26)Show that the straight lines x - 2y + 3 = 0 and 6x + 3y + 8 = 0 are perpendicular.
Answer: Slope of the straight line x - 2y + 3 = 0 is
\(m_1 = -\frac{\text{coeff of x}}{\text{coeff of y}} = -\frac{1}{-2} = \frac{1}{2}\)
Slope of the straight line 6x + 3y + 8 = 0 is
\(m_2 = -\frac{6}{3} = -2\)
Now, \(m_1 \times m_2 = \frac{1}{2} \times (-2) = -1\)
Hence, the two straight lines are perpendicular.