Question

If the point A(1,-2) B(2,3) C(-3,2) and D(-4,-3) are the vertices of a parallelogram ABCD,the taking AB aa the base,find the height of this parallelogram.

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Answer 1

Step-by-step explanation:

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

[ 1+a/2 ,−2+2/2 ]=[ 2−4/2 , 3−3/2 ]

[ a+1/2 ,0]=[ -2/2,0]

Since the mid points coincide, we have

1+a/2 =a

⇒a+1=−2

⇒a=−2−1

⇒a=−3

Now, area of ΔABC

= 1/2 |x1 (y2 −y3 )+x2 (y3 −y1)+x3(y1 −y2 )|

= 1/2 ∣1(3−2)+2(2−(−2))+(−3)(−2−3)∣

= 1/2 ∣1(1)+2(4)+(−3)(−5)∣

= 1/2∣1+8+15∣

= 24/2

=12 sq. units

ar(ABCD) parallelogram =2× Area of triangle

=2×12

=24 sq. units

Area of parallelogram =Base × Height

Area/base=height

So by the distance formula

= root(x2 −x1 ) ^2+(y2 −y1 ) ^2

= root (−3+4)^2+(2+3)^2

= root 1+25

= root 26

Thus height = 24/root 26

= 24/root 26 × root 26/ root 26

= 24 root26/ 26

= 12 root 26/13

Hope you like it.

Answer 2

Answer:

$height = \frac{12 \sqrt{26} }{13}$

Step-by-step explanation:

Plz refer to the attachment...