Answer:
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, \[{{(AB)}^{2}}+{{(BC)}^{2}}={{(AC)}^{2}}\] [By Pythagoras theorem] \[[{{(p-4)}^{2}}+{{(3-7)}^{2}}]+[{{(7-p)}^{2}}+{{(3-3)}^{2}}]=[{{(7-4)}^{2}}+{{(3-7)}^{2}}]\] \[{{(p-4)}^{2}}+{{(-4)}^{2}}+{{(7-p)}^{2}}={{(3)}^{2}}+{{(-4)}^{2}}\] \[{{p}^{2}}+16-8p+16+49+{{p}^{2}}-14p=9+16\] \[2{{p}^{2}}-22p+56=0\] \[{{p}^{2}}-11p+28=0\] \[{{p}^{2}}-7p-4p+28=0\] \[p(p-7)-4(p-7)=0\] \[p=4\] or \[7\] \[p\ne 7\] (As B and C will coincide) So, \[p=4\].
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