Straight Lines And Pair Of Straight Lines

Let's analyze each statement to determine which one is not correct.Statement A"If

l

passes through two points of

S

, then

m

must be rational."

S

consists of points

(a%2C%20b)

where both

a

and

b

are rational numbers. If

l

passes through two such points (

a_1%2C%20b_1

) and (

a_2%2C%20b_2

), the slope

m

of the line

l

would be:$

m%3D%5Cfrac%7Bb_2-b_1%7D%7Ba_2-a_1%7D

Since

a_1, a_2, b_1

%2C%20and

b_2

are%20rational%2C

must%20also%20be%20rational.%20So%2C%20statement%20A%20is%20correct.Statement%20C%22If

passes%20through%20exactly%20one%20point%20of

%2C%20then

must%20be%20irrational.%22If

were%20rational%20and

rational%2C%20then%20there%20could%20be%20infinitely%20many%20rational%20points%20on%20the%20line

because%20the%20set%20of%20rational%20points%20is%20dense%20in%20the%20set%20of%20real%20numbers.%20Therefore%2C%20if

passes%20through%20exactly%20one%20rational%20point%2C

must%20be%20irrational.So%2C%20statement%20C%20is%20correct.Statement%20D%22If

is%20irrational%20and

is%20rational%2C%20then

passes%20through%20exactly%20one%20point%20of

.%22For

(a, b)

on%20the%20line%20with

a, b \in \mathbb{Q}, b=m a+c

.%20If

is%20irrational%20and

is%20rational%2C%20for%20a%20specific%20rational

a, m a

is%20irrational%2C%20and%20adding%20rational

to%20irrational

m a

gives%20irrational

.%20Hence%2C%20there%20can%20only%20be%20at%20most%20one%20rational%20point%20on

%2C%20where%20both%20the%20numerator%20and%20denominator%20accidentally%20%22cancel%22%20out%20the%20irrationality%20(which%20is%20rare%2C%20but%20possible).So%2C%20statement%20D%20is%20correct.Statement%20B%22If

is%20rational%20and

is%20irrational%2C%20then

passes%20through%20at%20least%20one%20point%20of

.%22The%20line%20equation%20is

y=m x+c

.%20For

(a, b)

to%20be%20a%20point%20on%20the%20line%20where

a, b \in \mathbb{Q}

%2C%20we%20need

b=m a+c

.%20If

is%20rational%2C

m a

is%20rational%20for%20any%20rational

.%20However%2C%20if

is%20irrational%2C

m a+c

becomes%20irrational%2C%20which%20means

would%20be%20irrational%2C%20contradicting%20the%20condition%20that

b \in \mathbb{Q}

.%20Therefore%2C

cannot%20pass%20through%20any%20point%20in

.So%2C%20statement%20B%20is%20not%20correct.The%20incorrect%20statement%20is%3A

\mathbf{B}

.%20If

is%20rational%20and

is%20irrational%2C%20then

passes%20through%20at%20least%20one%20point%20of

S$.

To find the intersections of the given lines in the XY-plane, let's analyze each line and express them in the slope-intercept form

y%20%3D%20mx%20%2B%20c

.Lines and Their EquationsLine

L_1

:Given by the equation

5x%20-%202y%20%3D%201

.Rearranging, we get:

y%20%3D%20%5Cfrac%7B5%7D%7B2%7Dx%20-%20%5Cfrac%7B1%7D%7B2%7D

Line

L_2

:Passes through the points

(0%2C%201)

and

(100%2C%20101)

.Slope

(m)

is calculated as:

m%20%3D%20%5Cfrac%7B101%20-%201%7D%7B100%20-%200%7D%20%3D%201

Using point

(0%2C%201)

, the line equation is:

y%20%3D%20x%20%2B%201

Line

L_3

:Passes through

(1%2C%2011)

and is parallel to the vector

-%5Chat%7B%5Cimath%7D%20%2B%202%5Chat%7B%5Cjmath%7D

.The slope of the vector, and hence the line, is:

m%20%3D%20%5Cfrac%7B0%20-%202%7D%7B0%20-%20(-1)%7D%20%3D%20-2

Using point

(1%2C%2011)

, the line equation becomes:

y%20%3D%20-2x%20%2B%2013

Solving for IntersectionsNext, we find the points of intersection for each pair of these lines:Intersection of

L_1

and

L_2

:Solving:

%5Cfrac%7B5%7D%7B2%7Dx%20-%20%5Cfrac%7B1%7D%7B2%7D%20%3D%20x%20%2B%201

We find the common solution

(1%2C%202)

.Intersection of

L_2

and

L_3

:Solving:

x%20%2B%201%20%3D%20-2x%20%2B%2013

We find the common solution

(4%2C%205)

.Intersection of

L_3

and

L_1

:Solving:

-2x%20%2B%2013%20%3D%20%5Cfrac%7B5%7D%7B2%7Dx%20-%20%5Cfrac%7B1%7D%7B2%7D

We find the common solution

(3%2C%207)

.ConclusionThe set

A

, representing the union of all intersection points

(L_1%20%5Ccap%20L_2)%20%5Ccup%20(L_2%20%5Ccap%20L_3)%20%5Ccup%20(L_3%20%5Ccap%20L_1)

, contains the distinct points

(1%2C%202)

(4%2C%205)

, and

(3%2C%207)

. Therefore, there are a total of 3 distinct elements in set

A

High-Yield Trend

Chapter Questions
3 MCQs

Official Solution

Official Solution

Official Solution

Straight Lines And Pair Of Straight Lines

High-Yield Trend

Chapter Questions 3 MCQs

Official Solution

Official Solution

Official Solution

Chapter Questions
3 MCQs