Question

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

Answer 1

 kx − 5y + 4 = 0

${\displaystyle \Rightarrow }$ 5y = kx + 4

${\displaystyle \Rightarrow y=\frac{k}{5}x+\frac{4}{5}}$

Slope of this line = ${\displaystyle m_1=\frac{k}{5}}$

5x − 2y + 5 = 0

${\displaystyle \Rightarrow }$ 2y = 5x + 5

${\displaystyle \Rightarrow y=\frac{5}{2}x+\frac{5}{2}}$

Slope of this line = ${\displaystyle m_2=\frac{5}{2}}$

Since, the lines are perpendicular, m₁ × m₂ = –1

${\displaystyle \Rightarrow \frac{k}{5}\times \frac{5}{2}=-1}$

${\displaystyle \Rightarrow }$ k = –2