Question

2. a)Divide :-
6y2-y4+4y3-5y2-y-15 by 2y2-y+3​

Answer 1

Answer:

Consider the given points A(1,−2),B(2,3),C(0,2) and D(−4,−3)

Since ABCD form a parallelogram, the midpoint of the diagonal AC should coincide with the midpoint of BD.

Mid point of AC= Mid point of BD

[

2

1+a

,

2

−2+2

]=[

2

2−4

,

2

3−3

]

[

2

a+1

,0]=[

2

−2

,0]

Since the mid points coincide, we have

2

1+a

=a

⇒a+1=−2

⇒a=−2−1

⇒a=−3

Now, area of ΔABC

=

2

1

∣x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)∣

=

2

1

∣1(3−2)+2(2−(−2))+(−3)(−2−3)∣

=

2

1

∣1(1)+2(4)+(−3)(−5)∣

=

2

1

∣1+8+15∣

=

2

24

=12 sq. units

ar(ABCD) parallelogram =2× Area of triangle

=2×12

=24 sq. units

Area of parallelogram =Base × Height

Base

Area

=height

So by the distance formula

=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2