Calculate, correct to one decimal place, the acute angle between the lines 3x - 4y + 5 = 0 and 2x + 3y - 1 = 0.
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Correct Answer: Option A
Explanation:
\(\tan \theta = \frac{m_{1} - m_{2}}{1 - m_{1}m_{2}}\)
\(m_{1} = \text{slope of 1st line } 4y = 3x + 5 \implies y = \frac{3}{4}x + \frac{5}{4}\)
\(m_{1} = \frac{3}{4}\)
\(m_{2} = \text{slope of 2nd line} 3y = 1 - 2x \implies y = \frac{1}{3} - \frac{2}{3}x\)
\(m_{2} = -\frac{2}{3}\)
\(\tan \theta = \frac{\frac{3}{4} - (-\frac{2}{3})}{1 - ((\frac{3}{4})(-\frac{2}{3}))} = \frac{\frac{17}{12}}{\frac{1}{2}}\)
\(\tan \theta = \frac{17}{6}\)
\(\theta \approxeq 70.6°\)
\(\tan \theta = \frac{m_{1} - m_{2}}{1 - m_{1}m_{2}}\)
\(m_{1} = \text{slope of 1st line } 4y = 3x + 5 \implies y = \frac{3}{4}x + \frac{5}{4}\)
\(m_{1} = \frac{3}{4}\)
\(m_{2} = \text{slope of 2nd line} 3y = 1 - 2x \implies y = \frac{1}{3} - \frac{2}{3}x\)
\(m_{2} = -\frac{2}{3}\)
\(\tan \theta = \frac{\frac{3}{4} - (-\frac{2}{3})}{1 - ((\frac{3}{4})(-\frac{2}{3}))} = \frac{\frac{17}{12}}{\frac{1}{2}}\)
\(\tan \theta = \frac{17}{6}\)
\(\theta \approxeq 70.6°\)