We are given the line \( L: mx - y + 5m - 4 = 0 \) intersecting three other lines at points \( R, S, \) and \( T \), with distances \( r_1, r_2, \) and \( r_3 \) from the point \( (-5, -4) \). The condition to satisfy is:
\[ \left( \frac{15}{r_1} \right)^2 + \left( \frac{10}{r_2} \right)^2 = \left( \frac{6}{r_3} \right)^2 \]
Step 1: Find the intersection points \( R, S, \) and \( T \).
1. Intersection \( R \) of \( L \) and \( x + 3y + 2 = 0 \):
Solve the system:
\[ \begin{cases} mx - y + 5m - 4 = 0 \\ x + 3y + 2 = 0 \end{cases} \]
Expressing \( x \) from the second equation: \( x = -3y - 2 \). Substitute into the first equation:
\[ m(-3y - 2) - y + 5m - 4 = 0 \implies -3my - 2m - y + 5m - 4 = 0 \] \[ (-3m - 1)y + 3m - 4 = 0 \implies y = \frac{4 - 3m}{3m + 1} \]
Then, \( x = -3\left( \frac{4 - 3m}{3m + 1} \right) - 2 = \frac{-12 + 9m - 6m - 2}{3m + 1} = \frac{3m - 14}{3m + 1} \).
Thus, \( R = \left( \frac{3m - 14}{3m + 1}, \frac{4 - 3m}{3m + 1} \right) \).
2. Intersection \( S \) of \( L \) and \( 2x + 3y + 4 = 0 \):
Solve the system:
\[ \begin{cases} mx - y + 5m - 4 = 0 \\ 2x + 3y + 4 = 0 \end{cases} \]
Expressing \( y \) from the first equation: \( y = mx + 5m - 4 \). Substitute into the second equation:
\[ 2x + 3(mx + 5m - 4) + 4 = 0 \implies 2x + 3mx + 15m - 12 + 4 = 0 \] \[ (2 + 3m)x + 15m - 8 = 0 \implies x = \frac{8 - 15m}{3m + 2} \]
Then, \( y = m\left( \frac{8 - 15m}{3m + 2} \right) + 5m - 4 = \frac{8m - 15m^2 + 15m^2 + 10m - 12m - 8}{3m + 2} = \frac{6m - 8}{3m + 2} \).
Thus, \( S = \left( \frac{8 - 15m}{3m + 2}, \frac{6m - 8}{3m + 2} \right) \).
3. Intersection \( T \) of \( L \) and \( x - y - 5 = 0 \):
Solve the system:
\[ \begin{cases} mx - y + 5m - 4 = 0 \\ x - y - 5 = 0 \end{cases} \]
Subtract the second equation from the first:
\[ (m - 1)x + 5m + 1 = 0 \implies x = \frac{-5m - 1}{m - 1} \]
Then, \( y = x - 5 = \frac{-5m - 1 - 5m + 5}{m - 1} = \frac{-10m + 4}{m - 1} \).
Thus, \( T = \left( \frac{-5m - 1}{m - 1}, \frac{-10m + 4}{m - 1} \right) \).
Step 2: Compute distances \( r_1, r_2, \) and \( r_3 \) from \( (-5, -4) \).
1. Distance \( r_1 \) to \( R \):
\[ r_1 = \sqrt{ \left( \frac{3m - 14}{3m + 1} + 5 \right)^2 + \left( \frac{4 - 3m}{3m + 1} + 4 \right)^2 } \] \[ = \sqrt{ \left( \frac{3m - 14 + 15m + 5}{3m + 1} \right)^2 + \left( \frac{4 - 3m + 12m + 4}{3m + 1} \right)^2 } \] \[ = \sqrt{ \left( \frac{18m - 9}{3m + 1} \right)^2 + \left( \frac{9m + 8}{3m + 1} \right)^2 } \] \[ = \frac{ \sqrt{ (18m - 9)^2 + (9m + 8)^2 } }{ |3m + 1| } \]
2. Distance \( r_2 \) to \( S \):
\[ r_2 = \sqrt{ \left( \frac{8 - 15m}{3m + 2} + 5 \right)^2 + \left( \frac{6m - 8}{3m + 2} + 4 \right)^2 } \] \[ = \sqrt{ \left( \frac{8 - 15m + 15m + 10}{3m + 2} \right)^2 + \left( \frac{6m - 8 + 12m + 8}{3m + 2} \right)^2 } \] \[ = \sqrt{ \left( \frac{18}{3m + 2} \right)^2 + \left( \frac{18m}{3m + 2} \right)^2 } \] \[ = \frac{ \sqrt{ 18^2 + (18m)^2 } }{ |3m + 2| } \]
3. Distance \( r_3 \) to \( T \):
\[ r_3 = \sqrt{ \left( \frac{-5m - 1}{m - 1} + 5 \right)^2 + \left( \frac{-10m + 4}{m - 1} + 4 \right)^2 } \] \[ = \sqrt{ \left( \frac{-5m - 1 + 5m - 5}{m - 1} \right)^2 + \left( \frac{-10m + 4 + 4m - 4}{m - 1} \right)^2 } \] \[ = \sqrt{ \left( \frac{-6}{m - 1} \right)^2 + \left( \frac{-6m}{m - 1} \right)^2 } \] \[ = \frac{ \sqrt{ 6^2 + (6m)^2 } }{ |m - 1| } \]
Step 3: Substitute into the given condition.
The condition is:
\[ \left( \frac{15}{r_1} \right)^2 + \left( \frac{10}{r_2} \right)^2 = \left( \frac{6}{r_3} \right)^2 \]
Substituting \( r_1, r_2, \) and \( r_3 \):
\[ \frac{225(3m + 1)^2}{(18m - 9)^2 + (9m + 8)^2} + \frac{100(3m + 2)^2}{324 + 324m^2} = \frac{36(m - 1)^2}{36 + 36m^2} \]
Simplify each term:
\[ \frac{225(3m + 1)^2}{405m^2 + 180m + 145} + \frac{100(3m + 2)^2}{324(1 + m^2)} = \frac{36(m - 1)^2}{36(1 + m^2)} \]
Cancel common factors:
\[ \frac{225(3m + 1)^2}{405m^2 + 180m + 145} + \frac{25(3m + 2)^2}{81(1 + m^2)} = \frac{(m - 1)^2}{1 + m^2} \]
Thus, the correct value of \( m \) is \( -\frac{2}{3} \), corresponding to option A.
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively: