Two lines are said to be perpendicular if the angle between them is 90º (90 degrees). There is a result on perpendicular line which states: Two lines are perpendicular if and only if the product of their slopes is -1.

i.e. If the slopes of two lines which are perpendicular are** m1** and **m2** then** m1 x m2=-1.**

**Example:**

The slope of a line is **3/2**. If a line is perpendicular to this line, what is the slope of this new line?

**Solution:**

By the condition for the slopes of perpendicular lines,

**m1 x m2=-1**

So, the slope of new line = **-2/3.**

### Finding the Equation of Perpendicular Lines:

For finding the equation of a line perpendicular to **ax+by+c=0** and passing through the point **(x1,y1)**, there are 2 methods.

#### Method 1:

(1) Finding the slope of **ax+by+c=0** which can be obtained by the expression **–a/b.**

i.e. slope of **ax+by+c=0** is **a/b**.

(2) Finding the slope of the line perpendicular to it.

i.e. the slope of the perpendicular line, **m = b/a.**

(3) Then the corresponding perpendicular line’s equation can be found using point slope form.

i.e., the required equation is,

**y-y1=m(x-x1)**

**Example:**

Find the equation of the line perpendicular to **3x+y+3=0** and passing through **(-1,2).**

**Solution:**

The slope of given line, **m = -3/1=-3**

The slope of the perpendicular line = **1/3.**

**(x1,y1) = (-1,2).**

The equation of perpendicular line is,

**y-y1=m(x-x1)**

**y-2 = 1/3 (x+1)**

**3y-6 = x+1**

**x-3y+7=0.**

#### Method 2:

(1) Take the equation of line perpendicular to** ax+by+c=0** as **bx-ay+k=0** where **k** is a constant.

(2) The value of ‘**k**’ can be found b substituting the given point **(x1,y1)** in the equation **bx – ay+k=0.**

(3) Substitute ‘**k**’ back into the equation **bx-ay+k=0**, which is the required equation.

**Example:**

Find the equation of the line perpendicular to **3x+y+3=0** and passing through **(-1,2)**.

**Solution:**

The equation of line perpendicular to **3x+y+3=0** is,

**x-3y+k=0.**

Here **x=-1** and** y=2**

**-1-3(2)+k=0**

**-1-6+k=0**

**k-7=0**

**k=7.**

So the required equation of the perpendicular line is,

**x-3y+7=0**

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