Question

If a line ax + 2y = k forms a triangle of area 3 sq.units with the coordinate axis and is perpendicular to the line 2x - 3y + 7 = 0, then the product of all the possible values of k is 

• A. -36
• B. 36
• C. -64
• D. 64

Answer 1

The Correct Option is A

To find the product of all possible values of $k$ for the line $ax + 2y = k$ that is perpendicular to $2x - 3y + 7 = 0$ and forms a triangle with the coordinate axes of area 3, we proceed as follows:

1. Determining the Slope Condition:
Rewrite the line $ax + 2y = k$ in slope-intercept form:

$ y = -\frac{a}{2}x + \frac{k}{2} $
The slope is $-\frac{a}{2}$. Rewrite the line $2x - 3y + 7 = 0$:

$ y = \frac{2}{3}x + \frac{7}{3} $
The slope is $\frac{2}{3}$. Since the lines are perpendicular, the product of their slopes is -1:

$ \left( -\frac{a}{2} \right) \cdot \frac{2}{3} = -1 $
$ -\frac{a}{3} = -1 $
$ a = 3 $

2. Equation of the Line:
With $a = 3$, the line is $3x + 2y = k$. Rewrite in slope-intercept form:

$ y = -\frac{3}{2}x + \frac{k}{2} $

3. Finding the Intercepts:
The x-intercept occurs when $y = 0$:

$ 3x = k $
$ x = \frac{k}{3} $
The y-intercept occurs when $x = 0$:

$ 2y = k $
$ y = \frac{k}{2} $

4. Area of the Triangle:
The triangle formed by the x-intercept $\left( \frac{k}{3}, 0 \right)$ and y-intercept $\left( 0, \frac{k}{2} \right)$ with the coordinate axes has area:

$ \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot \frac{k}{3} \cdot \frac{k}{2} = \frac{k^2}{12} $
Given the area is 3:

$ \frac{k^2}{12} = 3 $
$ k^2 = 36 $
$ k = \pm 6 $

5. Product of Possible $k$ Values:
The possible values of $k$ are 6 and -6. Their product is:

$ 6 \cdot (-6) = -36 $

Final Answer:
The product of all possible values of $k$ is $-36$.