Question

If the angle between the lines $7x+3py+5=0$ and $3x-14py+11=0$ is $\frac{\pi}{2}$, then a value of $p$ is}

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{1}{\sqrt{2}}$

Hint:

Whenever two lines \[ A_1x+B_1y+C_1=0 \] and \[ A_2x+B_2y+C_2=0 \] are perpendicular, directly use \[ A_1A_2+B_1B_2=0. \] For this problem, \[ 7(3)+(3p)(-14p)=0 \] immediately gives \[ 21=42p^2 \] and hence \[ p=\pm\frac{1}{\sqrt{2}}. \] This method is much faster than converting both equations into slope-intercept form.

Answer 1

The Correct Option is D

Concept: The angle between two straight lines is determined by their slopes. When the angle between two lines is \[ \frac{\pi}{2}=90^\circ, \] the lines are perpendicular to each other. For two lines represented in the general form \[ A_1x+B_1y+C_1=0 \] and \[ A_2x+B_2y+C_2=0, \] the condition for perpendicularity is \[ A_1A_2+B_1B_2=0. \] This formula is obtained from the condition that the product of the slopes of two perpendicular lines is equal to $-1$.

Step 1: Identify the coefficients of the given lines. The first line is \[ 7x+3py+5=0. \] Comparing with the standard form, \[ A_1=7,\qquad B_1=3p. \] The second line is \[ 3x-14py+11=0. \] Comparing with the standard form, \[ A_2=3,\qquad B_2=-14p. \]

Step 2: Apply the condition for perpendicular lines. Since the angle between the lines is \[ \frac{\pi}{2}, \] the lines are perpendicular. Therefore, \[ A_1A_2+B_1B_2=0. \] Substituting the coefficients, \[ (7)(3)+(3p)(-14p)=0. \] \[ 21-42p^2=0. \]

Step 3: Solve for $p^2$. Transposing the second term to the right-hand side, \[ 21=42p^2. \] Dividing both sides by $42$, \[ p^2=\frac{21}{42}. \] \[ p^2=\frac{1}{2}. \]

Step 4: Determine the possible values of $p$. Taking square roots on both sides, \[ p=\pm\sqrt{\frac{1}{2}}. \] \[ p=\pm\frac{1}{\sqrt{2}}. \] Since the question asks for a value of $p$, one valid value is \[ \boxed{\frac{1}{\sqrt{2}}}. \]

Step 5: Verify with the given options. The available options are \[ \frac{1}{2},\quad \frac{1}{3},\quad \frac{2}{3},\quad \frac{1}{\sqrt{2}}. \] The value obtained from the perpendicularity condition is \[ \frac{1}{\sqrt{2}}, \] which matches option (D). Hence, the correct answer is \[ \boxed{\frac{1}{\sqrt{2}}}. \]