Question

the vertices of a triangle ABC are A(1,K),B(4,-3)and C(-9,7).AREA OF THE TRIANGLE IS 15 SQUARE UNITS .FIND THE ALTITUDE OF THE TRIANGLE with ab as the base

Answer 1

$\left[\begin{array}{ccc}1&k&1\\4&-3&1\\-9&7&1\end{array}\right]$*1/2=15

so, this matrix is equal to 30

solving this, 
1(-3-7)-k(4-(9))+1(28-27)=30
so -10+5k+1=30
      5k=39
        k=39/5

15=1/2 b*h
30=b*h
30/b=h
h=30/(AB)
  
now here u get ur h....