Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other.
Straight Line Eq
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Question

Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other.

KnowledgeBoat · Accepted Answer

Let slope of kx - 5y + 4 = 0 be m₁ and 5x - 2y + 5 = 0 be m₂.

Given,

⇒ kx – 5y + 4 = 0

⇒ 5y = kx + 4

⇒ y = $\dfrac{k}{5}x + \dfrac{4}{5}$ .

Comparing above equation with y = mx + c we get,

Slope of this line (m₁) = $\dfrac{k}{5}$

Given,

⇒ 5x – 2y + 5 = 0

⇒ 2y = 5x + 5

⇒ y = $\dfrac{5}{2}x + \dfrac{5}{2}$

Slope of this line (m₂) = $\dfrac{5}{2}$ .

As, the lines are perpendicular to each other so product of their slopes = -1.

⇒ m₁ x m₂ = -1

⇒ $\dfrac{k}{5} \times \dfrac{5}{2} = -1$

⇒ $\dfrac{k}{2}$ = -1

⇒ k = -2.

Hence, k = -2.

Mathematics