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Question

Find values of k if area of triangle is 4 sq. units and vertices are
(i) (k,0),(4,0),(0,2)
(ii) (2,0),(0,4),(0,k)

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Solution

Area of triangle having vertices as (x1,y1),(x2,y2),(x3,y3) is given as
=12∣ ∣ ∣x1y11x2y21x3y31∣ ∣ ∣

(i) Given, area of triangle = 4 sq.units
4=12∣ ∣k01401021∣ ∣

4=12|k(02)0(40)+1(80)|
|82k|=8
82k=±8
82k=8 and 82k=8
k=0 and 2k=16
k=0 and k=8

(ii) Given, area of triangle = 4 sq.units
4=12∣ ∣2010410k1∣ ∣

4=12|2(4k)0(00)+1(00)|
|82k|=8
82k=±8
82k=8 and 82k=8
k=0 and 2k=16
k=0 and k=8

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