Question

The area (in square units) of the triangle formed by the lines 6x² + 13xy + 6y² = 0 and x + 2y + 3 = 0 is:

• A. $\frac{9}{2}$
• B. $\frac{45}{4}$
• C. $\frac{9}{8}$
• D. $\frac{45}{8}$

Hint:

To find the area of a triangle formed by lines, find the intersection points and use the area formula.

Answer 1

The Correct Option is D

Step 1: Factorize the equation 6x² + 13xy + 6y² = 0.
6x² + 13xy + 6y² = 0
6x² + 9xy + 4xy + 6y² = 0
3x(2x + 3y) + 2y(2x + 3y) = 0
(3x + 2y)(2x + 3y) = 0
So, the two lines are 3x + 2y = 0 and 2x + 3y = 0.

Step 2: Find the intersection points of the lines.
Let the lines be:
L₁: 3x + 2y = 0
L₂: 2x + 3y = 0
L₃: x + 2y + 3 = 0
Intersection of L₁ and L₂:
3x + 2y = 0 and 2x + 3y = 0
Solving these, we get x = 0 and y = 0.
So, the intersection point is A(0, 0).
Intersection of L₁ and L₃:
3x + 2y = 0 and x + 2y + 3 = 0
Subtracting the equations:
2x - 3 = 0
x = \(\frac{3}{2}\)
2y = -3x = -\(\frac{9}{2}\)
y = -\(\frac{9}{4}\)
So, the intersection point is B(\(\frac{3}{2}\), -\(\frac{9}{4}\)).
Intersection of L₂ and L₃:
2x + 3y = 0 and x + 2y + 3 = 0
From x + 2y + 3 = 0, x = -2y - 3.
Substitute in 2x + 3y = 0:
2(-2y - 3) + 3y = 0
-4y - 6 + 3y = 0
-y = 6
y = -6
x = -2(-6) - 3 = 12 - 3 = 9
So, the intersection point is C(9, -6).

Step 3: Calculate the area of the triangle.
The area of the triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by:
Area = \(\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)
Area = \(\frac{1}{2} |0(-\frac{9}{4} + 6) + \frac{3}{2}(-6 - 0) + 9(0 + \frac{9}{4})|\)
Area = \(\frac{1}{2} |0 - 9 + \frac{81}{4}|\)
Area = \(\frac{1}{2} |\frac{-36 + 81}{4}|\)
Area = \(\frac{1}{2} \times \frac{45}{4} = \frac{45}{8}\)
Therefore, the area of the triangle is \(\frac{45}{8}\) square units.