 = - 2

⇒ 2 + x = - 4

⇒ x = -4-2 = - 6

And (3+y)/2 = 5

⇒ 3 + y = 10

⇒ y = 10 – 3

= 7

∴ Co-ordinates of other end are (-6, 7)

**6. If the mid-point of the line segment joining the points P (a, b–2) and Q (-2, 4) is R (2, -3), then the values of a and b are **

**(a) a = 4, b = -5 **

**(b) a = 6, b = 8 **

**(c) a = 6, b = -8 **

**(d) a = -6, b = 8 **

**Answer**

(c) a = 6, b = -8

The mid-point of the line segment joining the points P (a, b – 2) and Q (- 2, 4) is R (2, -3)

2 = (a – 2)/2

⇒ a – 2 = 4

⇒ a = 4 + 2 = 6

- 3 = (b – 2 + 4)/2 = (b + 2)/2

⇒ b + 2 = - 6

⇒ b = - 6 – 2 = - 8

∴ a = 6, b = - 8

**7. The point which lies on the perpendicular bisector of the line segment joining the points A (- 2, - 5) and B (2, 5) is **

**(a) (0, 0) **

**(b) (0, 2) **

**(c) (2, 0) **

**(d) (- 2, 0) **

**Answer**

(a) (0, 0)

The line segment joining the points A (- 2, - 5) and B (2, - 5), has mid-point = {(-2 + 2)/2, (-5 + 5)/2} = (0, 0)

(0, 0) lies on the perpendicular bisector of AB.

**8. The coordinates of the point which is equidistant from the three vertices of ∆AOB (shown in the given figure) are **

**(a)**

**(b) (y, x) **

**(c) (x/2, y/2) **

**(d) (y/2, x/2) **

**Answer**

(a) (x, y)

In the given figure, vertices of a ∆OAB are (0, 0), (0, 2y) and (2x, 0)

The point which is equidistant from O, A and B is the mid-point of AB.

∴ Coordinates are {(0 + 2x)/2, (2y + 0)/2} or (x, y)

**9. The fourth vertex D of a parallelogram ABCD whose three vertices are A (- 2, 3), B (6, 7) and C (8, 3) is**

**(a) (0, 1) **

**(b) (0, - 1) **

**(c) (- 1, 0) **

**(d) (1, 0) **

**Answer**

(b) (0, -1)

ABCD is a || gm whose vertices A (- 2, 3), B (6, 7) and C (8, 3)

The fourth vertex D will be the point on which diagonals AC and BD bisect each other at O.

∴ Co-ordinates of O are (-2+8)/2, (3+3)/2 or (6/2, 6/2) or (3, 3)

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

3 = (x + 6)/2 = 6

= x + 6

⇒ x = 6 – 6 = 0

And 3 = (y + 7)/2

⇒ y + 7 = 6

⇒ y = 6 – 7 = - 1

∴ Co-ordinates of D are (0, - 1)

**10. A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, - 5) is the mid-point of PQ, then the coordinates of P and Q respectively. **

**(a) (0, - 5) and (2, 0) **

**(b) (0, 10) and (- 4, 0) **

**(c) (0, 4) and (- 10, 0) **

**(d) (0, - 10) and (4, 0) **

**Answer**

(d) (0, - 10)

A line intersects y-axis at P and x-axis a Q.

R (2, - 5) is the mid-point

Let co-ordinates of P be (0, y) and of Q be (x, 0), then

2 = (0 + x)/2

⇒ x = 4

And – 5 = (y + 0)/2

⇒ y = - 10

∴ Co-ordinates of P are (4, 0) and of Q are (0, - 10)

**11. The points which divides the line segment joining the points (7, - 6) and (3, 4) in the ratio 1 : 2 internally lies in the **

**(a) I ^{st} quadrant **

**(b) IInd quadrant **

**(c) IIIrd quadrant **

**(d) IVth quadrant **

**Answer**

(d) IVth quadrant

A point divides line segment joining the points

A (7, - 6) and B (3, 4) in the ratio 1 : 2 internally.

Let (x, y) divides it in the ratio 1 : 2

∴ x = (mx_{2} + nx_{1})/(m + n)

= (1 × 3 + 2 × 7)/(1 + 2)

= (3 + 14)/3

= 17/3

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

= (1 × 4 + 2 × (-6)/(1 + 2)

= (4 – 12)/3

= -8/3

We see that x is positive and y is negative.

∴ It lies in the fourth quadrant.

**12. The centroid of the triangle whose vertices are (3, - 7), (- 8, 6) and (5, 10) is **

**(a) (0, 9) **

**(b) (0, 3)**

**(c) (1, 3)**

**(d) (3, 3) **

**Answer**

(b) (0,3)

Centroid of the triangle whose vertices are (3, -7), (-8, 6) and (5, 10) is {(3–8+5)/3, (-7+6+10)/3}

or (0, 9/3)

or (0, 3)

**Chapter Test**

**1. The base BC of an equilateral triangle ABC lies on y-axis. The co-ordinates of the point C are (0, - 3). If origin is the mid-point of the base BC, find the coordinates of the point A and B. **

**Answer**

Base BC of an equilateral ∆ABC lies on y-axis co-ordinates of point C are (0, - 3) origin (0, 0) is the mid-point of BC.

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

∴ 0 = (x + 0)/2

⇒ x/2 = 0

⇒ x = 0

(y – 3)/2 = 0

⇒ y – 3 = 0

⇒ y = 3

∴ Co-ordinates of B are (0, 3)

Again let co-ordinates of A be (x, 0) as it lies on x-axis.

∵ AB = AC = BC = 6 units

x^{2 }+ (- 3)^{2} = 6^{2}

x^{2 }+ 9 = 36

⇒ x^{2} = 36 – 9 = 27

⇒ x = *± 3**√**3*

∴ Co-ordinates of A will be (±3√3, 0)

**2. A and B have co-ordinates (4, 3) and (0, 1), find**

**(i) the image A’ of A under reflection in the y-axis.**

**(ii) the image of B’ of B under reflection in the line AA’. **

**(iii) the length of A’B’. **

**Answer**

**(i)**Co-ordinates of A’, the image of A (4, 3) reflected in y-axis will be (- 4, 3).

**(ii)** Co-ordinates of B’ the image of B (0, 1) reflected in the line AA’ will be (0, 5)

**(iii)**

**3. Find the co-ordinates of the point that divides the line segment joining the points P (5, - 2) and Q (9, 6) internally in the ratio of 3 : 1.**

**Answer**

Let R be the point whose co-ordinates are (x, y) which divides PQ in the ratio of 3 : 1.

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

= (3×9 + 1×5)/(3 + 1)

= (27+5)/4

= 32/4

= 8

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

= {3×6 + 1×(-2)}/(3+1)

= (18–2)/4

= 16/4

= 4

∴ Co-ordinates of R will be (8, 4)

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

**Answer**

Co-ordinates of A (3, 1) and B (- 2, 5)

P lies on AB such that

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

⇒ AP = 3PB

⇒ AP : PB = 3 : 1

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

∴ x = (mx_{2} + nx_{1})/(m + n) = {(3 × (-2) + 1 × 3}/(3 + 1)

= (-6 + 3)/4

= -3/4

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

= (3×5 + 1×1)/(3+1)

= (15+1)/4

= 16/4

= 4

∴ Co-ordinates of Pare (-3/4, 4)

**5. P and Q are the points on the line segment joining the points A (3, -1) and B (- 6, 5) such that AP = PQ = QB. Find the co-ordinates of P and Q. **

**Answer**

Given

AP = PQ = QB

∴ P divides AB in the ratio of 1 : 2 and Q divides it in 2 : 1.

Let co-ordinates of P will be (x_{1}, y_{1}) and of Q will (x_{2}, y_{2})

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

= {1×(-6) + 2×3}/(1+2)

= (- 6 + 6)/3

= 0/3

= 0

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

= {1×5 + 2(-1)}/(1+2)

= (5-2)/3

= 3/3

= 1

∴ Co-ordinates of P will be (0, 1)

Again

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

= {2×(-6) + 1×3}/(2+1)

= (-12+3)/3

= -9/3

= - 3

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

= {2×5 + 1×(-1)}/(2+1)

= (10–1)/3

= 9/3

= 3

∴ Co-ordinates of Q will be (-3, 3).

∴ Co-ordinates of Q will be (-3, 3).

**6. The center of a circle is (a + 2, a – 5). Find the value of a given that the circle passes through the points (2, - 2) and (8, -2). **

**Answer**

Let A (2, -2), B (8, - 2) and center of the circle be O (𝝰 + 2, 𝝰 – 5)

∵ OA = OB = radii of the same will

Squaring both sides,

𝝰^{2 }+ (𝝰 + 1)^{2} = (6 – 𝝰)^{2} = (1 + a)^{2}

⇒ a^{2} = (6 – 𝝰)^{2 }** [dividing by (𝝰 + 1) ^{2}]**

⇒ 𝝰^{2} = 36 - 12𝝰 + 𝝰^{2}

= 𝝰^{2 }– 𝝰^{2} + 12𝝰

= 36

⇒ 12𝝰 = 36

∴ = 36/12 = 3

**7. The mid-point of the line joining A (2, p) and B (q, 4) is (3, 5) . Calculate the numerical values of p and q.**

**Answer**

Given,

(3, 5) is the mid-point of A (2, p) and B (q, 4)

∴ 3 = (2 + q)/2

⇒ 2 + q = 6

⇒ q = 6 – 2 = 4

∴ q = 4

And 5 = (p + 4)/2

⇒ p + 4 = 10

⇒ p = 10 – 4 = 6

∴ p = 6

Hence, p = 6, q = 4

**8. The ends of a diameter of a circle have the co-ordinates (3, 0) and (- 5, 6). PQ is another diameter where Q has the coordinates (- 1, - 2). Find the co-ordinates of P and the radius of the circle. **

**Answer**

Let AB be the diameter where co-ordinates of A are (3, 0) and of B are (- 5, 6).

Co-ordinates of its origin O will be {(3 – 5)/2, (0 + 6)/2} or (-2/2, 6/2) or (- 1, 3)

Now PQ is another diameter in which co-ordinates of Q are (-1, -2).

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

Then co-ordinates of center O will be {(- 1 + x)/2, (-2 + y)/2}

∴ (-1+x)/2 = - 1

⇒ - 1+x = - 2

⇒ x = -2 + 1 = - 1

And (- 2 + y)/2 = 3

⇒ - 2 + y = 6

⇒ y = 6 + 2 = 8

∴ Co-ordinates of P will be (-1, 8)

**9. In what ratio does the point (- 4, 6) divide the line segment joining the points A (- 6, 10) and B (3, - 8) ?**

**Answer**

Let the point (- 4, 6) divides the line segment joining the points

A (- 6, 10) and B (3, - 8), in the ratio m : n

∴ - 4 = (mx_{2} + nx_{1})/(m + n) = {m × 3 + n(-6)}/(m + n)

- 4 = (3m – 6n)/(m + n)

⇒ - 4m – 4n = 3m – 6n

⇒ - 4n + 6n = 3m + 4m

⇒ 7m = 2n

⇒ m/n = 2/7

∴ Ratio = 2 : 7

**10. Find the ratio in which the point P (- 3, p) divides the line segment joining the points (- 5, - 4) and (-2, 3). Hence find the value of p. **

**Answer**

Let P (- 3, p) divides AB in the ratio of m_{1} : m_{2} coordinates of A (- 5, - 4) and B (- 2, 3)

∴ - 3 = (m_{1}x_{2} + m_{2}x_{1})/(m_{1} + m_{2})

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

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

⇒ - 3m_{1} – 3m_{2} = - 2m_{1 }– 5m_{2 }

⇒ - 3m_{1} + 2m_{1} = - 5m_{2} + 3m_{2}

⇒ - m_{1} = - 2m_{2}

⇒ 2m_{2} = m_{1}

⇒ m_{1}/m_{2} = 2/1

⇒ m_{1}: m_{2} = 2 : 1

Again,

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

= {(2×3 + 1×(-4)}/(2 + 1)

= (6 – 4)/3

= 2/3

Hence, p = 2/3

**11. In what ratio is the line joining the points (4, 2) and (3, - 5) divided by the x-axis? Also find the co-ordinates of the point of division. **

**Answer**

Let the point P which is on x-axis, divides the line segment

Joining the points A (4, 2) and B (3, - 5) in the ratio of m_{1} : m_{2}.

and let co-ordinates of P be (x, 0)

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

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

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

⇒ - 5m_{1} + 2m_{2} = 0

⇒ - 5m_{1} = - 2m_{2}

⇒ 5m_{1} = 2m_{2 }

⇒ m_{1}/m_{2 }= 2/5

⇒ m_{1} : m_{2} = 2 : 5

Again,

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

= {2×3 + 5×4}/(2+5)

= (6 + 20)/7

= 26/7

∴ Co-ordinates of P will be (26/7, 0)

**12. If the abscissa of a point P is 2, find the ratio in which it divides the line segment joining the points (- 4, - 3) and (6, 3). Hence, find the co-ordinates of P. **

**Answer**

Let co-ordinates of A be (- 4, 3) and of B (6, 3) and of B (6, 3) and P be (2, y)

Let the ratio in which the P divides AB be m_{1} : m_{2}

_{1}x

_{2}+ m

_{2}x

_{1})/(m

_{1}+ m

_{2})

⇒ 2 = (m_{1}×6 + m_{2}×(-4)}/(m_{1}+m_{2})

⇒ 2 = (6m_{1} – 4m_{2})/(m_{1} + m_{2})

⇒ 2m_{1} + 2m_{2} = 6m_{1 }– 4m_{2}

⇒ 6m_{1} – 2m_{1} = 2m_{2} + 4m_{2}

⇒ 4m_{1} = 6m_{2}

⇒ m_{1}/m_{2} = 6/4 = 3/2

∴ m_{1} : m_{2} = 3 : 2

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

= (3×3 + 2×3)/(3+2)

= (9 + 6)/5

= 15/5

= 3

∴ Co-ordinates of P will be (-2, 3)

**13. Determine the ratio in which the line 2x + y – 4 = 0 divide the line segment joining the points A (2, - 2) and B (3, 7). Also find the co-ordinates of the point of division. **

**Answer**

Points are given A (2, - 2), B (3, 7)

And let the line 2x + y – 4 = 0 divides AB in the ratio m_{1} : m_{2 }

At P and let co-ordinates of

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

= (m_{1}×3 + m_{2}×2)/(m_{1} + m_{2})

= (3m_{1} + 2m_{2})/(m_{1} + m_{2})

And y = {m_{1}×7 + m_{2}×(-2)}/(m_{1}+m_{2})

= {7m_{1} – 2m_{2}}/(m_{1}+m_{2})

∴ P lies on the line 2x + y – 4 = 0, then

2(3m_{1} + 2m_{2})/(m_{1}+m_{2}) + (7m_{1} – 2m_{2})/(m_{1}+m_{2}) – 4 = 0

⇒ 6m_{1 }+ 4m_{2 }+ 7m_{1} – 2m_{2} – 4m_{1 }– 4m_{2} = 0

⇒ 9m_{1} – 2m_{2} = 0

⇒ 9m_{1} = 2m_{2}

⇒ m_{1}/m_{2} = 2/9

or m_{1} : m_{2} = 2 : 9

∴ x = (2×3 + 2×9)/(2 + 9)

= (6 + 18)/11

= 24/11

And y = (2 × 7 – 2 × 9)/(2 + 9)

= (14 – 18)/11

= -4/11

∴ Co-ordinates of P will be (24/11, -4/11)

**14. The point A (2, -3) is reflected in the y-axis onto the point A’. Then the point A’ is reflected in the line x = 4 onto the point A’’. **

**(i) Write the coordinates of A’ and A’’. **

**(ii) Find the ratio in which the line segment AA’’ is divided by the x-axis. Also find the coordinates of the point of division. **

**Answer**

A’ is the reflection of A (2, -3) in the x-axis

**(i)** ∴ Co-ordinates of A’ will be (2, 3)

Draw a line x = 4 which is parallel to y-axis

A’’ is the reflection of A’ (2, 3)

∴ Co-ordinates OA’’ will be (6, 3)

**(ii)** Join AA’’ which intersect x-axis at P whose co-ordinates are (4, 0)

Let P divide AA’’ in the ratio in m_{1} : m_{2}

_{1}y

_{2}+ m

_{2}y

_{1})/(m

_{1}+ m

_{2})

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

⇒ 3m_{1 }– 3m_{2} = 0

⇒ 3m_{1} = 3m_{2}

⇒ m_{1}/m_{2} = 3/3 = 1/1

∴ m_{1 }: m_{2} = 1 : 1

Hence P (4, 0) divides AA’’ in the ratio 1 : 1

**15. ABCD is a parallelogram. If the coordinates of A, B and D are (10, - 6), (2, - 6) and (4, - 2) respectively, find the co-ordinates of C. **

**Answer**

Let the co-ordinates of C be (x, y) and other three vertices of the given parallelogram are A (10, - 6), B (2, - 6) and D (4, - 2)

∴ ABCD is a parallelogram

Its diagonals bisect each other.

Let AC and BD intersect each other at O.

∴ O is mid-point of BD

∴ Co-ordinates of O will be

{(2+4)/2, (-6–2)/2) or (6/2, -8/2) or (3, -4)

Again, O is the mid-point of AC then

3 = (10 + x)/2

⇒ 10 + x = 6

⇒ x = 6 – 10 = - 4

And – 4 = (-6 + y)/2

⇒ 6 + y = - 8

⇒ - 8 + 6

∴ y = - 2

Hence Co-ordinates of C will be (- 4, - 2).

**16. ABCD is a parallelogram whose vertices A and B have co-ordinates (2, - 3) and (- 1, - 1) respectively. If the diagonals of the parallelogram meet at the point M (1, - 4), find the co-ordinates of C and D. Hence, find the perimeter of the parallelogram. Find the perimeter of the parallelogram. **

**Answer**

ABCD is a ||gm, m which co-ordinates of A are (2, - 3) and B (- 1, -1)

Its diagonals AC and BD bisect each other at M (1, - 4)

∴ M is the midpoint of AC and BD

Let co-ordinates of C be (x_{1}, y_{1}) and D be (x_{2}, y_{2})

When M is the midpoint of AC then

∴ l = (2 + x_{1})/2 and – 4 = (-3 + y_{1})/2

⇒ 2 + x_{1} = 2

⇒ x_{1} = 2 – 2 = 0

And –8 = -3 + y_{1}

⇒ y_{1} = -8 + 3 = -5

∴ Co-ordinates of C are (0, -5)

Again M is mid-point of BD, then

1 = (-1 + x_{2})/2, -4 = (-1 + y_{2})/2

⇒ -1 + x_{2} = 2

⇒ x_{2} = 2 + 1 = 3

And –1 + y_{2} = -8

⇒ y_{2} = -8 + 1 = -7

∴ Co-ordinates of D are (3, -7)

**17. In the adjoining figure, P (3, 1) is the point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of A and B.**

**Answer**

A lies on x-axis and

B lies on y-axis

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

And P (3, 1) divides it in the ratio of 2 : 3.

∴ 3 = (m_{1}x_{2} + m_{2}x_{1})/(m_{1} + m_{2})

= (2×0 + 3×x)/(2+3)

= (0 + 3x)/5

⇒ 3x = 15

⇒ x = 15/3 = 5

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

= (2×y + 3×0)/(2 + 3)

= (2y + 0)/5

= 2y/5

⇒ 2y = 5

⇒ y = 5/2

∴ Co-ordinates of A will be (5, 0) and of B will be (0, 5/2)

**18. Given, O (0, 0), P (1, 2), S (- 3, 0) P divides OQ in the ratio of 2 : 3 and OPRS is a parallelogram. Find : **

**(i) the co-ordinates of Q. **

**(ii) the co-ordinates of R. **

**(iii) the ratio in which RQ is divided by y-axis. **

**Answer**

**(i)** Let co-ordinates of Q be (x’, y’) and of R (x’’, y’’)

Point P (1, 2) divides OQ in the ratio of 2 : 3

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

= (2x’ + 3×0)/(2+3)

⇒ (2x’+0)/5 = 1

⇒ 2x’ = 5

⇒ x’ = 5/2

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

= (2y’ + 3×0)/(2+3)

⇒ 2y’/5 = 2

⇒ 2y’ = 10

⇒ y’ = 5

∴ Co-ordinates of Q will be (5/2, 5)

∵ the diagonals of a parallelogram bisect each other

∴ In ||gm OPRS, diagonals OR and PS bisect each other at M.

∵ M is the mid-point of PS

∴ Co-ordinates of M will be = {(-2+1)/2, (0+2)/2}

or (-2/2, 2/2)

or (-1, 1)

**(ii)** ∵ M is the mid-point of OR also

∴ - 1 = (0 + x’’)/2

⇒ x’’ = - 2

And 1 = (0 + y’’)/2

⇒ y’’ = - 2

∴ Co-ordinates of R will be (- 2, 2)

**(iii)** RQ is dividing by y-axis in N

Let the ratio in which N divides RQ in m_{1} : m_{2}

∵ N lies on y-axis

∴ its abscissa (x) = 0

0 = (m_{2}x’ + m_{2}x”)/(m_{1} + m_{2})

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

⇒ (5m_{1}/2 – 2m_{2})/(m_{1}+m_{2}) = 0

⇒ 5/2 m_{1} – 2m_{2} = 0

⇒ 5/2 m_{1} = 2m_{2}

⇒ m_{1}/m_{2} = (2×2)/5 = 4/5

∴ m_{1} : m_{2} = 4 : 5

**19. If A (5, -1), B (- 3, - 2) and C (- 1, 8) are the vertices of a triangle ABC, find the length of the median through A and the co-ordinates of the centroid of triangle ABC.**

**Answer**

A (5, - 1), B (- 3, -2) and C (- 1, 8) are the vertices of ∆ABC D, E and F are the midpoints of sides BC, CA and AB respectively

And G is the centroid of the ∆ABC

∵ D is the midpoint of BC

∴ Co-ordinates of D will be {(x_{1} + x_{2})/2}, (y_{1 }+ y_{2})/2} or {(-3 –1)/2 , (-2+8)/2} or (-4/2, 6/2) or (-2, 3)

∵ G is the centroid

∴ Co-ordinates of G will be {(x_{1} + x_{2} + x_{3})/3 , (y_{1} + y_{2} + y_{3})/3}

Or {(5 – 3 – 1)/3, (-1–2 + 8)/3} or (1/3, 5/3)

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