Question

The area of the triangle formed by the line L with the coordinate axes is 12 sq. units. If L passes through the point (12,4) and the product P of X-intercept of L and square of the Y-intercept of L is negative, then P =

• A. -48
• B. -24
• C. -192
• D. -72

Hint:

When a problem involves a line's intercepts and area, the intercept form \(x/a + y/b = 1\) is the most convenient starting point. The area formed with the axes is simply \( \frac{1}{2}|ab| \).

Answer 1

The Correct Option is A

Let the equation of the line L in intercept form be \( \frac{x}{a} + \frac{y}{b} = 1 \).
The area of the triangle formed with the axes is \( \frac{1}{2}|ab| = 12 \), which means \( |ab| = 24 \).
The line passes through the point (12, 4), so this point must satisfy the equation.
\( \frac{12}{a} + \frac{4}{b} = 1 \). Multiplying by \(ab\) gives \( 12b + 4a = ab \).
We have two cases for \(ab\). Case 1: \(ab = 24\). Case 2: \(ab = -24\).
Case 1: \(ab = 24\). The equation becomes \( 12b + 4a = 24 \). Since \(b=24/a\), we have \( 12(24/a) + 4a = 24 \).
\( 288/a + 4a = 24 \implies 288 + 4a^2 = 24a \implies a^2 - 6a + 72 = 0 \). The discriminant is \( (-6)^2 - 4(72)<0 \), so no real solutions.
Case 2: \(ab = -24\). The equation becomes \( 12b + 4a = -24 \). Since \(b=-24/a\), we have \( 12(-24/a) + 4a = -24 \).
\( -288/a + 4a = -24 \implies -288 + 4a^2 = -24a \implies a^2 + 6a - 72 = 0 \).
Solving for a: \( a = \frac{-6 \pm \sqrt{36-4(1)(-72)}}{2} = \frac{-6 \pm \sqrt{324}}{2} = \frac{-6 \pm 18}{2} \).
This gives two possible values for a: \( a_1 = \frac{12}{2} = 6 \) and \( a_2 = \frac{-24}{2} = -12 \).
If \(a=6\), then \(b=-24/6 = -4\). If \(a=-12\), then \(b=-24/(-12) = 2\).
We need to find P, which is defined as \( P = a \cdot b^2 \).
For the pair (a=6, b=-4): \( P = 6 \cdot (-4)^2 = 6 \cdot 16 = 96 \).
For the pair (a=-12, b=2): \( P = -12 \cdot (2)^2 = -12 \cdot 4 = -48 \).
The problem states that P is negative, so we choose the second case.
Thus, P = -48.