Question

A vertices of a triangle ABC , right angled at B are A(-2,1),B(2,-2),C(k,2). Find the value of k

Answer 1

${ab}^{2} = 25$
${bc}^{2} = 20 - 4k + {k}^{2}$
${ca}^{2} = {k}^{2} + 4k + 5$
${ab}^{2} + {bc}^{2} = {ca}^{2}$
hence after inserting the values and simplifying we will get k=5