Question

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

Answer 1

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

.Here triangle ABC is a right angle triangle,

AB² + BC² = AC² (p - 4)² + (3 - 7)² + (7 - p)² + (3 - 3)² = (7 - 4)² + (3 - 7)²p² + 16 - 8p + 16 + 49 + p² - 14p = 9 + 16 2p² - 22p + 56 = 0 p² - 11p + 28 = 0 (p - 4)(p - 7) = 0 so, p = 4 and 7.




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