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(A) Let the slope of the required line be $m$. The given lines are $3x - 4y - 7 = 0$ and $12x - 5y + 6 = 0$. The slopes of these lines are $m_1 = \fra

Vedclass · Accepted Answer

$9x - 7y = 1$ and $7x + 9y = 73$

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(A) Let the slope of the required line be $m$. The given lines are $3x - 4y - 7 = 0$ and $12x - 5y + 6 = 0$. The slopes of these lines are $m_1 = \frac{3}{4}$ and $m_2 = \frac{12}{5}$. Since the required line makes equal angles with the given lines,the angle between the required line and the given lines is equal. Using the formula $	an 	heta = |\frac{m - m_1}{1 + m m_1}| = |\frac{m - m_2}{1 + m m_2}|$,we have: $|\frac{m - 3/4}{1 + 3m/4}| = |\frac{m - 12/5}{1 + 12m/5}|$ $|\frac{4m - 3}{4 + 3m}| = |\frac{5m - 12}{5 + 12m}|$ $(4m - 3)(5 + 12m) = \pm (5m - 12)(4 + 3m)$ Case $1$: $(4m - 3)(5 + 12m) = (5m - 12)(4 + 3m)$ $20m + 48m^2 - 15 - 36m = 20m + 15m^2 - 48 - 36m$ $33m^2 = -33 \Rightarrow m^2 = -1$ (No real solution). Case $2$: $(4m - 3)(5 + 12m) = -(5m - 12)(4 + 3m)$ $48m^2 - 16m - 15 = -(15m^2 - 16m - 48)$ $48m^2 - 16m - 15 = -15m^2 + 16m + 48$ $63m^2 - 32m - 63 = 0$ Solving for $m$: $m = \frac{32 \pm \sqrt{1024 - 4(63)(-63)}}{126} = \frac{32 \pm \sqrt{1024 + 15876}}{126} = \frac{32 \pm 130}{126}$ $m_1 = \frac{162}{126} = \frac{9}{7}$ and $m_2 = \frac{-98}{126} = -\frac{7}{9}$. Using $y - 5 = m(x - 4)$: For $m = 9/7$: $7y - 35 = 9x - 36 \Rightarrow 9x - 7y = 1$. For $m = -7/9$: $9y - 45 = -7x + 28 \Rightarrow 7x + 9y = 73$.

Find the equation of the line passing through the point $(4, 5)$ and making equal angles with the lines $3x = 4y + 7$ and $5y = 12x + 6$.

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