If the points A(3, 0), B(x, 5), C(-1, 4) and D(-2, -1) are the vertices of a rhombus, taken in order, find the value of x.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
Since A, B, C, D are vertices of a rhombus taken in order, all sides are equal.
So,
$AB = BC$
Step 1: Find AB
$A(3,0),\; B(x,5)$
$AB = \sqrt{(x-3)^2 + (5-0)^2} = \sqrt{(x-3)^2 + 25}$
Step 2: Find BC
$B(x,5),\; C(-1,4)$
$BC = \sqrt{(-1-x)^2 + (4-5)^2} = \sqrt{(x+1)^2 + 1}$
Step 3: Equate $AB = BC$
$(x-3)^2 + 25 = (x+1)^2 + 1$
Expand:
$x^2 - 6x + 9 + 25 = x^2 + 2x + 1 + 1$
$x^2 - 6x + 34 = x^2 + 2x + 2$
Cancel $x^2$:
$−6x+34=2x+2$
$-8x = -32$
$x=4$