# Parallel and Perpendicular Lines in Graphs of Linear Equations

A **Linear equation** is an equation in which the highest exponent of the variable present in the equation is one. When we draw the **graph of the linear equation**, it **forms a straight** **line.**

If any two lines in the plane are drawn, they are either parallel or intersecting.

### How do you know if a line is parallel?

**Two lines are parallel if their slopes are equal**.

Hence the lines *y* = *m*_{1}*x *+ *c*_{1} and* y* = *m*_{2}*x *+ *c*_{2} are parallel if *m*_{1} = *m*_{2}.

In fact, **two parallel lines differ by a constant**.

Hence if the equation of line is *y* = *mx* + *c*, then, equation of a line parallel to it is *y* = *mx* + *k*, where *k* is a constant. To find the particular line, we require a unique value of *k*. For this additional condition is given.

** Example 1: **Write the equation of a line parallel to

*y*= 7

*x*+ 10.

Using the above concept, the equation of a line parallel to above line is *y* = 7*x* + *k*, where *k* is any real number.

** Example 2: **Write the equation of a line parallel to

*y*= 7

*x*+ 10 which passes from (1, 1).

The equation of a required line parallel to above line is* y* = 7*x* + *k*, where *k* is any real number.

Since the above line passes from (1, 1), hence it satisfies the required equation* y* = 7*x* + *k*.

Hence, 1 = 7(1) + *k* and *k* = 1 – 7 = -6.

Hence, the required equation is *y* = 7*x* – 6.

### How do you know if a line is perpendicular?

**Two lines are perpendicular if the product of their slopes is -1**.

Hence the lines *y* = *m*_{1}*x *+ *c*_{1} and* y* = *m*_{2}*x *+ *c*_{2} are perpendicular if *m*_{1}*m*_{2} = -1 or *m*_{1 = }

** Example 3:** Find the equation of a line perpendicular to

*y*= 2

*x*+ 5.

Comparing with the slope intercept form, *y* = *mx* + *c*, the slope of given line is *m *= 2. Hence, the slope of a line perpendicular to given line is –= – .

Equation of line perpendicular to given line is *y= – **x* + *k = – **x+k*, where *k* is any real number.

** Example 4: **Find the equation of a line perpendicular to

*y*= 3

*x*+ 5 passing from (1, 2).

As, done in the previous example slope of given line is *m* = 3.

Hence, the slope of a line perpendicular to given line is –* = –*.

Equation of line passing from (1, 2) and slope – is (*y* – 2) = –* (x-1)*

3(*y* – 2) = -(*x* – 1)

3*y* – 6 = –*x* + 1

*x* + 3*y* = 7

## Note:

## (1) If two lines are parallel,then their slopes are equal.

## (2) If two lines are perpendicular,then product of their slopes is -1.

## To find the slope of a line perpendicular to given line,We take the reciprocal and change its sign.

## Hence,slope of a line perpendicular to a given line with slope m is -1/m.

** Example 5:** Check whether the lines

*x*+

*y*= 10 and

*x*+

*y*= 100 are parallel or perpendicular.

Comparing with the slope intercept form,* y* = *mx* + *c*

The slope of first line, *y* = (-1)*x* + 10 is *m*_{1} = -1.

The slope of second line, *y* = (-1)*x* + 100 is* m*_{2} = -1. ** **

*m*_{1} = *m*_{2} = -1. Hence the given lines are parallel.

** Example 6:** Check whether the lines

*x*+

*y*= 10 and

*x*–

*y*= 100 are parallel or perpendicular.

Comparing with the slope intercept form,* y* = *mx* + *c*

The slope of first line, *y* = (-1)*x* + 10 is *m*_{1} = -1.

The slope of second line, *y* = (1)*x* – 100 is *m*_{2} = 1.

*m*_{1}* m*_{2} = -1, so the given lines are perpendicular.

**Check Point**

- Find the equation of the line passing through (-1, 5) & parallel to the line
*y*= 5*x*+ 1. - Find the equation of the line passing through (-1, 5) & perpendicular to the line
*y*= 5*x*+ 1. - Check whether the lines, 3
*x*+*y*= 15 and 21*x*+ 7*y*= 28 are parallel or perpendicular. - Check whether the lines 3
*x*+*y*= 15 and*x*– 3*y*= 28 are parallel or perpendicular. - Find the equation of the line parallel to the line,
*y*= 7*x*+ 51. - Find the equation of the line perpendicular to the line
*y*= 7*x*+ 51.

##### Answer Key

- The required parallel line is
*y*= 5*x*+ 10. - The required perpendicular line is y = –
*x +*or*x*+ 5*y*= 26. - Since, slopes of the two lines are equal which is -3. Hence the given lines are parallel.
- Since, product of the slopes of the two lines is -1. Hence the given lines are perpendicular.
- The equation of the line parallel to the line
*y*= 7*x*+ 51 is*y*= 7*x*+*k*, where*k*is any real - The equation of the line perpendicular to the line
*y*= 7*x*+ 51 is*y = – x + k*, where*k*is any real

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