Question

Points $ P $ and $ Q $ lie on the lines: $$ 3x + 4y - 4 = 0 \quad \text{and} \quad 5x - y - 4 = 0 $$ The midpoint of segment $ PQ $ is $ (1, 5) $. Find the slope of the line $ PQ $.

• A. \( \frac{83}{35} \)
• B. \( \frac{63}{35} \)
• C. \( -\frac{3}{4} \)
• D. \( \frac{3}{4} \)

Hint:

Use midpoint constraints to form equations. Solve using substitution and simplify carefully to find coordinates and slope.

Answer 1

The Correct Option is A

Let \( P = (x_1, y_1) \in 3x + 4y - 4 = 0 \), and \( Q = (x_2, y_2) \in 5x - y - 4 = 0 \) Midpoint: \[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (1, 5) \Rightarrow x_1 + x_2 = 2,\quad y_1 + y_2 = 10 \] Now express:
From \( P \in 3x + 4y = 4 \Rightarrow y_1 = \frac{4 - 3x_1}{4} \)
From \( Q \in 5x - y = 4 \Rightarrow y_2 = 5x_2 - 4 \) Now use: \[ y_1 + y_2 = 10 \Rightarrow \frac{4 - 3x_1}{4} + (5x_2 - 4) = 10 \Rightarrow \frac{4 - 3x_1 + 20x_2 - 16}{4} = 10 \Rightarrow \frac{-3x_1 + 20x_2 - 12}{4} = 10 \Rightarrow -3x_1 + 20x_2 = 52 \quad \text{(1)} \] Also: \( x_1 + x_2 = 2 \Rightarrow x_1 = 2 - x_2 \) Sub into (1): \[ -3(2 - x_2) + 20x_2 = 52 \Rightarrow -6 + 3x_2 + 20x_2 = 52 \Rightarrow 23x_2 = 58 \Rightarrow x_2 = \frac{58}{23} = \frac{58}{23} \Rightarrow x_1 = 2 - \frac{58}{23} = \frac{46 - 58}{23} = -\frac{12}{23} \] Now find \( y_1, y_2 \): \[ y_1 = \frac{4 - 3x_1}{4} = \frac{4 + \frac{36}{23}}{4} = \frac{\frac{92 + 36}{23}}{4} = \frac{128}{92} = \frac{32}{23} \] \[ y_2 = 5x_2 - 4 = \frac{290}{23} - 4 = \frac{198}{23} \] Now slope: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{198 - 32}{23}}{\frac{58 + 12}{23}} = \frac{166}{70} = \boxed{ \frac{83}{35} } \]