Let A(h, k) , B(1, 1) and C(2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the

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Let A(h, k) , B(1, 1) and C(2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1 , then the set of values which k can take is given by:

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![image](https://pub-7faa453bd1ec4bf0a04cc031c6f058de.r2.dev/images/27bfbb94-2415-4299-ab00-fc8245878188.png) For AC to be the hypotenuse, B must be 90^ .Wait, checking slopes: B = (1, 1) and C = (2, 1) . The line BC is horizontal ( y=1 ).For B = 90^ , AB must be vertical. So the x -coordinate of A must be 1 .Thus, A = (1, k) .Area of ABC = 1/2 × base × height = 1 .Base BC = √((2-1)^2 + (1-1)^2) = 1 .Height AB = √((1-1)^2 + (k-1)^2) = |k-1| . 1/2 × 1 × |k-1| = 1 |k-1| = 2 k - 1 = 2 k = 3 k - 1 =

Let $A (h, k)$ , $B (1, 1)$ and $C (2, 1)$ be the vertices of a right angled triangle with $A C$ as its hypotenuse. If the area of the triangle is $1$ , then the set of values which $k$ can take is given by:

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Let A(h,k), B(1,1) and C(2,1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1, then the set of values which k can take is given by: