For what value of k the area of triangle with vertices ki, 2, 3, 2 and 256 square units
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For what value of k the area of triangle with the verticies (k,1), (3k,1)an
For what value of k the area of triangle with the verticies (k,1), (3k,1)and (2,4)is 6square units
A)
Area of the triangle is =1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
The vertices are (k,1)(3k,1)(2,4)
Given area=6 square units
1/2[k(1-4)+3k(4-1)+2(1-1)]
6=1/2[k(1-4)+3k(4-1)+2(1-1)]
12=k(-3)+12k-3k+0
12=-3k+12k+3k
12=12k
k=1
therefore
vertices are
(1,1),(3,1) and (2,4)
For what value of k the area of triangle with the verticies (k,1), (3k,1)and (2,4)is 6square units
A)
Area of the triangle is =1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
The vertices are (k,1)(3k,1)(2,4)
Given area=6 square units
1/2[k(1-4)+3k(4-1)+2(1-1)]
6=1/2[k(1-4)+3k(4-1)+2(1-1)]
12=k(-3)+12k-3k+0
12=-3k+12k+3k
12=12k
k=1
therefore
vertices are
(1,1),(3,1) and (2,4)
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