Find the equation of straight line passing through (2, 3) and perpendicular to the line \(3x + 2y + 4 = 0\)
\(3x + 2y + 4 = 0\)
Rearrange:
\(2y = -3x - 4\)
Divide both sides by 2
y = \(\frac {-3 \times - 4}{2}\)
y = \(\frac {-3}{2} \times - 2\)
∴ the gradient of the line 3x + 2y + 4 = 0 is \(\frac {-3}{2}\)
If two lines are perpendicular to each other ∴ \(m_1 x m_2\) = -1
Let \(m_1 = \frac {-3}{2} \therefore m_2 = \frac {-1}{m_1} = \frac {-1}{-3/2} = \frac {2}{3}\)
From the equation of a line which is given as m = \(\frac {y - y_1}{x - x_1} where (x_1, y_1) = (2,3)\)
\(\therefore \frac {2}{3} = \frac {y - 3}{x - 2}\)
=3(y - 3) = 2(x - 2)
=3y - 9 = 2 x -4
=3y = 2 x -4 + 9
∴ 3y = 2x + 5
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