Question

let A(h,k),B(1,1)andC(2,1) be the vertices of a right angled triangle with AC as it's hypotenuse.if the area of the triangle is 1, then the set of values which K can take is given by​

Answer 1

Answer:

∵A(h,k),B(1,1) and C(2,1) are the vertices of a right angled triangle ABC

Now, AB=

(1−h)

2

+(1−k)

2

or BC=

(2−1)

2

+(1−1)

2

=1

or CA=

(h−2)

2

+(k−1)

2

Now, pythagorus theorem

AC

2

=AB

2

+BC

2

4+h

2

−4h+k

2

+1−2k

=h

2

+1−2h+k

2

+1−2k+1

5−4h=3−2h

⇒h=1

Now, given that area of the triangle is 1,

Then, area (△ABC)=

2

1

×AB×BC

1=

2

1

×

(1−h)

2

+(1−k)

2

×1⇒2=

(1−h)

2

+(1−k)

2

...(2)

Putting h=1 from equation (1), we get 2=

(k−1)

2

Squaring both the sides , we get

4=k

2

+1−2k or k

2

−2k−3=0

or (k−3)(k+1)=0

So, k=−1,3

Thus the set of values of k is {−1,3}

Step-by-step explanation:

I hope it help you