Question

A pair of lines drawn through the origin forms a right-angled isosceles triangle with right angle at the origin with the line \( 2x + 3y = 6 \). The area (in square units) of the triangle thus formed is:

• A. \( \frac{36}{13} \)
• B. \( \frac{32}{13} \)
• C. \( \frac{18}{5} \)
• D. \( \frac{25}{9} \)

Hint:

For isosceles right-angled triangles, use the perpendicular line intercept method to find the base and height, then apply the area formula.

Answer 1

The Correct Option is A

To solve this problem, we need to find the area of a right
-angled isosceles triangle formed by a pair of lines through the origin and the line \( 2x + 3y = 6 \), with the right angle at the origin. Step 1: Understand the problem
The triangle is right
-angled at the origin, and it is isosceles, meaning the two legs of the triangle are equal in length.
The two lines forming the legs of the triangle pass through the origin, and the hypotenuse is the line \( 2x + 3y = 6 \). Step 2: Find the distance from the origin to the line \( 2x + 3y = 6 \)
The distance \( d \) from the origin \((0, 0)\) to the line \( 2x + 3y = 6 \) is given by the formula: \[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}, \] where \( Ax + By + C = 0 \) is the equation of the line, and \((x_0, y_0)\) is the point. For the line \( 2x + 3y = 6 \), rewrite it in standard form: \[ 2x + 3y
- 6 = 0. \] Here, \( A = 2 \), \( B = 3 \), and \( C =
-6 \). Substituting \((x_0, y_0) = (0, 0)\): \[ d = \frac{|2(0) + 3(0)
- 6|}{\sqrt{2^2 + 3^2}} = \frac{6}{\sqrt{13}}. \] Step 3: Relate the distance to the triangle
Since the triangle is right
-angled and isosceles, the two legs of the triangle are equal in length. Let the length of each leg be \( L \). The hypotenuse of the triangle is the line \( 2x + 3y = 6 \), and its length is twice the distance from the origin to the line (because the triangle is isosceles). Thus: \[ \text{Hypotenuse} = 2d = \frac{12}{\sqrt{13}}. \] For a right
-angled isosceles triangle, the relationship between the legs and the hypotenuse is: \[ \text{Hypotenuse} = L\sqrt{2}. \] Substitute the value of the hypotenuse: \[ L\sqrt{2} = \frac{12}{\sqrt{13}}. \] Solve for \( L \): \[ L = \frac{12}{\sqrt{13} \cdot \sqrt{2}} = \frac{12}{\sqrt{26}}. \] Step 4: Compute the area of the triangle
The area \( A \) of a right
-angled isosceles triangle is: \[ A = \frac{1}{2} \cdot L^2. \] Substitute \( L = \frac{12}{\sqrt{26}} \): \[ A = \frac{1}{2} \cdot \left(\frac{12}{\sqrt{26}}\right)^2 = \frac{1}{2} \cdot \frac{144}{26} = \frac{72}{26} = \frac{36}{13}. \] Final Answer: The area of the triangle is: \[ \boxed{\frac{36}{13}} \]