Problems on Condition of Perpendicularity

Here we will solve various types of problems on condition of perpendicularity of two lines. 

1. Prove that the lines 5x + 4y = 9 and 4x – 5y – 1 = 0 are perpendicular to each other.

Solution:

Equation of the 1st line 5x + 4y = 9.

Now we need to express the above equation in the form y = mx + c.

5x + 4y = 9

4y = -5x + 9

y = -\(\frac{5}{4}\)x + \(\frac{9}{4}\)

Therefore, the slope (m \(_{1}\)) of the 1st line = -5/4

Equation of the second line 4x - 5y - 1 = 0

Now we need to express the above equation in the form y = mx + c.

4x – 5y – 1 = 0

⟹ -5y = -4x + 1

⟹ y = \(\frac{4}{5}\)– \(\frac{1}{5}\) 

Therefore, the slope (m \(_{2}\)) of the 2nd line = \(\frac{4}{5}\)

Now,

m \(_{1}\) × m \(_{2}\) = \(\frac{-5}{4}\)  ×  \(\frac{4}{5}\)= -1

Therefore, the given lines are perpendicular to each other.


2. Find the value of k if the lines 7y = kx + 4 and x + 2y = 3 are perpendicular.

Solution:

The slope of the lines can be found by comparing the equations with y = mx + c.

Equation of the first straight line 7y = kx + 4

Now we need to express the given equation in the form y = mx + c.

7y = kx + 4

⟹ y = \(\frac{k}{7}\)x + \(\frac{4}{7}\)

Therefore, the slope (m\(_{1}\)) of the given line = \(\frac{k}{7}\)

Equation of the second line x + 2y = 3

Now we need to express the given equation in the form y = mx + c.

x + 2y = 3

⟹ 2y = -x + 3

⟹ y = -\(\frac{1}{2}\)x + \(\frac{3}{2}\)

Therefore, the slope (m\(_{2}\)) of the given line = -\(\frac{1}{2}\)

Now according o the problem the two given lines are perpendicular.

i.e., m\(_{1}\) × m\(_{2}\) = -1

⟹ \(\frac{k}{7}\) × -\(\frac{1}{2}\) = -1

⟹ -\(\frac{k}{14}\) = -1

⟹ k = 14

Therefore, the value of k = 14







10th Grade Math

From Problems on Condition of Perpendicularity to HOME




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.