Multiple choice

The quadratic equation whose roots are the $x$ and $y$ intercepts of the line passing through $(1,1)$ and making a triangle of area $A$ with the axes, may be

  1. $ x^{2}+Ax+2A=0 $
  2. $ x^{2}-2Ax+2A=0 $
  3. $ x^{2}-Ax+2A=0 $

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let intercepts be a and b. Line is x/a + y/b = 1. Passes through (1,1) => 1/a + 1/b = 1 => (a+b)/ab = 1 => a+b = ab. Area A = 0.5 * |ab| = A => ab = 2A. Since a+b = ab, a+b = 2A. The quadratic equation with roots a and b is x^2 - (a+b)x + ab = 0, which is x^2 - 2Ax + 2A = 0.

AI explanation

Let the line intersect the x-axis at (p, 0) and the y-axis at (0, q). The equation of the line in intercept form is x/p + y/q = 1, and since it passes through (1, 1), we have 1/p + 1/q = 1, or p + q = pq. The area of the right triangle formed with the axes is A = pq/2, which means pq = 2A. The quadratic equation whose roots are p and q is x^2 - (p + q)x + pq = 0. Substituting p + q = 2A and pq = 2A gives the equation x^2 - 2Ax + 2A = 0.

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