Question

Thepoint 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

Answer 1

Hey,
Mate here's the Solution.....



B is right angle so, AC is hypotenous 
now use Pythagoras theorem 
AC^2=BC^2+AB^2
(-3)^2+(4)^2=(p-7)^2+(3-3)^2+(4-p)^2+(4)^2

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

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

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

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

p=7,4

Answer 2

$\huge\mathbb{SOLUTION:-}$

$\Delta$ ABC is right angled at B.

$\blue{AC {}^{2} = AB {}^{2} + BC {}^{2} \: .............(1)}$

$\blue{Also, \: A = (4,7) \:,\: B = (p,3) \: and \: C = (7,3)}$

$\blue{Now, \: AC {}^{2} = (7 - 4) {}^{2} + (3 - 7) {}^{2} = (3) {}^{2} + ( - 4) {}^{2} = 9 + 16 = 25}$

$\blue{AB {}^{2} = (p - 4) {}^{2} + (3 - 7) {}^{2} = p {}^{2} - 8p + 16 + ( - 4) {}^{2}}$

$=\blue{p {}^{2} - 8p + 16 + 16}$

$=\blue{ p {}^{2} - 8p + 32}$

$\blue{BC {}^{2} = (7 - p) {}^{2} + (3 - 3) {}^{2} = 49 - 14p + p {}^{2} + 0}$

$=\blue{p {}^{2} - 14p + 49}$

$\green{From \: (1), \: We \: have}$

$\blue{25 = (p {}^{2} - 8p + 32) + (p {}^{2} - 14p + 49)}$

$\blue{25 = 2p {}^{2} - 22p + 81}$

$= \blue{ 2p {}^{2} - 22p + 56 = 0}$

$= \blue{p {}^{2} - 11p + 28 = 0}$

$= \blue{p {}^{2} - 7p - 4p + 28 = 0}$

$=\blue{ p(p - 7) - 4(p - 7) = 0}$

$=\blue{ (p - 7) \: (p - 4) = 0}$

$=\red{p = 7 \: and \: p = 4}$