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Find the equation of straight line passing through (2, 3) and perpendicular to the line ...

Find the equation of straight line passing through (2, 3) and perpendicular to the line \(3x + 2y + 4 = 0\)
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  • A 3y = 5x - 2
  • B Y = \(\frac {5}{3} \times - 2\)
  • C None of these
  • D 3y = 2x + 5
Correct Answer: Option D
Explanation:
\(3x + 2y + 4 = 0\)

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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