Question

prove that (1, -2), (2,3), (-3,2), and (-4,-3) are the vertices of a rhombus​

Answer 1

Answer:

we have to compare the quadratic equation, 2y² = 9y with the standard form and find the values of a, b and c.

solution : quadratic equation is 2y² = 9y

⇒2y² - 9y + 0 = 0

standard form of quadratic equation is ay² + by + c = 0

on comparing

⇒2y² + (-9)y + 0 = ay² + by + c

here , coefficient of y² of 2y² + (-9y) + 0 = coefficient of y² of ay² + ay + c

i.e., 2 = a

coefficient of y of 2y² + (-9)y + 0 = coefficient of y of ay² + by + c

i.e., -9 = b

constant of 2y² + (-9)y + 0 = constant of ay² + by + c

i.e., 0 = c

Therefore a = 2 , b = -9 and c = 0