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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  m1 = \frac{-1}{m_{2}}

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

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 xy = 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

  1. Find the equation of the line passing through (-1, 5) & parallel to the line y = 5x + 1.
  2. Find the equation of the line passing through (-1, 5) & perpendicular to the line y = 5x + 1.
  3. Check whether the lines, 3x + y = 15 and 21x + 7y = 28 are parallel or perpendicular.
  4. Check whether the lines 3x + y = 15 and x – 3y = 28 are parallel or perpendicular.
  5. Find the equation of the line parallel to the line, y = 7x + 51.
  6. Find the equation of the line perpendicular to the line y = 7x + 51.
Answer Key
  1. The required parallel line is y = 5x + 10.
  2. The required perpendicular line is y = – \frac{1}{5}x + \frac{26}{5} or x + 5y = 26.
  3. Since, slopes of the two lines are equal which is -3. Hence the given lines are parallel.
  4. Since, product of the slopes of the two lines is -1. Hence the given lines are perpendicular.
  5. The equation of the line parallel to the line y = 7x + 51 is y = 7x + k, where k is any real
  6. 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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