The Missing Mail | Class IX - Interact in English

Q 1. Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square.

Solution: - Using Distance Formula

Diagonal AC = diagonal BD

A, B, C and D are vertices of a square

Q 2. Determine the ratio in which the point P (m, 6) divides the join of A(-4, 3) and B(2, 8). Also find the value of m.

Solution:-

Let AP : PB = K : 1

Co-ordinates of P are

Burt co-ordinates of P are given (m, 6)

Q 3. The coordinates of the mid-point of the line joining the points (2p + 2, 3) and (4, 2q + 1) are (2p, 2q). Find the value of p and q.

Solution:-

Let the given points be A (2p + 2, 3), B(4, 2q + 1) and C (2p, 2a)

Co-ordinates of C are

Q 4. Find the ratio in which the line-segment joining the points (6, 4) and (1, -7) is divided by x-axis.

Solution:-

Let the two points A (6, 4) and B (1, -7), be divided by x-axis at C (x, 0)

Let AC : CB = K : 1

Co-ordinates of C are

As C lies on x-axis

Q 5. Prove that coordinates of the centroid of a triangle ABC, with vertices (x1 y1), (x2, y2) and (x3, y3) are given by

Solution:-

Let the vertices of triangles be A(x1, y1), B(x2, y2) and C(x3, y3)

Let D be the mid-point of BC.

Co-ordinates of D are

Let centroid of triangle ABC be G.

Co-ordinates of G are

Q 6. Find the value of m for which the points with co-ordinates (3, 5), (m, 6) and (1/2, 15/2) are collinear.

Solution :- Let the three points be A (3, 5) B (m, 6) and (1/2, 15,2)

As points A, B and care collinear

Q7. To construct a tangent to a circle at a point P on it without using the centre of the circle.

Procedure: -

1. A chord PQ through P is drawn.
2. Any point R is taken on the major arc PQ. PR and QR are joined.
3. Equal to is constructed.
   PX is required tangent to the circle at P.

Q7. To construct a tangent to circle from a point P outside the circle using its centre O.

Procedure:-

1. OP is joined and is bisected at M.
2. Taking M as centre and MO as radius a semicircle is drawn which intersect the given circle at Q.
3. PQ is the required tangent from P to the circle.

Q8. To construction a tangent to a circle from a point out side the circle without using its centre.

Procedure:-

1. A secant PAB to the circle is drawn.
2. PB is bisected at M.
3. Taking M as a centre and PM as a radius, a semicircle is drawn.
4. Through A is drawn perpendicular to AB which intersect the semicircle at C.
5. Taking P as centre and PC as radius, arcs are drawn to intersect the given circle at Q and R.
6. PQ and PR are joined which is the required tangent.

Q9. To construct incircle of a triangle ABC whose sides are BC = a, CA = b and AB = c.

Procedure:-

1. Triangle ABC in which BC = a, CA = b and AB = c is constructed.
2. BM and CN is constructed angle bisectors of which intersect at I.is drawnTaking I as centre and IL as radius, circle is drawn. This is the required incircle.

Q 10. To construct a circumcircle of a triangle ABC where a = BC, b = CA and c = AB.

Proceture:-

1. A triangle ABC is constructed with BC = a, CA = b and AB = c.
2. Perpendicular bisector PQ of BC and RS of CA is constructed. They intersect at O.
3. Taking O as centre and OC as a radius circle is drawn which passes through A, B and C.

Q 11. An unbiased dice is tossed.

1. Write the sample space of the experiment
2. Find the probability of getting a number greater than 4.
3. Find the probability of getting a prime number.

Solution:-

1. Sample space = {1, 2, 3, 4, 5, 6}

n(s) = 6

2. E = event of getting a number greater than 4
   
   = {5, 6}

n (E) = 2

P (> 4) = Probability of a number greater than 4
= n(E)/n(S) = 2/6 = 1/3

3. E = Event of getting a prime number
   
   = {2, 3, 5}

n (E) = 3

P(Prime number) = Probability of a prime number

Q 12. Out of 400 bulbs in a box, 15 bulbs a defective. One ball is taken out at random from the box. Find the probability that the drawn bulb is not defective.

Solution:- Total number of bulbs = 400

Total number of defective bulb = 15

Total number of non-defective bulbs = 400-15

= 385

P (not defective bulb) = 385/400

= 77/80

Q 13. Find the probability of getting 53 Fridays in a leap year.

Solution:- No. of days in a leap year = 366

366 days = 52 weeks and 2 days.

A leap year must has 52 Fridays

The remaining two days can be

(a) Sunday an Monday
(b) Monday and Tuesday
(c) Tuesday and Wednesday
(d) Wednesday and Thursday
(e) Thursday and Friday
(f) Friday and Saturday
(g) Saturday and Sunday

Out of 7 case, 2 cases have Friday

P (53 Friday) = 2/7

Q 14. Three unbiased coins are tossed simultaneously. What is the probability of getting exactly two heads?

Solution: - When three coins are tossed simultaneously, the sample space is

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.

n(S) = 8

E = Set of cases favorable to the event

= {HHT, HTH, THH}

n(E) = 3

P (exactly two heads) =

Q 15. A dice is thrown twice. Find the probability of getting (a) doublets (b) prime number on each die.

Solutions: - Sample space =

S = { (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }

n (S) = 36

(i) E = Events getting doublet

= {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}

n (E) = 6

P(doublet) =

(ii) E = Events getting prime number on each die.

= {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)}

n(E) = 9

P (getting prime number on each die)

= n(E)/n(S) = 9/36 = 1/4

Q 16. Solve the system of equation:

Solution :

Here ,

and a1b2 - a2b1

=1 x 2 - 3 x (-4)

= 2 + 12

=

The system has unique solution.

Or,

Or,

Or, x = 28/14 = 2,

Or, y = 56/14 = 4

Q 17. Find the value of p for which the following system of eqation have exactly one solution:

Solution: - here we have

as the system of equation have exactly one solution

Hence, the system of equations have exactly one solution for all value of a except 6.

Q 18. Find the value of p for which the following system of eqation have exactly one solution:

Solution: - here we have

as the system of equation have exactly one solution

Hence, the system of equations have exactly one solution for all value of a except 6.

Q 18. Solve the following system of equation:

Solution: multiplying equation (1) by 4 and equation (2) by 7 we get

Subtracting (3) fr0m (4) we get

Or, 19/v = 38

Or, 36 v = 19

Or, v = 19/38

Or, v = 1/2

Putting v = 1/2 in (1) we get

Or, 22 - 7/u = 1

Or, 21 = 7/u

Or, 21u = 7

Or, u = 7/21

Or, u = 1/3

Q 19. Find the solution of following system of equation such that :

Solution: Dividing both sides of the given equations by uv we get

Multiplying equation (2) by 2 and then subtracting equation (1) we get

Or, 15u = 15

u = 1

Putting u = 1 in equation on (1)

Or, 2/v = (7-3)/3

Or. 2/v = 4/3

Or, 4v = 6

Or, v = 6/4

Or, v = 3/2

Q 20. Find the solution of following system of equation such that :

Solution: Dividing both sides of the given equations by uv we get

Multiplying equation (2) by 2 and then subtracting equation (1) we get

Or, 15u = 15

u = 1

Putting u = 1 in equation on (1)

Or, 2/v = (7-3)/3

Or. 2/v = 4/3

Or, 4v = 6

Or, v = 6/4

Or, v = 3/2

Q 21. Solve the following system of equation:

Solution:-

Multiplying equation (1) by 4 and (2) by 3 and then adding we get

Putting x = 3 in (i) we get

Q 22 . Find the value of a and b for which the following system of equations has infinitely many solution; [2002 C, D]

Solution:-

For infinitely many solutions

Taking 1st and 3rd ratio

Taking last two retio

5b - 10 = -9

5b = -9 + 10 = 1

b = 1/5

Q 23. Determine the value of k so that the following linear equations have no solution:

Solution:-

For no solution,

Or, (k - 2) (3k + 2) = 3(k2 + 1)

Or, 3k2 - 5k - 2 = 3k2 + 3

Or, -5k = 5

Or, k = -1

Q 24 . Solve graphically the system of equations:

Solution: To draw the graphic of equations (1) and (2) we find three solutions of each of the equations (1) and (2)

Or,

x	1	-2	4
y	-1	-3	1

Now we plot the print (1, -1) (-2, -3) and (4, 1). After joining we get a straight line.

And 3x + 4y + 1 = 0

x	1	-3	5
y	-1	2	-4

Again we plot the point (1, -1), (-3, 2) and ((5, -4) on the same graph paper and join then.

Now, we get see that the lines of these two equations intersect at (1, -1).
. Is the solution of the system of equations.

Q 25 . Solve graphically the system of linear equations:

Solution:- Let us take

When

When

When

x	0	3	-3
y	1.5	-0.5	3.5

We plot the point (0, 1.5) (3, -0.5) and (-3, 3.5) on a graph paper and join them which is a straight line.

And 2x +3y = -11

When

When

When

x	-1	-4	>2
y	-3	-1	-5

Again we plot the points (-1, -3), (-4, -1) and (2, -5) on the same graph paper and join then we get a straight line which is parallel to previous line. The lines do not intersect and hence we get no solution.