The points A(4,7) ,B(p,3) and C(7,3) are the vertices of a right angled triangle, right-angled at B. Find the value of p.

The given points are A(4,7), B(p,3) and C(7,3).

Since A,B and C are the vertices of a right angled triangle then, $(AB{)}^2+(BC{)}^2=\left(A{C}^2\right)$

[By Pythagoras theorem]

$\left[\right(p-4{)}^2+(3-7{)}^2]+\left[\right(7-p{)}^2+(3-3{)}^2]=\left[\right(7-4{)}^2+(3-7{)}^2]$

$(p-4{)}^2+(-4{)}^2+(7-p{)}^2=(3{)}^2+(-4{)}^2$

$p^2+16-8p+16+49+p^2-14p=9+16$

$2p^2-22p+56=0$

$p^2-11p+28=0$

$p^2-7p-4p+28=0$

$p(p-7)-4(p-7)=0$

$(p-7)(p-4)=0$

$p=4or7$

$p\ne 7$ (As B and C will coincide)

So, $p=4$