In this post we will understand two formulas which are important for Grade 11 NCERT and CBSE Math Syllabus. Along with the formulas we have also provided solved examples for your better understanding.

**Distance of a Point from a Line**

The distance (d) of a point P( x_{1} ,\ y_{1}) from a line Ax + By + C =0 is given by the below mentioned formula –

d\ =\ \frac{|\ Ax_{1} \ +\ By_{1} \ +\ C\ |}{\sqrt{A^{2} \ +\ B^{2}}}

**How to solve distance of a point questions**

**Step 01**: Write the equation of the straight line in the form of general equation as shown below –

⟹ Ax + By + C = 0

**Step 02**: Find the distance ‘d’ of the point ( x_{1} ,\ y_{1}) from the straight line line Ax + By + C = 0 by using the formula –

d\ =\ \frac{|\ Ax_{1} \ +\ By_{1} \ +\ C\ |}{\sqrt{A^{2} \ +\ B^{2}}}

**Distance of a Point solved problems**

**(Q1) Find the distance of the point ( -1, 1) from the line 12( x + 6) = 5( y – 2) **

**Solution**:

The equation of the straight line in the general form

⟹ 12( x + 6) = 5 ( y – 2)

⟹ 12x + 72 = 5y – 10

⟹ 12x – 5y + 82 = 0

Here,

A = 12,

B = -5,

C = 82

Then the distance of the point ( -1 ,1) from the line 12x – 5y + 82 =0 is –

d =\ \frac{|\ Ax_{1} \ +\ By_{1} \ +\ C\ |}{\sqrt{A^{2} \ +\ B^{2}}}\\\ \\ d\ =\ \frac{|\ ( 12)( -1) \ +\ ( -5)( 1) \ +\ 82\ |}{\sqrt{( 12)^{2} \ +\ ( -5)^{2}}} \\\ \\ d\ =\ \frac{|\ -12 \ -5 \ +\ 82\ |}{\sqrt{144\ +\ 25}}\\\ \\ d\ =\ \frac{|\ 82\ -\ 17\ |}{\sqrt{169}}\\\ \\ d\ = \ \frac{65}{13} \ =\ 5 \ units \\\ \\ Hence,\ the\ distance\ of\ the\ point\ ( -1,\ 1) \\ \\ is\ 5\ units\ from\ the\ given\ straight\ line.

**(Q2) Find the distance of the point ( 3, -5) from the line 3x – 4y – 26 =0**

**Solution**

the equation of the straight line in the general form

⟹ 3x -4y -26 = 0

Here,

A = 3,

B = -4,

C = -26

Then the distance of the point ( 3 , -5) from the line 3x – 4y – 26 = 0 is :

d\ =\ \frac{|\ Ax_{1} \ +\ By_{1} \ +\ C\ |}{\sqrt{A^{2} \ +\ B^{2}}}\\\ \\ d\ =\ \frac{|\ ( 3)( 3) \ +\ ( -4)( -5) \ +\ ( -26) \ |}{\sqrt{( 3)^{2} \ +\ ( -4)^{2}}}\\\ \\ d\ =\ \frac{|\ 9 \ + \ 20\ -26\ |}{\sqrt{9\ +\ 16}}\\\ \\ d\ =\ \frac{|\ 29\ -\ 26\ |}{\sqrt{25}}\\\ \\ d\ =\ \frac{3}{5}\\\ \\ Hence,\ the\ distance\ of\ the\ point\ ( 3,\ -5) \\ \\ is\ \frac{3}{5} \ from\ the\ given\ straight\ line.**(Q3) Find the points on the x-axis, whose distances from the line \frac{x}{3} \ +\frac{y}{4} \ =1 are 4 units**

**Solution**.

The equation of the straight line in the general form

\frac{x}{3} \ +\frac{y}{4} \ =1 \\\ \\ \frac{4x\ +\ 3y}{12} \ =\ 1 \\\ \\ 4x\ +\ 3y\ =\ 12 \\\ \\ 4x\ +\ 3y\ -\ 12\ =\ 0 \\\ \\

Here,

A = 4,

B = 3,

C = -12

We have to find the distance of any point on x-axis, for which the distance are 4 units.

so, the point must have co-ordinates ( x, 0)

Then the distance of the point ( x , 0) from the line 4x + 3y – 12 = 0 is

d\ =\ \frac{|\ Ax_{1} \ +\ By_{1} \ +\ C\ |}{\sqrt{A^{2} \ +\ B^{2}}}\\\ \\ 4\ =\ \frac{|\ ( 4)( x) \ +\ ( 3)( 0) \ +\ ( -12) \ |}{\sqrt{( 4)^{2} \ +\ ( 3)^{2}}}\\\ \\ 4\ =\ \frac{|\ 4x\ +0\ -12\ |}{\sqrt{16\ +\ 9}}\\\ \\ 4\ =\ \frac{|\ 4x\ _{\ } -\ 12\ |}{\sqrt{25}}\\\ \\ 4\ =\pm \ \frac{4x\ -\ 12}{5} \\\ \\ 20\ =\ \pm \ ( 4x\ -\ 12)\\\ \\{Case I}

⟹ 20 = 4x – 12

⟹ 20 + 12 = 4x

⟹ x = 8

{Case II}

⟹ 20 = – (4x – 12)

⟹ 20 = – 4x +12

⟹x = -2

Hence, the points on the x-axis are ( -2 , 0) or ( 8 , 0) for which the distance are 4 units

**Distance Between Two Parallel Lines**

If the two lines are given in form of general equation i.e., Ax + By + C_{1} = 0 \ and\ Ax + By +C_{2} = 0, then the distance between the two parallel lines can be obtained using the below mentioned formula –

d\ =\ \frac{|\ C_{1} \ -C_{2} \ |}{\sqrt{A^{2} \ +\ B^{2}}} \\\ \\But if the two parallel lines are given in the slope-intercept form i.e., y = mx\ + c_{1} and y\ =\ mx\ +c_{2} then the distance between the two parallel lines can be obtained using the below mentioned formula –

d\ =\ \frac{|\ c_{1} \ -c_{2} \ |}{\sqrt{1\ +\ m^{2}}}**Distance between Straight Lines Questions**

**(Q1) Find the distance between the two parallel lines**

⟹{15x\ +\ 8y\ -\ 34\ =\ 0}

⟹{15x\ +\ 8y\ +\ 31\ =\ 0}

**Solution**

The equation of the two parallel straight line in the general form –

⟹15x + 8y – 34 = 0

⟹15x + 8y + 31 = 0

Here,

A = 15;

B = 8;

C_{1} = -34

C_{2} = 31

**(Q2) Find the distance between the two parallel lines** –

⟹ l ( x + y ) + p = 0

⟹ l ( x + y ) – r = 0

**Solution**

The equation of the two parallel straight line in the general form –

⟹ l (x + y) + p = 0

⟹ lx + ly +p = 0

⟹ l ( x + y) – r = 0

⟹ lx + ly – r = 0

Here,

A = l;

B = l;

C_{1} = p

C_{2} = – r

Hence, the distance between the two parallel lines is \frac{|p+r|}{l\sqrt{2}}

**(Q3) Find the distance between the two parallel lines**

⟹ y = 4x + 7

⟹ y = 4x – 10

**Solution:**

the equation of the two parallel straight line in the general form –

⟹ y = 4x + 7

⟹ y = 4x – 10

Here,

m =4;

c_{1} = 7

c_{2} = -10

the distance between the two parallel lines can be obtained using the below mentioned formula:

d\ =\ \frac{|\ c_{1} \ -c_{2} \ |}{\sqrt{1\ +\ m^{2}}}\\\ \\ Putting\ values \\\ \\ d\ =\ \frac{|\ 7\ -( -10) \ |}{\sqrt{1\ +\ ( 4)^{2}}}\\\ \\ d\ =\ \frac{|\ 7\ +10\ |}{\sqrt{1\ +\ 16}} \\\ \\ =\ \frac{|17|}{\sqrt{17}} \\\ \\ =\ \frac{17}{\sqrt{17}} \ =\ \sqrt{17}\\\ \\ Hence,\ the\ distance\ between\\ \\ the\ two\ parallel\ line\ is\ \ \sqrt{17}