Find k, if the area of the triangle with vertices at P(3, -5), Q(-2, k), R(1, 4) is 33/2 sq. units.
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P(3,-5) Q(-2,k) R(1,4)
area of triangle=1/2{x1(y2-y3)+x2(y3-y1)+x3(y1-y2)}
33/2=1/2{3(k-4)-2(4+5)+1(-5-k)}
33/2×2={3(k-4)-2(9)+1(-5-k)}
33={3(k-4)-18+1(-5-k)}
33=(3k-12-18-5-k)
33=2k-35
33+35=2k
68=2k
68/2=k
34=k
Value of k is 34
if this is the right answer then mark it as brainliest please.
area of triangle=1/2{x1(y2-y3)+x2(y3-y1)+x3(y1-y2)}
33/2=1/2{3(k-4)-2(4+5)+1(-5-k)}
33/2×2={3(k-4)-2(9)+1(-5-k)}
33={3(k-4)-18+1(-5-k)}
33=(3k-12-18-5-k)
33=2k-35
33+35=2k
68=2k
68/2=k
34=k
Value of k is 34
if this is the right answer then mark it as brainliest please.
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