Let A (1, 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 1square unit, then the set of values which 'k' can take is given by

Question

Let A (1, 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 1square unit, then the set of values which 'k' can take is given by (A) -1, 3 (B) -3, -2 (C) 1, 3 (D) 0, 2

Tardigrade Admin · Accepted Answer

Given: The vertices of a right angled triangle A(l, k), B(1, 1) and C(2, 1) and Area of ΔABC = 1 square unit We know that, area of right angled triangle = ( 1/2) × BC × AB = 1 = (1/2)(1)|(k-1)| ⇒ ±(k-1) = 2 ⇒ k = - 1, 3

Q. Let A (1, 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 1square unit, then the set of values which 'k' can take is given by

{-1, 3}

{-3, -2}

{1, 3}

{0, 2}

Given : The vertices of a right angled triangle A(l, k), B(1, 1) and C(2, 1) and Area of $Δ$ ABC = 1 square unit
We know that, area of right angled triangle
$= \frac{1}{2} \times BC \times A B = 1 = \frac{1}{2} (1) ∣ (k - 1) ∣$
$\Rightarrow \pm (k - 1) = 2 \Rightarrow k = - 1, 3$

Q. Let A (1, 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 1square unit, then the set of values which 'k' can take is given by

{-1, 3}

{-3, -2}

{1, 3}

{0, 2}

Given : The vertices of a right angled triangle A(l, k), B(1, 1) and C(2, 1) and Area of ΔABC = 1 square unit We know that, area of right angled triangle =21​×BC×AB=1=21​(1)∣(k−1)∣ ⇒±(k−1)=2⇒k=−1,3

Given : The vertices of a right angled triangle A(l, k), B(1, 1) and C(2, 1) and Area of $Δ$ ABC = 1 square unit
We know that, area of right angled triangle
$= \frac{1}{2} \times BC \times A B = 1 = \frac{1}{2} (1) ∣ (k - 1) ∣$
$\Rightarrow \pm (k - 1) = 2 \Rightarrow k = - 1, 3$