Question

if a (-2,1) , b(9,0) , c(4,b) and d (1,a) are vertices of a parallelogram ,abcd . find the values of a and b and hence find the length of each side

Answer 1

ABCD is parallelogram. It is possible only when AB || CD and AD || BC .
Means slope of AB = slope of CD
slope of AB = (0 - 1)/(9 +2) = -1/11
slope of CD = (a - b)/(1 - 4) = (a - b)/-3
e.g., -1/11 = (a - b)/-3
3 = 11(a - b) ------(1)

Similarly slope of AD = slope of BC
slope of AD = (a - 1)/(1 + 2) = (a - 1)/3
slope of BC = (b - 0)/(4 - 9) = b/-5
e.g., b/-5 = (a - 1)/5
5b = -5a + 5
5(a + b) = 5 ⇒ a = 1 - b , put it in above equation
3 = 11( 1 - b - b) = 11 - 22b
-8 = -22b ⇒ b = 4/11
So, a = 1 - b = 1 - 4/11 = 7/11

Hence, a = 4/11 and b = 7/11
so, Length of AB = $\sqrt{(-2-9)^2+(1-0)^2}=\sqrt{122}$
Similarly you can find length of BC , CD and DE by using distance formula.

Answer 2

ABCD is parallelogram. It is possible only when AB || CD and AD || BC .
Means slope of AB = slope of CD 
slope of AB = (0 - 1)/(9 +2) = -1/11 
slope of CD = (a - b)/(1 - 4) = (a - b)/-3
e.g., -1/11 = (a - b)/-3 
3 = 11(a - b) ------(1)

Similarly slope of AD = slope of BC 
slope of AD = (a - 1)/(1 + 2) = (a - 1)/3 
slope of BC = (b - 0)/(4 - 9) = b/-5 
e.g., b/-5 = (a - 1)/5 
5b = -5a + 5 
5(a + b) = 5 ⇒ a = 1 - b , put it in above equation 
3 = 11( 1 - b - b) = 11 - 22b 
-8 = -22b ⇒ b = 4/11 
So, a = 1 - b = 1 - 4/11 = 7/11 

Hence, a = 4/11 and b = 7/11 
so, Length of AB = \sqrt{(-2-9)^2+(1-0)^2}=\sqrt{122}√​(−2−9)​2​​+(1−0)​2​​​​​=√​122​​​ 
Similarly you can find length of BC , CD and DE by using distance formula.