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

 Equation of a Straight Line










10th Grade Math

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