Question

If the points A (1,-2) , B (2,3) , C (-3,2) and D (-4,-3) are the vertices of paralleogram ABCD, then taking AB as the base, find the height?

whenever solving the solution why do we double the area can't we just take the height of triangle formed please clear my doubt

Answer 1

Step-by-step explanation:

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

=

(−3+4)

2

+(2+3)

2

=

1+25

=

26

Thus height =

26

24

=

26

24

×

26

26

=

26

24

26

=

13

12

26

.