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How to answer the following question?

Ramesh is a cashier at Canara Bank. He has notes of denominations of Rs 100. 50 and 10 respectively. The ratio of the number of these notes are is  2: 3: 5 respectively.

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What is the multiplicative inverse of rational numbers



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Identify an irrational number among the following numbers: 0.13 ,0.13overline{ 15}0.overline{ 1315} , 0.3013001300013.....

0.13 is a terminating number. So, it is not an irrational number.

0.13overline{15} = 0.131515......, 15 is repeating continuously so it is not an irrational number.

0.overline{ 1315} = repeating continuously so it is not an irrational number.

0.3013001300013....., non terminating and non recurring decimal. Hence, it is an irrational number. So, 0.3013001300013 is an irrational number.

Find six rational numbers between 3 and 4.

let  a = 3 , and b = 4

Here, we find six rational numbers i,e n = 6

 d = frac{b - a}{n + 1} = frac{4 - 3}{6 + 1} = frac{1}{7}

1st rational number =   a + d = 3 + frac{1}{7} = frac{22}{7}

2nd rational number =  a + 2d = 3 + frac{2}{7} = frac{23}{7}

3 rd rational number =  a + 3d = 3 + frac{3}{7} = frac{24}{7}

4th rational number  =  a + 4d = 3 + frac{4}{7} = frac{25}{7}

5th rational number  =  a + 5d = 3 + frac{5}{7} = frac{26}{7}

6th rational number  =   a + 6d = 3 + frac{6}{7} = frac{27}{7}

So six rational numbers are frac{22}{7} , frac{23}{7} , frac{24}{7} , frac{25}{7} , frac{26}{7} , frac{27}{7}

Rationalize the denominator of   frac{30}{5sqrt{3}-3sqrt{5}}

 frac{30}{5 sqrt{3}-3sqrt{5}} times frac{5sqrt{3 }+3sqrt{5}}{5sqrt{3 }+3sqrt{5}} =frac{30(5(sqrt{3 }+3sqrt{5})}{(5sqrt{3 })^{2}-(3sqrt{5})^{2}}

                                                    = frac{30(5sqrt{3}+3sqrt{5})}{30}

                                                    = 5sqrt{3}+3sqrt{5

Alternative Method

frac{30}{5sqrt{3}-3sqrt{5}} = frac{30}{(5sqrt{3}-3sqrt{5})}times frac{(5sqrt{3}+3sqrt{5})}{(5sqrt{3}+3sqrt{5})}

                      = frac{30(5sqrt{3}+3sqrt{5})}{(5sqrt{}3)^{2}-(3sqrt{5})^{2}}

                      = frac{30(5sqrt{3}+3sqrt{5})}{75-45} =frac{30(5sqrt{3}+3sqrt{5})}{30}

                      = 5sqrt{3 }+3sqrt{5}


Rationalize the denominator of frac{1}{(sqrt{2}+sqrt{3})-sqrt{4}}

frac{1}{(sqrt{2}+sqrt{3})-sqrt{4}}times frac{(sqrt{2}+sqrt{3})+sqrt{4}}{(sqrt{2}+sqrt{3})+sqrt{4}}  =frac{sqrt{2}+sqrt{3}+sqrt{4}}{(sqrt{2}+sqrt{3})^{2}-{4}} = frac{sqrt{2}+sqrt{3}+sqrt{4}}{(2+3+2sqrt{6})-4}

                                                                                 =frac{sqrt{2}+sqrt{3}+sqrt{4}}{1+2sqrt{6}}times frac{1-2sqrt{6}}{1-2sqrt{6}}

                                                                                 =frac{sqrt{2}+ sqrt{3}+sqrt{4}-2sqrt{12}-2sqrt{18}-4sqrt{6}}{1^{2}-(2sqrt{6})^{2}}




if  x =3-2sqrt{2}  then find the value of  x^{4}-frac{1}{x^{4}}

Given x: = 3-2sqrt{2}

 frac{1}{x}= frac{1}{(3-2sqrt{2})}times frac{(3+ 2sqrt{2})}{(3+2sqrt{2})}: : (: rationalizing: )


frac{1}{x}= frac{(3+2sqrt{2})}{9-8}= 3+2sqrt{2}


frac{1}{x^{2}}= (3+2sqrt{2})^{2}=9+8+12sqrt{2}=17+12sqrt{2}


x^{2}= (3-2sqrt{2})^{2}=9+8-12sqrt{2}=17-12sqrt{2}


Now , x^{4}-frac{1}{x^{4}}=left ( x^{2}-frac{1}{x^{2}} right )left ( x^{2}+frac{1}{x^{2}} right )


                          =left [ (17-12sqrt{2})-(17+12sqrt{2}) right ] left [ (17-12sqrt{2}) +(17+12sqrt{2})right ]

                         = 17-12sqrt{2}-17-12sqrt{2}left ( 17-12sqrt{2}+17+12sqrt{2} right )

                         =(-24sqrt{2})times 34=-816sqrt{2}

(i)Find the six rational  numbers bbetween 3 and 4 

(ii) Which mathematical concept is used in this problem

(iii) Which value is depticted in this question 


(i) We known that between two rational numbers  x and y such that x< y there is a rational number frac{x+y}{2}.

               ie,       3< frac{7}{2}< 4

Now a rational number between 3 and   frac{7}{2}< 4   is :

frac{1}{2}left ( 3+frac{7}{2} right )=frac{1}{2}times left ( frac{6+7}{2} right )=frac{13}{4}

A rational nummber berween  frac{7}{2}: : and : : 4 : : is

frac{1}{2}left ( frac{7}{2}+4 right )=frac{1}{2}times left ( frac{7+8}{2} right )=frac{15}{4}

 3< frac{13}{4}< frac{7}{2}< frac{15}{4}< 4

Further a rational number between 3 and  frac{13}{4}: : is :

frac{1}{2}left ( 3+frac{13}{4} right )=frac{1}{2}left ( frac{12+13}{4} right )=frac{25}{8}

A rational number between  frac{15}{4}: : and: : 4: : is :

frac{1}{2}left ( frac{15}{4}+4 right )=frac{1}{2}times frac{15+16}{4}=frac{31}{8}

A rational number between  frac{31}{8}: : and: : 4: : is

frac{1}{2}left ( frac{31}{8}+4 right ) = frac{1}{2}times left ( frac{31+32}{8} right )=frac{63}{16}

therefore 3< frac{25}{8}< frac{13}{4}< frac{7}{2}< frac{15}{4}< frac{31}{8}< frac{63}{16}< 4

Hence , six rational numbers between 3 and 4 are 

frac{25}{8}, frac{13}{4},frac{7}{2}, frac{15}{4}, frac{31}{8}: : and: : frac{63}{16}

(ii) Number syatem 

(iii) Rationality is always welcomed 

Show that 5sqrt{6} is an irrational number.


Let  5sqrt{6}  be a rational number, which can be put in the form frac{a}{b} , where   b neq 0 , a and b   are co-prime



Rightarrow sqrt{6}= rational 

But, we know that sqrt{6} is an irrational number.

Thus, our assumption is wrong. 

Hence, 5sqrt{6} is an irrational number.


Prove that sqrt{2} is an irrational number  

Let sqrt{2} be a rational number

therefore sqrt{2}=frac{p}{q}

Where p and q are co-prime integers and  q neq 0

Rightarrow 2 =frac{p^{2}}{q^{2}}

Rightarrow p^{2} = 2q^{2}

Rightarrow P^{2} is divisible by 2.

therefore  P is divisible by 2. 

Let p = 2r for some positive integer r

Rightarrow p^{2} = 4r^{2}

therefore 2q^{2} = 4r^{2}

Rightarrow q^{2} = 2r^{2}

Rightarrow q^{2} is divisible by 2.

therefore q is divisible by 2.

From (i) and (ii), p and q are divisible by 2, which contradicts the fact that p and q are co-primes. 

Hence, our assumption is false.

So, sqrt{2} is irrational




Prove that sqrt{3}  is an irrational number.

Let sqrt{3}  be a rational number

sqrt{3}= frac{a}{b}

(a and b are integers and co-primes )

On squaring both the sides, 

3 = frac{a^{2}}{b^{2}}

Rightarrow 3b^{2}= a^{2}

Rightarrow a^{2} is divisible by 3

therefore a  is divisible by 3

We can write  a = 3c for some integer

Rightarrow a^{2}=9c^{2}

Rightarrow 3b^{2}=9c^{2}

Rightarrow b^{2}=3c^{2}

b^{2} is divisible by 3

b is divisible by 3.

From equations (i) ad (ii),  we get 3 as a factor of "a"  and  "b" 

Which is contradicting the fact that a and b are co-primes. Hence our assumption that  sqrt{3} is a rational number, is false. So  sqrt{3} is an irrational number. 




if p is a prime number, then prove that sqrt{p} is irrational.

Let p be a prime number and if possible, let sqrt{p} be rational.

Let  sqrt{p}= frac{m}{n}  where m and n are  integers having no

common factor other than 1 and nneq 0

Then,  sqrt{p}= frac{m}{n} 

Squaring on both sides, we get 

frac{(sqrt{p})^{2}}{1}= left ( frac{m}{n} right )^{2}

Rightarrow frac{p}{1} = frac{m^{2}}{n^{2}}

Rightarrow pn^{2} = m^{2}

therefore P divides  m2 and p  divides m . [ because p divides pn2 ]

[because P is prime and p divides m2 Rightarrow p divides m]

Let              m = pq  for some integer q

 on putting  m = pq  [in eq.(i) we get ]

                pn2 = p2q2

                 n2 = pq2

 p divides n2

and p divides n.

[ because p is prime and p divides nRightarrow p divides n ]

Thus p is a common factor of m and n but this contraficts  the fact that m and n have no common factor other than 1.

The contradiction arises by assuming that sqrt{p} is a rational .

Hence, sqrt{p} is irrational.





Prove that   3+sqrt{5} is an irrational number. 

Let 3+sqrt{5} is  a rational number 

therefore 3+sqrt{5} = frac{p}{q}, qneq 0

   3+sqrt{5} = frac{p}{q}

Rightarrow sqrt{5} = frac{p}{q} -3

Rightarrow sqrt{5} = frac{p-3q}{q}

sqrt{5} is irrational  and   frac{p-3q}{q}  is a rational 

But the rational number cannot be equal to an irrational number. 

therefore 3+sqrt{5} is an irrational number.

Prove that  sqrt{3}  is an irrational number. Hence show that 7+2sqrt{3}  is also an irrational number 

If possible let sqrt{3}  be a rational number.

(i)   therefore frac{a}{b} = sqrt{3}, where a  and b are integers and co -primes

squaring both sides, we have 

       frac{ a^{2}}{b^{2}} = 3

Rightarrow a^{2} = 3b^{2}

Rightarrow a^{2} is divisible by 3

therefore a is divisible by 3    .............................(1)

We can write a = 3c for some c (integer)

(3c)2 = 3b2

Rightarrow 9c^{2} = 3b^{2}

Rightarrow b^{2} = 3c^{2}

Rightarrow b^{2} is divisible by  3.

 Rightarrow b is divisible by 3   ...........................(2)

From eq (i) and (ii) we have,

3 is a factor a and b which is contradicting the fact that 'a' and 'b' are co-prime.

Thus our assumption that  sqrt{3}  is rational numbers is wrong.

Hence, sqrt{3} is an irrational number.

(ii) Let us assume to the contrary that 7+2sqrt{3}  is a rational number.

7+2sqrt{3}=frac{p}{q},: q: neq 0: and : p: ,: qepsilon : Integer





p- 7q and  2q both are integers hence sqrt{3}  is a rational number. 

But this contradicts the fact that is sqrt{3} is  irrational number. Hence is 7+2sqrt{3} is an irrational number .


PQRS is a square. N and M are the midpoints of siders SR and QR respectively. O is a point on diagonal PR such that OP = OR. Show that ONRM is a  square. Also, find the ratio of ar (ORM ) and ar (PQRS).

 Since OP = OR

therefore  O is the midpoint of PR.  

   In,, Delta , SRP

O and N  are midpoints of siders PR and SR respectively 

therefore by midpoint theorem,  

ON = frac{1}{2}, , SP, , and, , ONparallel PQ, , , ................., (1) 

Similarly, , , OM left | PQ, , , , ...........(2)

Using (1) and (2)  we get 

ONRM, , is , , a, left | ^{gm}

Now, , , ON = frac{1}{2}, , SP

                 = frac{1}{2} SR

                 = NR 

In, , left | ^{gm} , ,ONRM, ,a, ,pair, , of, , adjacent, , sides, , ON, , and, , NR, , are, , equal, , and, , angle, , S = angle N = 90^{0}    (Corresponding angle as ON II PS)

ONRM , , is , , a , , square

Since OR is diagonal of a square

ar(ORM) = frac{1}{2}, , ar(ONRM ) ...............(3)

because  diagonal of an IIgm divides it into two congruent triangles) 

ar(ONRM ) NR times RM

                     = frac{1}{2}SRtimes frac{1}{2}RQ

                     = frac{1}{4}(SRtimes RQ)

                    = frac{1}{4}(SR )^{2}

                   = frac{1}{4}, ar(PQRS ) .............. (5)

Using (3) and (4) we get

frac{ ar(ORM)}{ar(PQRS)} = frac{frac{1}{2}ar(ONRM)}{4ar(PQRS)} = frac{1}{8} 

therefore The ratio of  ar (Delta ORM) : ar (PQRS) = 1:8

Construct a triangle whose angles are in the ratio 1:3:5 and length of sides included by first and last angles is 6 cm.

Let angles x, 3x, and 5x

    therefore x+3x+5x = 180^{0}           (angles of sum property)

                     9x = 180^{0}

                       x = 20^{0}

    therefore First angle  =  20^{0}

   Second angle  =  600

    Third angle    = 1000

Steps of construction:

(i)  Draw a line segment BC  = 6 cm 

(ii)  At point A draw angle XAB = 20^{0}  and at point B, draw angle YAB = 100^{0}

(iii) XA and YB intersect at point C. 

       Thus, ABC is the required triangle.


The radius of a spherical balloon increases from 7 cm to 14 cm as air is pumped into it. Find the ratio of surface area of the balloon in two cases.

Original , , surface , , area, , of , , S_{1}, = 4, pi , r_{1}^{2}

                                                     = 4pi times 7times 7 cm^{2}

              New , , surface, , area, , S_{2}, = 4, pi , r_{1}^{2}

                                              = 4pi times 14times 14cm^{2}

                                  S_{1}: S_{2} = frac{4pi times 7times 7}{4pitimes 14times 14}

                                             = frac{1}{4}

                                             = 1:4.

Divide a line segment of 7 cm length and divide externally in the ratio 3:5

Given : AB is a line segment of length 7 cm. 

To construct: To divide a line segment of 7 cm length externally in the ratio 3: 5 

Steps of construction :

1. Draw a line segment AB = 7 cm 

2. Draw ray BX making an acute 

3. Along BX, Mark off B1,B2,B3,B4 & B5. Join B2 to A.

4. Through B5 draw B5P II B2A intersecting BA at P.

5. The point P so obtained is the required point which divides AB externally in 3: 5.

A ball is rolled up a slanted plank at 18m/second and the speed decreases every second by  a3m/s. When the speed becomes  0. It will start rolling downwards with speed increasing at the same rate.

At what time does the speed become 0? If the speed is too be obtained as a non-negative number at all times,

a) What is the algebraic equation connecting speed and time till 6 seconds?

b) For the journey after 6 seconds, what is the algebraic equation connecting speed and time?

c) If the speed downwards is taken as a negative number, what is the speed-time equation which holds for all times?

a) V =  u +at

1 sec         2sec       3sec        4sec       5 sec

18 m/s     15 m/s    12m/s       9m/s        6m/s

v = 18 + a x 6 

v - 6a = 18 

V is the find velocity a is the acceleration.

b) V = 3 + at

after 6 seconds 

V = 3 + 6a

ie., V - 6a = 3

c) V = u+at

 a is negative hence 

v = u - at 

Ramu and Venu started a business. Ramu inested 50000 rupees and Venu invested 1, 50000 rupees. If they got 20% profit in a year, then 

a. What is the total profit 

b. What is the ratio of their investments?

c.If the profit is dividing according to the ratio of their investments, how much amount each will get?

Ramu invested = RS. 50000/-;

Venu invested = Rs.1,50000/-.

Profit = 20% 

a. Total investment  = 50000 + 1,50000 = 2,00000.

Total profit  = 200000 x 20/100 = 40000

b. Ratio of their investment = 50000 : 150000

                                           = 5 : 15

                                           = 1: 3

c. Ramu will get 40000 x 1/4 =  Rs. 10000/-

   Venu will get  40000 x 3/4 =  Rs. 30000/-

In the figure, the centres A, B, C of the semicircles are on a line. The radii of the unshaded semicircles are in the ratio 1:3. The radius of the smallest semicircle is 2 centimetre.

i) Find the area of the semicircle with centre at A.

ii) Find the area of the semicircle with centre at B. 

iii) Find the area of the shaded region. 

Radius of the semicircle = 2 cm 

Radius of the semicircle with centre C = 3 x 2 = 6 cm 

Radius of the semicircle with centre B  = 2 x 6 = 12 cm 

i. Area of the semicircle with centre A  = frac{1}{2}, , pi times 2times2 = 2 pi , , sq. cm

ii. Area of the semicircle with centre B = frac{1}{2}, , pi times 12times 12 = 6times 12 pi , , = 72pi , , sq. cm

iii. Area of the shaded portion =72pi - (2pi +frac{1}{2}, , pi times 6times 6)

                                            =72pi - (2pi +18pi ) = 72pi - 20pi

                                            =52pi , , sq, , , cm

Prove that sqrt{5} is an irrational number and hence show that 2-sqrt{5} is also an irrational number.

Let sqrt{5} be a rational number 

therefore sqrt{5} = frac{a}{b}

(a, b are co -prime integers and bneq 0)

Rightarrow a = b sqrt{5}

Rightarrow a ^{2}= 5b^{2}

Rightarrow 5 is a factor of a^{2}

Rightarrow 5 is a factor of a

let  a = 5c, 

Rightarrow a^{2} = 25c^{2}

Rightarrow 5b^{2} = 25c^{2}

Rightarrow b^{2} = 5c^{2}

Rightarrow 5 is a factor of b^{2}

Rightarrow 5 is a factor of b

5 is a common factor of a, b 

But this contradicts the fact that a, b are co-primes 

therefore sqrt{5} is irrational 

Let 2- sqrt{5} be a rational

therefore 2- sqrt{5} = a

Rightarrow 2- a = sqrt{5}

2 - a is rational, so is sqrt{5}

But sqrt{5} is not rational contradiction 

therefore 2-sqrt{5} is irrational.

Prove that sqrt{2} is an irrational number. Hence show that frac{3}{sqrt{2}} is alos an irrational number.

Let sqrt{2} be a rational number 

sqrt{2} = frac{a}{b}

(a, b are co-prime integers and bneq 0)

a = sqrt{2}b

a^{2} = 2b^{2}


Rightarrow 2 divides a2

Rightarrow 2 divides a

So we can write a = 2c for some integer c,, substitute for a, 2b^{2} = 4c^{2} , b^{2} = 2c^{2}

This means 2 divides b2, so 2 divides b.

a and b have '2' as a common factor

But this contradicts that a, b have no commom factor other than 1.

Our assumption is wrong.

Hence, sqrt{2} is irrational.

Let frac{3}{ sqrt{2}} be rational 

frac{3}{ sqrt{2}} = frac{a}{b} where a and b are integers, b neq 0

Rightarrow 3b = sqrt{2a}

Rightarrow sqrt{2a} = frac{3b}{a}

frac{3b}{a} is rational but sqrt{2} is not rational

Our assumption is wrong

frac{3}{ sqrt{2}} is rational.

Show that there is no positive integer n, for which sqrt{n-1}+sqrt{n+1} is rational.

Let us assume that there is a positive integer n for  whichsqrt{ n-1} + sqrt{ n + 1} is rational and equal to  frac{P}{Q},where  P and Q are positive integers (Qneq 0)

sqrt{ n-1} + sqrt{ n + 1} = frac{P}{Q}................(i)

Rightarrow frac{P}{Q} = frac{1}{sqrt{n-1} + sqrt{n+1}}

= frac{sqrt{n-1} -sqrt{n+1}}{(sqrt{n-1} + sqrt{n+1}) (sqrt{n-1} - sqrt{n+1})}

= frac{sqrt{n-1} -sqrt{n+1}}{(n-1) - (n+1) } = frac{sqrt{n-1} - sqrt{n+1}}{-2}

Rightarrow sqrt{n +1} - sqrt{n-1} = frac{2Q}{P}................(ii)

Apply (i) + (ii) we get

2 sqrt{n+1} = frac{P}{Q} + frac{2Q}{P} = frac{P^{2} + 2Q^{2}}{PQ}

Rightarrow sqrt{n+1} = frac{P^{2}+2Q^{2}}{2PQ}.............(iii)

Apply (i) and (ii) we get 

sqrt{n-1} = frac{P^{2} + 2Q^{2} }{2PQ}.................(iv)

From (iii) and (iv), we can say sqrt{n +1} and sqrt{n -1} both are rational because P and Q both are  rational. But it is possible only when (n+1) and (n-1)  both  are perfect squares. But they differ by 2 and two perfect square never differ by 2. So both (n+1) and (n-1) cannnot be perfect squares, hence there is no positive integer n for which sqrt{n -1} + sqrt{n+1} is rational.


The coordinates of the vertices of a triangle are A(1,1), B(5,5), C(2,5).

a. Write the coordinates of the midpoint D of AB.

b. What is the length of CD ?

c. What are the coordiaates of the point dividing the line CD in the ratio 2:1 ?

a. A(1,1), b(5,5), c(2,5)

Coordinates of D=left ( frac{1+5}{2},frac{1+5}{2} right )=(3,3)

b. Cd=sqrt{(2-3)^2 +(5-3)^2 }=sqrt{1+4}=sqrt{5}

c. x coordinates  of the point dividing the line joining c(2,5) and D(3,3) in the ratio 2:1


                          =2+frac{2}{3}times 1=2+frac{2}{3}=frac{6}{3}+frac{2}{3}=frac{8}{3}

y coordinates =y_{1}+frac{p}{p+q}(y_{2}-y_{1})

                      =5+frac{2}{3}(3-5)=5+frac{2}{3}times -2


Coordinates of this point =left ( frac{8}{3} ,frac{11}{3}right )

In the figure the radius of the circle centred at 0 is 6 units. Line Ab touches the circle at P and angle OAB=30^0.

a. Find the coordinates of the point A and P.

b. Find teh equation ao AB.

a. In Delta OPA, angle A=30^0 ,angle P=90^0 and angle O=60^0 .

    Since the angles are 30^0 ,60^ 0 and 90^0 , sides are in the  ratio 1:sqrt{3}:2

    Since OP =6

    PA =6sqrt{3}, OA=12

   Since OA =12, coordinates of A =(12,0)

   Draw PQ perpendicular to OA.

   Angles of Delta OPQ are 30^0 ,60^0 and 90^0.

   Since OP=6,  OQ=3,  PQ=3sqrt{3}

   Coordinates of P=(3,3sqrt{3})

b. Slope of AB=frac{y_{2}-y_{1}}{x_{2}-x_{1}}=frac{3sqrt{3}-0}{3-12}

                     =frac{3sqrt{3}}{-9}=frac{sqrt{3}}{-3}=frac{sqrt{3}}{-sqrt{3}times sqrt{3}}


   Let (x,y) be a point on the line AB.

   Then slope of the line joining (x,y) and (12,0) will also be frac{-1}{sqrt{3}}

    frac{0-y}{12-x}=frac{-1}{sqrt{3}},-1(12-x)=sqrt{3}times -y



    x+sqrt{3}y=12.This is the equation of the line.

In the figure, equation of the line joining the points A and B is x+2y=10. P and Q are points on this line.

a. Find the coordinates of the points A and B.

b. Find the coordinates of the points P that divides the line AB in the ratio 2:3.

c. If AQ: BQ =2:3, find the coordinates of the point Q.

a. Equation of the line AB is x+2y=10

    Since the y coordinates of the point A is zero, x+2times 0=10,x=10

    Coordinates of A = (10,0)

    Since the x coordinates of the point B is sero, 0+2y=10,y=5

    Coordinates of B =(0,5)

b. AP : PB =2:3

   x coordinates of P =x_{1}+frac{p}{(p+q)}(x_{2}-x_{1})

                                =10+frac{2}{5}(0-10)=10+frac{2}{5}times -10


    y coordinates of p =y_{1}+frac{p}{p+q}(y_{2}-y_{1})


    Coordinates of P=(6,2)

c. AQ: BQ=2:3

   Then BA: AQ=1:2

   if x is the x coordinates of Q.



   x-0=10times 3=3 0,x=30

   If y is the y coordinates of Q.


  3times 5+frac{3(y-5)}{3}=3times 0


  Coordinates of Q =(30,-10)