The area of triangle with vertices ( ( K , 0),(4,0),(0,2) ) is ( 4 ) square units, then value of ( K ) is

Step-by-step solution

Given vertices of a triangle is, ( ( K , 0),(4,0),(0,2) ) and area of triangle is ( 4 sq ). units. We know that, area of triangle with vertices ( (x₁, y₁ ), (x₂, y₂ ) (x₃, y₃ ) ) is given by [ aligned A=& | array ccc x₁ & x₂ & x₃ y₁ & y₂ & y₃ z₁ & z₂ & z₃ array | A=& 1 2 (x₁ (y₁-y₃ )+x₂ (y₃-y₂ )+x₃ (y₁-y₂ ) ) aligned ] So, [ array l A= 1 2 |K(0-2)+4(2-0)-0(0-0)|=4 1 2 |-2 K+8|=4 8-2 K= 8 =8+8,60=8 array ] For ( +8 ), we have ( 8-2 K=8 K=0 ) For ( -8 ), we have ( 8-2 K=-8 K=8 )