Question

I am stuck at this question​

Answer 1

Answer:

Step-by-step explanation:

B(2,10); D(6,2)

Midpoint(x1+x2/2, y1+y2/2) = M ( 2+6/2, 10+2/2) = M(8/2, 12/2) = M(4,6)

Rhombus all sides are equal.

AB = BC = CD =AD

distance = √(x2-x1)² + (y2- y1)²

As A lies on x-axis, it y-co ordinate = 0; Let its x-co ordinate be x

A(X,0)

AB = AD

√(2-x)² + (10-0)² = √(6-x)² + (2-0)²

√(2-x)² + (10)² =  √(6-x)² + (2)²

√x² -4x +4 + 100 =  √x²-12x+36 + 4

√x² -4x + 104 =  √x²-12x+40

square both sides,

x² -4x + 104 =  x²-12x+40

x² -4x - x²+ 12x = 40 - 104

             8x = -64

               x = -64/8

               x = -8

A(-8,0)

Let C(a,b)

M is AC midpoint

(-8+a/2, 0 + b/2)  = M(4,6)

     (-8+a/2, b/2)  = M(4,6)

Comparing;  

-8+a/2 = 4          ; b/2 = 6

  -8+a = 4*2       ; b = 6*2

  -8+a = 8          ; b = 12

        a = 8 +8

       a = 16

Hence, C(16,12)