Tag: basic geometrical concepts and shapes

Let A = $(0,0)$ , B$(0,1)$ , C$(2k,3k)$ , D$(1,0)$

As we can see

$AD \perp AB$

$\angle A = 90$

$\angle C = 90$

$\implies m _{BC} \times m _{DC} = -1$

$\dfrac{3k - 1}{2k} \times {3k}{2k - 1} = -1$

$\implies k(9k - 3) = -(4k - 2)k$

$k = 0 , 9k + 4k = 2+ 3 \implies 13k = 5 \implies k = \dfrac{5}{13}$

But if $k = 0$, C will be $(0,0)$ which is A

$k = \dfrac{5}{13} $ only one point.