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

Question

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 (A)  1, 3  (B)  0, 2  (C)  -1, 3  (D)  -3, -2

Tardigrade Admin · Accepted Answer

(1/2)×1(k-1)=±1 k â 1 = ± 2 k=3 k=-1

Q. 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

${1, 3}$

${0, 2}$

${- 1, 3}$

${- 3, - 2}$

$\frac{1}{2} \times 1 (k - 1) = \pm 1$
$k - 1 = \pm 2$
$k = 3$
$k = - 1$

Q. 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

{1,3}

{0,2}

{−1,3}

{−3,−2}