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
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