# Forms of Two-Variable Linear Equations – Straight Lines | Class 11 Maths

Line is the simplest geometrical shape. It has no endpoints and extends in both directions till infinity. The word “**straight**” simply means without “**bend**”. The gradient between any two point on the line is same. Hence,** **we can say that, if the gradient between any two points on the line is same, then the locus is a** straight line**. So let’s see what are the different ways in the line can be represented.

## Different Forms Of a Line

- Point Slope Form
- Two – Point Form
- Slope – Intercept Form
- Intercept Form
- Normal Form

**Point – Slope Form**

The equation of the line passing through the point (x_{1},y_{1}) and slope ‘m’ can be written as :

**Proof:**

Let the point that lies on the curve be

(xand any general point can be denoted by_{1}, y_{1})(h, k).So, the slope of the line can be written as

Slope=(k-y

_{1}) / (h-x_{1})Also, we know that the slope of line is

m.Hence,

(k-y

_{1}) / (h-x_{1})= mReplace k by y and h by x

And, we get y – y

_{1 }= m(x – x_{1})

**Example 1.Write the equation of the line passing through (5,6) and slope equal to 3**

**Solution:**

Putting the value of (x1,y1) as (5,6) and m = 3, we get

y – 6 = 3* (x -5)

y – 6 = 3x – 15

y = 3x – 9

**Example 2.Write the equation of the line passing through (0,0) and slope equal to 1**

**Solution:**

Putting the value of (x1,y1) as (0,0) and m=1, we get

y – 0 = 1* (x -0)

y = x

**Two – Point Form**

The equation of the line passing through the point (x_{1},y_{1}) and (x_{2},y_{2}) can be written as :

**Proof:**

Since we know any two points on the line, we can write the slope of the line as

m = (y

_{2}-y_{1}) / (x_{2}-x_{1})And we also know from point slope form that

y-y

_{1 }= m(x-x_{1})Substituting the value of m in the above equation, we get

y-y

_{1 }= ((y_{2}-y_{1}) / (x_{2}-x_{1}) * (x-x_{1})

**Example 1.Write the equation of the line passing through (5,6) and (6,7)**

**Solution:**

Putting the value of (x1,y1) as (5,6) and (x2,y2) as (6,7) , we gety – 6 = (7-6)/(6-5) * (x -5)

y – 6 = 1* (x – 5)

y = x + 1

**Example 2.Write the equation of the line passing through (0,5) and (5,5)**

**Solution:**

Putting the value of (x1,y1) as (0,5) and (x2,y2) as (5,5), we get

y – 5 = (5-5)/(5-0) * (x -0)

y = 0

## Slope – Intercept Form

The equation of line with slope ‘m’ and cutting a intercept ‘c’ on y-axis can be written as:

**Proof:**

Since y-intercept=c

Equation of line passing through (0,c) is given by

y – c = m(x-0)

y = mx + c

**Example 1.Write the equation of the line having slope =5 and y-intercept as 3**

**Solution:**

Putting the value of m=5 and c=3 in Equation y = mx + c

y = 5x+3

**Example 2.Write the equation of the line having slope =1 and y-intercept as 1**

**Solution:**

Putting the value of m=1 and c=1 in Equation y = mx + c

y = x+1

**Intercept Form**

The equation of the line cutting intercept ‘a’ on x-axis and ‘b’ on y-axis can be written as:

**Proof:**

So we know that the line passes through the point (a,0) and (0,b), we can write the equation using two point form

y – y

_{1 }= (y_{2 }– y_{1}) / (x_{2 }– x_{1}) * (x – x_{1})Putting x

_{1 }= a, y_{1 }= 0, x_{2 }= 0 , y_{2 }= bWe get,

y = (b/(-a) * (x-a)

Dividing both sides by b and simplify RHS we get

y/b = -(x/a)+1

⇒ x/a + y/b =1

**Example 1.Write the equation of the line having x-intercept as 5 and y-intercept as 3 **

**Solution:**

Putting the value of a=5 and b=3 in Equation x / a + y / b = 1

x/5 + y/3 =1

**Example 2.Write the equation of the line having x-intercept as 1 and y-intercept as 1**

**Solution:**

Putting the value of a=1 and b=1 in Equation x / a + y / b = 1

x/1 + y/1 =1

**Normal Form**

The equation of the straight line upon which the length of the perpendicular from the origin is ‘p’ and this perpendicular makes an angle ‘∝’ with positive direction of x-axis is :

**Proof:**

Let the line AB be such that the length of the perpendicular OQ from the origin O to the line be p and <XOQ =∝.

From the diagram, using the intercept form we get

Equation of line AB is

x/p sec∝ + y/p cosec∝ =1

or

xcos∝ + ysin∝ =p

**Example 1.Write the equation of the line for which the length of the perpendicular from the origin is 5 units and this perpendicular makes an angle of 45 ^{0 }with positive direction of x-axis.**

**Solution:**

So, basically we are given the value of p=5 and ∝=45

^{0}Putting the values in the above equation we get,

x /√2 + y√2 = 5

x + y = 5√2

**Example 2.Write the equation of the line for which the length of the perpendicular from the origin is 1 units and this perpendicular makes an angle of 60 ^{0 }With positive direction of x-axis.**

**Solution:**

So, basically we are given the value of p=1 and ∝=60

^{0}Putting the values in the above equation we get,

x * (1/2) + y * (√3/2) = 1

x + y√3 = 2

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