The coordinates of the vertices of a parallelogram ABCD are A(0,0) B(b,y), C(a+b,y) and D(a,0). If ABCD is a rhombus, how would you express y in terms of a and b?

Question

Maverick Smith · Accepted Answer

#y = sqrt{a^2 - b^2}# #|AB|^2 = b^2 + y^2# #|BC|^2 = a^2# #|CD|^2 = b^2 + y^2# #|DA|^2 = a^2 = b^2 + y^2#

Savannah Aguilar · Answer

undefined If ABCD is a rhombus, then the opposite sides are equal in length. Therefore, the distance between points A and B must be equal to the distance between points C and D, and the distance between points B and C must be equal to the distance between points A and D.

Using the distance formula $ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $, we can set up equations for the distances between the vertices:

1. Distance between points A and B: 
$$ \sqrt{(b - 0)^2 + (y - 0)^2} = \sqrt{b^2 + y^2} $$

2. Distance between points C and D:
$$ \sqrt{(a - (a + b))^2 + (0 - y)^2} = \sqrt{b^2 + y^2} $$

3. Distance between points B and C:
$$ \sqrt{((a + b) - b)^2 + (y - y)^2} = \sqrt{a^2} = a $$

4. Distance between points A and D:
$$ \sqrt{(a - 0)^2 + (0 - 0)^2} = \sqrt{a^2} = a $$

Since ABCD is a rhombus, all these distances are equal:

$$ \sqrt{b^2 + y^2} = a $$

$$ b^2 + y^2 = a^2 $$

Therefore, to express $ y $ in terms of $ a $ and $ b $, we rearrange the equation:

$$ y^2 = a^2 - b^2 $$

$$ y = \pm \sqrt{a^2 - b^2} $$

Since the vertices of the rhombus are in the first and fourth quadrants (based on the given coordinates), we take the positive root:

$$ y = \sqrt{a^2 - b^2} $$