Question

Find the value of X such that the triangle whose vertices are (-4,0) (0,4) and (x,-x) is EQUILATERAL

Answer 1

Ello user !!!!!

Here is your answer,

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Let A(x,y), B(-4,0), C(4,0)

Distance between BC = 8 (Calculate it)

Distance between AB = √ (-4-x)^2 + (0-y)^2

Distance between AC = √ (4-x)^2 + (0-y)^2

Given that the triangle is equilateral. So, AB=BC= AC

AB=AC

√(-4-x)^2+ (0-y)^2 = √ (4-x)^2+ (0-y)^2

(-4-x)^2+ (0-y)^2 = (4-x)^2+ (0-y)^2

(-4-x)^2= (4-x)^2

x^2+8x+16= x^2-8x+16

8x+8x= 16-16

x= 0 ..........(1)

Again, AC = BC

√(4-x)^2+ (0-y)^2 = 8

(4-x)^2+ (0-y)^2= 64

(4-0)^2+ (0-y)^2 { substituting 1}= 64

4^2 + y^2 = 64

y^2 = 64-16=38

y = +√38 or -√38

Therefore, x=0 and y = +√38 or -√38

Third vertex is (0, √38) or (0,-√38)

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HOPE THIS HELPS YOUU :)

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