Math, asked by Sid234, 1 year ago

hello frNds plss solve these Qsn 28,29...?

Attachments:

Answers

Answered by GauravGumber
1
x=a
x²- 3y²=0
(x-√3y)(x+√3y)=0
x=√3y.......(1)
x=-√3y.......(2)
x=a............(3)
Solve (1) and (2)√3y=-√3y, 2√3y=0,y=0
for y=0 x=0,

point is A (0,0)

solve (2) and (3)
y=-a/√3,

point B(a,-a/√3)

solve (3) and (1)
point C(a,a/√3)
Distance AB=√(a²+a²/3)=√4a²/3=2a/√3
BC=√(0²+4a²/3)=2a/√3
CA=√(a²+a²/3)=√(4a²/3)=2a/√3

SO ,AB =BC =CA EQUILATERAL TRIANGLE

MARK AS BRAINIEST



Similar questions