Intersection of Two Straight Lines

Two lines will intersect at a point. The point will have a pair of values as (x1, y1). The straight lines are represented by equation with two or one variable in x and y.  If those equations are solved to get the value of x and y, will represent the point intersection of those two lines. To solve the set of lines to get the value of x and y , we can use either the method of elimination or substitution. Now let us discuss few problems on this topic intersection of two straight lines.

Example Problems on Intersection of Two Straight Lines

Ex 1: Find the point of intersection of the following two lines

2x + 3y = 10; 2x + y = 6.

Sol: Given: 2x + 3y = 10 --------------(1)

2x + y = 6  --------------(2)

The point intersection of above two lines can be obtained by solving them as follows:

(1) – (2) implies: 2x + 3y = 10

2x + y = 6

We get, 2y = 4

y = 2.

From (2), 2x + (2) = 6

2x = 6 – 2 = 4

2x = 4

x = 2.

Therefore, the point of intersection is (2, 2).

Ex 2: Find the point of intersection of the following two lines

x + 2y = 1; 5x + 4y = -7.

Sol: Given: x + 2y = 1 ------------------(1)

5x + 4y = -7 -----------------(2)

The point of intersection of above two lines can be obtained by solving the as follows:

(1) x 5 – (2) implies: 5x + 10y = 5

5x + 4y = -7

We get, 6y = 12

y = 2.

Therefore, from (1), x + 2(2) = 1

Implies, x = 1 – 4 = -3.

Therefore the point of intersection is (-3, 2).

Ex 3: Find the point of intersection of the following two lines

3x + y = 10; y = 7.

Sol: Given: 3x + y = 10 -------------(1)

y = 7 -----------(2)

Since, y = 7 is one of the line, the value of the y coordinate will be 7.

Therefore, from (1), we get , 3x + 7 = 10

3x = 10 – 7 = 3

x = 1.

Therefore, the point of intersection is ( 1, 7).

Practice Problems on Intersection of Two Straight Lines

1. Find the point of intersection of lines 3x + 2y = 7 and x + y = 3.

[ Answer: (1, 2)]

2. Find the point of intersection of lines 3x - 2y = -2 and x - 2y = 6.

[ Answer: (-4, -5)]

3. Find the point of intersection of lines 4x - 3y = -10 and 3x + y = -1.

[ Answer: (-1, 2)]

Solving Binomials by Factoring

Introduction: 

In algebra, the polynomials which have two terms are called binomials. To factor binomials, we need to follow the following methods:

(i) 2a + ab = a(2 + b ) [ Here the given expression has two terms, where a is the common value]    
     = a( 2 + b)         
(ii) a² – b² = ( a + b) ( a – b)   [ This is the standard form]
(iii) (a + b)² = ( a + b)(a +b)
(iv) (a - b)² = ( a - b)(a - b)

Product of two polynomials will give three terms.
( a + b)² = ) a² +b² + 2ab
( a - b)² = ) a² +b² - 2ab

Let us see few problems on this topic solving binomials by factoring.

Example Problems on Solving Binomials by Factoring

Ex 1: Solve (x + 3) (x – 4) = 0

Soln: Given: (x + 3) (x – 4) = 0
This implies: x + 3 = 0 or x – 4 = 0
That is : x = -3, 4.
Therefore the solution is { -3, 4}.

Ex 2: Solve x + y = 7 and xy = 12, find x and y.

Soln: Given : x + y = 7 -----------(1)
                            xy = 12 ---------(2)
Therefore, x – y = sqrt [(x+y)² – 4xy]
                             = sqrt[ 7² – 4(12)]
                             = 1
Therefore, x – y = 1 ---------------(3)
From (1) and (3), we get:
x + y = 7 -----------(1)
x – y = 1 -----------(3)
2x = 8, this implies that x = 4.
Therefore, (1) implies 4 + y = 7
Hence y =3.
Therefore the solution is {4,3}.

Ex 3: Solve x - y = 5 and xy = 24, find the value of x + y.

Soln: x + y = sqrt [(x-y)² + 4xy]
                   = sqrt[ 5² – 4(24)]
                   = 11.
Therefore from, x + y = 11
                             x – y = 5,
We get 2x = 16.
Therefore, x = 8.
Hence from x + y = 11. y = 3.
Therefore the solution is { 8, 3}.

Ex 4: Solve x + y = 11 and xy = 24, find the value of x² – y².

Soln: x – y = sqrt [(x+y)² – 4xy]
                   = sqrt[ 11² – 4(24)]
                   = 5
Therefore, x² – y² = ( x + y )( x – y)
                             = 11 x 5 = 55.

Practice Problems on Solving Binomials by Factoring

1. If a + b = 9  and ab = 36, find a - b
[Ans: a – b = 5]

2. If a – b = 4 and ab = 12, find a² – b².
[Ans: a² – b² = 32]

Graph of Sinx

 The coordinate graph is called the Cartesian coordinate plane. The graph contains a couple of the vertical lines are called coordinate axes. The vertical axis of the y axis value and the horizontal axis value is the x axis value. The points of the intersection of those two axes values are called the origin of coordinate graphing pictures.  The trigonometry graph is a sin or cos waves. In this graph equation is in the form of y = mx + c. m is nothing but a sin or cos. In this article we shall discuss graph of sin x.

Sample Problem for Graph of Sin X:

Graph of sin x problem 1:

Solve the given trigonometry functions 3sin 5x - y = 0 and draw the graph for the given function.

Solution:
In the first step we find the plotting point of the given trigonometry functions. The given function is
                                      3sin 5x - y = 0
We are going to find out the plotting points for a given equation. In the first step we are going to change equation in the form of y = mx + c, we get the following term
                                  3sin 5x – y = 0
                                                y = 3sin 5x
In the next step we are find out the plotting points of the above equation.
In the above equation we put x = -5 we get
           y = 3sin 5(-5)
           y = -0.4
In the above equation we put x = -4 we get
            y = 3sin 5(-4)    
           y = -2.7
In the above equation we put x = 0 we get
            y = 3sin 5(0)
           y = 0
In the above equation we put x = 2 we get
            y = 3sin 5(2)
           y = -1.6
From equation (1) we get the following value

X	-5	-4	0	2
y	0.4	-2.7	0	-1.6

Graph:

Graph of Sin X Problem 2:

Solve the given trigonometry functions y = 2sin 3x and the draw graph for the given function.
Solution:
In the first step we find the plotting point of the given trigonometry functions. The given function is
                      y = 2sin 3x
We are going to find out the plotting points for a given equation. In the first step we are going to change equation in the form of y = mx + c, we get the following term
                        2sin 3x – y = 0
                              2sin 3x = y                                   
In the above equation we put x = -4 we get
           y = 2sin 3(-4)
           y = 1
In the above equation we put x = -3 we get
            y = 2sin 3(-3)   
           y = -0.82
In the above equation we put x = 0 we get
            y = 2sin 0
           y = 0
In the above equation we put x = 3 we get
            y = 2sin 3(3)
           y = 0.82
From equation (1) we get the following value

x	-4	-3	0	3
y	1	-0.82	0	0.82

Graph: