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Question: If the slope is \(\dfrac{7}{3}\), how do you find the perpendicular slope and the parallel slope ?...

If the slope is 73\dfrac{7}{3}, how do you find the perpendicular slope and the parallel slope ?

Explanation

Solution

To find the perpendicular and the parallel slope, use the formula of finding angle between two lines which is given as tan⁡x=(m2−m1)(1+m1m2),  where  x,  m1  and  m2\tan x = \dfrac{{({m_2} - {m_1})}}{{(1 + {m_1}{m_2})}},\;{\text{where}}\;x,\;{m_1}\;{\text{and}}\;{m_2} are angle between the two lines, slope of the first line and slope of the second line respectively. Perpendicular lines have a right angle between them and parallel lines have zero angle between them.

Complete step by step answer:
In order to find the perpendicular slope and the parallel slope if the slope of the line is given, we will remember the formula of finding angle between two lines that is
tan⁡x=(m2−m1)(1+m1m2),  where  x,  m1  and  m2\tan x = \dfrac{{({m_2} - {m_1})}}{{(1 + {m_1}{m_2})}},\;{\text{where}}\;x,\;{m_1}\;{\text{and}}\;{m_2} are angle between the two lines, slope of the first line and slope of the second line respectively.

Now first finding slope for the perpendicular line.We know that perpendicular lines have an angle of 900{90^0} between them, so using this information and above formula we can write
tan⁡900=(m2−73)(1+73m2) ⇒∞=(m2−73)(1+73m2) \tan {90^0} = \dfrac{{({m_2} - \dfrac{7}{3})}}{{(1 + \dfrac{7}{3}{m_2})}} \\\ \Rightarrow \infty = \dfrac{{({m_2} - \dfrac{7}{3})}}{{(1 + \dfrac{7}{3}{m_2})}} \\\
For this to be true, denominator of right hand side should be equals to zero,
1+73m2=0 ⇒73m2=−1 ⇒m2=−37 1 + \dfrac{7}{3}{m_2} = 0 \\\ \Rightarrow \dfrac{7}{3}{m_2} = - 1 \\\ \Rightarrow {m_2} = - \dfrac{3}{7} \\\
Now finding slope for parallel line.We know that parallel lines have zero angle between them, so we can write
tan⁡00=(m2−73)(1+73m2) ⇒0=(m2−73)(1+73m2) \tan {0^0} = \dfrac{{({m_2} - \dfrac{7}{3})}}{{(1 + \dfrac{7}{3}{m_2})}} \\\ \Rightarrow 0 = \dfrac{{({m_2} - \dfrac{7}{3})}}{{(1 + \dfrac{7}{3}{m_2})}} \\\
For this to be true, numerator of right hand side should be equals to zero,
m2−73=0 ∴m2=73 {m_2} - \dfrac{7}{3} = 0 \\\ \therefore {m_2} = \dfrac{7}{3} \\\
Therefore slope of perpendicular and parallel lines are 73  and  (−37)\dfrac{7}{3}\;{\text{and}}\;\left( { - \dfrac{3}{7}} \right) respectively.

Note: We have solved this question in a general way, you can directly solve for slope of perpendicular and parallel lines by using their property that the product of slopes of two perpendicular lines equals negative one and slopes of two parallel lines are always equal.