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 $1$ square unit, then the set of values which $k$ can take is given by

{- 1, 3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

{- 3, - 2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{1, 3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{0, 2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A ${- 1, 3}$
$\frac{1}{2} \times | k - 1 | \times 1 = 1$
$\Rightarrow k = - 1, 3$

Suggest Corrections

Similar questions

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 square unit, then the set of values which

^{'} k^{'}

can take is given by

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

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

sq units, then the set of values which

k

can take is given by

Let k be an integer such that the triangle with vertices (k,-3k), (5, k) and (-k, 2) has area 28 sq units. Then, the orthocentre of this triangle is at the point

Join BYJU'S Learning Program

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 square unit, then the set of values which k can take is given by

The correct option is A {−1,3}12×|k−1|×1=1 ⇒k=−1,3

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 $1$ square unit, then the set of values which $k$ can take is given by

The correct option is A ${- 1, 3}$
$\frac{1}{2} \times | k - 1 | \times 1 = 1$
$\Rightarrow k = - 1, 3$