# 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

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