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 $1$ sq units, then the set of values which $k$ can take is given by

Answer 1

Let A, B, C are the vertices of a right angled triangle with AC as its hypotenuse.

Using Pythagoras theorem we get

$A{C}^2=A{B}^2+B{C}^2$

$(h-2{)}^2+(k-1{)}^2=(h-1{)}^2+(k-1{)}^2+1$

$h^2+4-4h=h^2+1-2h+1$

$2h=2$

$[h=1]$

Now area of triangle $=\frac{1}{2}\times AB\times BC$

$1=\frac{1}{2}\times 1\times \sqrt{(h-1{)}^2+(k-1{)}^2}$

$(h-1{)}^2+(k-1{)}^2=4$

$h=1$

$(k-1{)}^2=4$

$k-1=\pm 2$

$k=3,-1$

Hence k can take values $(-1,3)$.