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 = m₁x + c₁ and y = m₂x + c₂ are parallel if m₁ = m₂.

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 = 7x + 10.

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

Example 2: Write the equation of a line parallel to y = 7x + 10 which passes from (1, 1).

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

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

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

Hence, the required equation is y = 7x – 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 = m₁x + c₁ and y = m₂x + c₂ are perpendicular if m₁m₂ = -1 or m_{1 =}

Example 3: Find the equation of a line perpendicular to y = 2x + 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 – $\frac{1}{m}$ = – $\frac{1}{2}$ .

Equation of line perpendicular to given line is y= – $\frac{1}{m}$ x + k = – $\frac{1}{2}$ x+k, where k is any real number.

Example 4: Find the equation of a line perpendicular to y = 3x + 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 – $\frac{1}{m}$ = – $\frac{1}{3}$ .

Equation of line passing from (1, 2) and slope – $\frac{1}{3}$ is (y – 2) = – $\frac{1}{3}$ (x-1)

3(y – 2) = -(x – 1)

3y – 6 = –x + 1

x + 3y = 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 m₁ = -1.

The slope of second line, y = (-1)x + 100 is m₂ = -1.

m₁ = m₂ = -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 m₁ = -1.

The slope of second line, y = (1)x – 100 is m₂ = 1.

m₁ m₂ = -1, so the given lines are perpendicular.

Check Point

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

Answer Key

The required parallel line is y = 5x + 10.
The required perpendicular line is y = – $\frac{1}{5}$ x + $\frac{26}{5}$ or x + 5y = 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 = 7x + 51 is y = 7x + k, where k is any real
The equation of the line perpendicular to the line y = 7x + 51 is y = – $\frac{1}{7}$ x + k , where k is any real

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