## GRE: COORDINATE GEOMETRY

Two real number lines that are perpendicular to each other and that intersect at their respective zero points define a rectangular coordinate system. Often called the x-y coordinate system or x-y plane. The horizontal number line is called the x-axis and the vertical number line is called the y-axis. The point where the two axes intersect is called the origin, denoted by O. The positive half of the x-axis is to the right of the origin, and the positive half of the y-axis is above the origin. The two axes divide the plane into four regions called quadrants IIIIII, and IV, as shown in the figure below.

Each point P in the x-y plane can be identified with an ordered pair (x, y) of real numbers and is denoted by P (x, y). The first number is called the x-coordinate, and the second number is called the y-coordinate. A point with coordinates (x, y) is located lxl units to the right of the y-axis if x is positive or to the left of the y-axis if x is negative. Also, the point is located lyl units above the x-axis if y is positive or below the x-axis if y is negative. If x = 0, the point lies on the y-axis, and if y = 0, the point lies on the x-axis. The origin has coordinates (0, 0).

In the figure above, the point P (4, 2) is 4 units to the right of the y-axis and 2 units above the x-axis, and the point P’’’ (-4, -2) is 4 units to the left of the y-axis and 2 units below the x-axis.

Distance Formula:

If Two points P and Q are such that they are represented by the points (x1, y1) and (x2, y2) on the x-y plane, then the distance between two pints P and Q = ((x– x2)2 + (y– y2)2)0.5

Section Formula:

If any point C (x, y) divides the line segment joining the points A (x1, y1) and B (x2, y2) in the ratio m: n internally,

X = (mx2 + nx1)/ (m + n)

Y = (my+ my2)/ (m + n)

If any point C (x, y) divides the line segment joining the points A (x1, y1) and B (x2, y2) in the ratio m: n externally,

X = (mx2 – nx1)/ (m + n)

Y = (my– my2)/ (m + n)

Slope of a line:

The slope of a line joining two points A (x1, y1) and B (x2, y2) is denoted by m and is given by m = (y– y1)/ (x– x1) = tan , where  is the angle that the line makes with the positive direction of x-axis.

Equation of line:

Slope Intercept form:

The equation of a straight line passing through the point A (x1, y1) and having slope m is given by

(y – y1) = m (x – x1)

Intercept form:

The equation of a straight line making intercepts a and b on the axes of x and y respectively is given by

x/a + y/b = 1

Perpendicularity and parallelism:

Conditions for two lines to be parallel:

Two lines are said to be parallel if their slopes are equal.

Conditions for two lines to be perpendicular:

Two lines are said to be perpendicular if the product of slopes of two lines is equal to -1.

### Area of triangle:

The area of a triangle whose vertices are $$A({ x }_{ 1 },{ y }_{ 1 }),\quad B({ x }_{ 2 },{ y }_{ 2 }),\quad C({ x }_{ 3 },{ y }_{ 3 })$$  is given by

$$=\frac { 1 }{ 2 } ({ x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }))$$

Since the area can’t be negative, we have to take the modulus value given by the above equation.

Problem 1) Find the area of the triangle whose vertices are (1. 3), (-7, 6) & (5, 1)?

solution.  Vertices are A(1,3), B(-7,6) and C(5,-1)

Area of triangle =

$$=\frac { 1 }{ 2 } ({ x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }))$$ $$=\frac { 1 }{ 2 } (1(6+1)-7(-1-3)+5(3-6))\\ =\frac { 1 }{ 2 } (7-7(-4)+5(-3))\\ =\frac { 1 }{ 2 } (7+28-15)\\ =\frac { 1 }{ 2 } (28-8)\\ =10\quad Sq.\quad units$$

Problem 2) What is the area of a triangle with vertices (-2, 4), (2, 4) & (-6, 6) in the coordinate plane?

solution.

Let point A(-6, 6), point B(-2, 4) & C(2,4).

Line BC is parallel to x-axis.

Slope of BC = $$(\frac { 4-4 }{ 2+2 } )=0$$

We can use line BC as the Base of triangle.

Drop a height vertically from (-6, 6).

Height = different of y coordinates of points A & B = 6-4 = 2

Length of Base (BC) = difference between the $$x$$ coordinates of point B & C = 2 – ( -2) = 4

Area of triangle = $$\frac { 1 }{ 2 } \times 4\times 2=4\quad Sq.\quad units$$

Problem 3) What will be the reflection of the point (4, 5) in the 3rd quadrant?

solution.

Point (-4,-5) is reflection of the point (4,5) in the third quadrant.

Problem 4) If the origin gets shifted to (2, 2), then what will be the new coordinates of the point (4, -2)?

a) (-2, 4)

b) (2, 4)

c) (4, 2)

d) (2, -4)

solution.

If origin gets shifted to (2, 2), horizontal distance of point A from y-axis becomes (2) & vertical distance of point. A from x-axis becomes (4) in the downward direction.

Therefore, new co-ordinate of the point will be (2, -4)

Problem 5) What will be the equation of the straight line that passes through the intersection of the straight lines $$2x-3y+4=0$$ & $$3x+4y-5=0$$  and is perpendicular to the straight lime $$3x-4y-5=0$$?

solution.  Line $$2x-3y+4=0$$ & $$3x+4y-5=0$$  intersect at point $$(\frac { -1 }{ 17 } ,\frac { 22 }{ 17 } )$$

Slope of straight line $$3x-4y-5=0$$ is $$\frac { 3 }{ 4 }$$.

Slope of line perpendicular to this line = $$\frac { -4 }{ 3 }$$

Equation of line having slope $$\frac { -4 }{ 3 }$$ & passing through the point $$(\frac { -1 }{ 17 } ,\frac { 22 }{ 17 } )$$ is

$$(y-\frac { 22 }{ 17 } )=\frac { -4 }{ 3 } (x+\frac { 1 }{ 17 } )\\ 3y-\frac { 66 }{ 17 } =-4(x+\frac { 1 }{ 17 } )\\ 51y-66=-4(17)(x+\frac { 1 }{ 17 } )\\ 51y-66=-68x-4\\ 68x+51y=-4+66=62\\ Equation={ (68x+5y=62) }^{ Ans }$$

Problem 6) Line m & n are perpendicular, neither line is vertical or line m passes through the origin.

Quantity A:         The product of the slopes of lines m & n

Quantity B:         The product of the x-intercepts of lines m & n

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

solution.  The product of slope of perpendicular lines = -1

(The only exception is when one of the lines has an undefined slope because its vertical, but that case has been specifically excluded).

If line m passes through the origin, its x-intercept is 0, So regardless of the x-intercept of line n.

Quantity B is 0.

Therefore, quantity B is greater.

Problem 7) Which of the following could be the slope of a line that passes through the point (-2, -3) & crosses the y-axis above the origin?

a) $$-\frac { 2 }{ 3 }$$

b) $$\frac { 3 }{ 7 }$$

c) $$\frac { 3 }{ 2 }$$

d) $$\frac { 5 }{ 3 }$$

e) $$\frac { 9 }{ 4 }$$

f) 4

solution.  Slope of the line m passes through the origin & through the point (-2,-3) = $$\frac { -3 }{ -2 } =\frac { 3 }{ 2 }$$

Let angle formed by line m with positive x-axis be $$\theta$$, $$\tan { \theta =\frac { 3 }{ 2 } }$$

Angle formed by all the lines passing through the point (-2,-3) & crossing the y-axis above the origin lines between $$\theta \quad and\quad { 90 }^{ o }$$.

$$\tan { \theta =\frac { 3 }{ 2 } }$$ $$\tan { { 90 }^{ 0 }=\infty }$$

Therefore, slope of all lines passing through the point (-2,-3) & crossing the y-axis above the origin lies between $$\frac { 3 }{ 2 } \quad and\quad \infty$$.

Therefore, $$\frac { 5 }{ 3 } ,\quad \frac { 9 }{ 4 }$$ & 4 could be the slope of such lines.

## GRE: PROBABILITY

Probability means the chance of the occurrence of an event. In layman terms, we can say that it is the likelihood that something- that is defined as the event will or will not occur.

In probability, our first approach is to define the event and then we approach to find out the probability of an event or we approach to find out the chance of the occurrence of an event.

Sample space This is defined with respect to a random experiment and denotes the set representing all the possible outcomes of the random experiment. For example: – Sample space when a coin is tossed is S = (Head, Tail). Sample space when a dice is thrown is S = (1, 2, 3, 4, 5, 6). Sample space when two dices are thrown is 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)

Event The set representing the desired outcome of a random experiment is called the event. Consider the experiment of tossing a coin two times. An associated sample space is S = {HH, TT, TH, HT}. Suppose, getting two heads is the desired outcome. We find that (HH) is the only element corresponds to the desired outcome.

Therefore, event E = {HH}

Consider the experiment of throwing a dice. An associated sample space is S = {1, 2, 3, 4, 5, 6}. Suppose, we want to get a number less than 4. We find that (1, 2, 3) are the only element of S corresponds to our requirement. Therefore, event E = {1, 2, 3}.

Event (E) is always a subset of sample space (S).

When the sets A and B are two events associated with a sample space, then event ‘either A or B or both’ is represented by A ∪ B.

Event ‘A and B’ is represented by A ∩ B.

Mutually exclusive events A set of events is mutually exclusive when the occurrence of any one of them means that the other event cannot occur. Consider the experiment of throwing a dice. An associated sample space is S = {1, 2, 3, 4, 5, 6}. Consider events A ‘an odd number appears’ and B ‘an even number appears’

A = {1, 3, 5}      B = {2, 4, 6}

Occurrence of event A means that the other event B cannot occur.

A ∩ B = f (null)

Therefore, A and B are mutually exclusive events.

Equally likely events If two events have the same probability they are called equally likely events. In a toss of a coin, the chance of getting head is equal to chance of getting tail. Therefore, getting head and getting tail are equally likely events.

Exhaustive set of events A set of events that includes all the possibilities of the sample space is said to be an exhaustive set of events. Let us define the following events

A: ‘a number less than 3 appears’

B: ‘3 appears’

C: ‘a number more than three appears’

A = {1, 2}

B = {3}

C = {4, 5, 6}

A U B U C = {1, 2} U {3} U {4, 5, 6} = {1, 2, 3, 4, 5, 6} = S

Therefore, A, B and C are called exhaustive set of events.

Independent events An event is described as such if the occurrence of an event has no effect on the probability of the occurrence of another event. (If the first child of a couple is a boy, there is no effect on the chances of the second child being a boy.)

Formula for finding probability

Let A be an event. Probability of event A is denoted by P(A).

Probability of an event A = P(A) =

(Number of outcomes favorable to event)/ (Total number of outcomes possible)

Probability of an event ‘not A’ is denoted by P(A’).

P(A’) = 1 – P(A)

Some other Formulae:

Let A and B be two events associated with a random experiment. Then,

P (either A or B) = P(A) + P(B) – P (A and B)

or

P (A U B) = P(A) + P(B) – P (A ∩ B)

P (A U B)’ = P (A’ ∩ B’)

Let A, B and C be three events associated with a random experiment. Then,

P (A U B U C) = P(A) + P(B) + P(C) – P (A ∩ B) – P (A ∩ B) – P (A ∩ B) + P (A ∩ B ∩ C)

Problem 1) In a throw of two dice, find the probability of getting one prime and one composite number?

Sol. Sample space when two dices are thrown is 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)

Let A denote the event of getting a prime and a composite number

A = (1, 2) (1, 4) (1, 6)

(2, 1) (2, 3) (2, 5)

(3, 2) (3, 4) (3, 6)

(4, 1) (4, 3) (4, 5)

(5, 2) (5, 4) (5, 6)

(6, 1) (6, 3) (6, 5)

No. of outcomes favorable to A = 18

Total no. of outcomes = 36

Probability = (18)/ (36) = 1/ 2

Problem 2) The letters of the word LUCKNOW are arranged among themselves. Find the probability of always having NOW in the word?

Sol.  Number of arrangements of word LUCKNOW = 7!

Number of arrangements of word LUCKNOW having NOW = 5!

Probability = (Total number of outcomes favorable to event)/ (Total number of outcomes)

= (5!)/ (7!)

Use of conjunction AND If A AND B are two independent events, and if the probability of their occurrence is P(A) and P(B) respectively, then the probability that A and B occur is equal to

P(A) x P(B)

Use of conjunction OR If A AND B are two independent events, and if the probability of their occurrence is P(A) and P(B) respectively, then the probability that A or B occur is equal to

P(A) + P(B)

Problem 3) If we have the probability of A winning a race is 1/3 and that of B winning the race is 1/ 2, then the probability that either A or B win a race is given by?

Sol.  Probability that either A or B wins a race is given by P(A) + P(B)

= 1/3 + 1/ 2 = 5/6

Problem 4) One card is drawn from a well-shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be

a) a diamond

b) not an ace

c) a black card

d) not a diamond

e) not a black card

Sol.  Let A be the event ‘the card drawn is a diamond’ = (Number of outcomes favorable to event)/ (Total number of outcomes possible) = P (A) = 13/52 = 1/ 4

b) Let B be the event ‘the card drawn is an ace’

P(B) = (Number of outcomes favorable to event)/ (Total number of outcomes possible)

= 4/ 52

P(B)’ = 1 – 4/52 = 48/52

c) Let C be the event ‘the card drawn is a black card’

P(C) = (Number of outcomes favorable to event)/ (Total number of outcomes possible)

= 26/52 = 1/2

d) Let D be the event ‘the card drawn is a diamond’

P(D) = (Number of outcomes favorable to event)/ (Total number of outcomes possible)

= 13/52 = 1/ 4

P(D)’ = 1 – 1/4 = 3/4

e) Let E be the event ‘the card drawn is a black card’

P(E) = (Number of outcomes favorable to event)/ (Total number of outcomes possible)

= 26/52

P(E)’ = 1 – 1/2 = 1/2

Problem 5) Find the probability that a year chosen at random will have 53 Sundays provided that year was not leap year?

Sol.  A Year has 365 days. Year has 52 complete weeks and one day. One day maybe a Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or Sunday.

Probability that this day will be a Sunday = 1/7

Therefore, probability that a year chosen at random will have 53 Sundays = 1/7

Problem 6) If two dice are thrown, what is the probability that the sum of the numbers is not less than 10?

Sol.  Let A be an event ‘the sum of numbers is equal to 10’

Let B be an event ‘the sum of numbers is equal to 11’

Let C be an event ‘the sum of numbers is equal to 12’

Probability that the sum of numbers is not less than 10 = P(A) + P(B) + P(C)

Sample space when two dices are thrown is 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)

A = {(4, 6), (6, 4), (5, 5)}

B = {(5, 6), (6, 5)}

C = {(6, 6)}

P(A) = 3/36

P(B) = 2/36

P(C) = 1/36

Probability that the sum of numbers is not less than 10 = 1/36 + 2/36 + 3/36 = 6/36 = 1/6

Problem 7) There are two bags containing white and black balls. In the first bag, there are 8 white and 6 black balls and in the second bag, there are 4 white and 7 black balls. One ball is drawn at random from any of these two bags. Find the probability of this ball being black?

Sol.  Let A be an event ‘Selection of the first bag’

Let B be an event ‘Selection of second bag’

One bag is selected at random out of two bags. Therefore, P(A) = P(B) = 1/2

Let C be an event ‘Black ball drawn from the first bag’

P(C) = 6/14

Let D be an event ‘Black ball drawn from the first bag’

P(D) = 7/14

Probability of this ball being black = P(A). P(C) + P(B). P(D) = (1/2). (6/14) + (1/2). (7/14)

= (1/2). (13/14) = 13/28

Problem 8) The probability that Ajeet will solve a problem is 1/5. What is probability that

a) he does not solve a single problem out of ten problems

b) he solves at least one problem out of ten problems

Sol.  Let A be an event ‘Ajeet will solve a problem’

P(A) = 1/5

P(A)’ = 4/5

He solves no problems i.e. he does not solve the first problem and he does not solve the second problem……….. and he does not solve the tenth problem.

Probability that he does not solve a single problem out of ten problems = (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) = (1/5)10

b) Let B be an event ‘Ajeet does not solve a single problem out of ten problems’

P(B) = (1/5)10

P(B)’ = 1 – (1/5)10

Problem 9) In a four-game series between Radha and Anand, the probability that Anand wins a particular game is 2/5 and that of Radha winning a game is 3/5. Assuming that there is no probability of a draw in an individual game, what is the chance that the series is drawn (2-2)?

Sol.  Let R be an event ‘Radha will win a particular match’

Let A be an event ‘Anand will win a particular match’

P(A) = 2/5 and P(R) = 3/5

The event definition for the series to end in a draw can be described as

(R AND R AND A AND A) OR (R AND A AND R AND A) OR (R AND A AND A AND R) OR (A AND A AND R AND R) OR (A AND R AND R AND A) OR (A AND R AND A AND R)

= (2/5)2(3/5)+ (2/5)2(3/5)+ (2/5)2(3/5)+ (2/5)2(3/5)+ (2/5)2(3/5)+ (2/5)2(3/5)2

= (2/5)2(3/5)x (6)

= (2/5)2(3/5)x (4C2)

Where 4C2 gives us the number of ways in which Radha can win two games and Anand can win two games.

If E1, E2, E3 ……..En are mutually exclusive events of an experiment. Then,

P (EU EU E3………En) = P(E1) + P(E2) + P(E3)…………….+ P(En)

If E1, E2, E3 ……..En are mutually exclusive and exhaustive events of an experiment. Then,

P(E1) + P(E2) + P(E3)…………….+ P(En) = 1

Problem 10) If P(A) = 1/3, P(B) = 1/2, P (A∩ B) = 1/4  then find the P(A’ U B’)

Sol.  P (A U B)’ = P (A’ ∩ B’)

Replacing A with A’ and B with B’

P (A’ U B’)’ = P (A ∩ B)

P (A ∩ B) = 1/4

P (A’ U B’) = 1 – 1/4 = 3/4

Problem 11) A and B are two mutually exclusive events of an experiment. If P(A’) = 0.65 and P (A U B) = 0.65 and P(B) = p, find the value of p.

Sol.  P(A’) = 0.65

P(A) = 0.35

P (A U B) = 0.65

A and B are mutually exclusive events.

Therefore, P (A U B) = P(A) + P(B)

0.65 = 0.35 + p

P = 0.30

Problem 12) A committee of two persons is selected from two men and two women. What is the probability that the committee will have

a) no man

b) one man

c) two men?

Sol.  Total number of persons = 4. Out of these 4 persons, two can be selected in 4Cways.

a)There will be two women in the committee.

Out of two women, two can be selected in 2C= 1 way

Probability (no man) = 2C2/4C2

b)One man and one woman can be selected in 2C1 x 2C1

Probability (One man and one woman) = (2C1 x 2C1)/ (4C2)

c)Out of two men, two can be selected in 2C= 1 way

Probability (two-man) = 2C2/4C2

Problem 13) A carton contains 25 bulbs, 8 of which are defective. What is the probability that if a sample of 4 bulbs is chosen, exactly two of them will be defective?

Sol.  4 bulbs out of 25 bulbs can be drawn in 25Cways.

Two bulbs out of 8 defective bulbs can be drawn in 8Cways.

Two bulbs out of 17 non-defective bulbs can be drawn in 17Cways.

Two defective and two non-defective bulbs out of 25 bulbs can be drawn in (8C2) x (17C2)

Probability that if a sample of 4 bulbs is chosen, exactly two of them will be defective = ((8C2) x (17C2))/ (25C4)

Problem 14) From a bag containing 8 green and 5 red balls, three are drawn one after the another. Find the probability of all three balls being green if balls are drawn without replacement?

Sol.  3 balls out of 13 balls can be drawn in 13Cways

3 green balls out of 8 green balls can be drawn in 8Cways

probability of all three balls being green if balls are drawn without replacement = (8C3)/(13C3)

Conditional Probability

It is the probability of the occurrence of an event A given that the event B has already occurred. This is denoted by P(A/B).

P(A/B) = (Number of elementary events favorable to A ∩ B)/ (Number of elementary events which are favorable to B)

P(A’/B) = 1 – P(A/B)

Problem 15) If P(A) = 7/ 13, P(B) = 9/13 and P (A∩ B) = 4/13, evaluate P(A/B).

Sol.  P(A) = 7/13

P(B) = 9/13

P (A ∩ B) = 4/13

P(A/B) = P (A ∩ B)/ P(B) = (4/13)/ (9/13) = 4/9

Problem 16) Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Sol.  Let A be ‘number on the drawn card is even number’

Let B be ‘number on drawn card is more than 3’

P(A/B) = P (A ∩ B)/ P(B)

= (4/10)/ (7/10) = 4/7