Question

find the value of" A" such that triangle DEF with vertices D (a,1) , E(10,7) and F(7 ,11) is a right angle at E?​

Answer 1

a=2

Answer:

(m1) slope of DE=(7-1)/(10-a)=6/(10-a)

(m2) slope of EF=(11-7)/(7-10)=4/-3=-4/3

since line are perpendicular

m1m2=-1

6/(10-a) ×-4/3=-1

6/(10-a)=3/4

6×4=3(10-a)

24=30-3a

3a=30-24

a=6/3=2