Find the value of k, if area of triangle is 4 sq unit and vertices are
(k,0),(4,0)(0,2)
(−2,0),(0,4),(0,k)
Given, 12∣∣
∣∣k01401021∣∣
∣∣=4
⇒ |k(0−2)+1(8−0)|=8⇒k(0−2)+1(8−0)=±8
On taking positive sign −2k+8=8
⇒ −2k=0⇒k=0
On taking negative sign −2k+8=−8
⇒ −2k=−16⇒k=8∴ k=0,8
Given, (12∣∣
∣∣−2010410k1∣∣
∣∣=4⇒ |−2(4−k)+1(0−0)|=8
⇒−2(4−k)+1(0−0)=±8⇒[−8+2k]=±8
On taking positive sign, 2k−8=8⇒2k=16⇒k=8
On taking negative sign, 2k+8=−8⇒2k=0⇒k=0
∴ k=0,8