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Determinants

Question
CBSEENMA12034498

Find the equation of the line joining A(l, 3) and B(0, 0) using determinants and find k if D(A, 0) is a point such that area of triangle ABD is 3 sq. units.

Solution
Let P(1, 3) be any point on the line joining A(l, 3) and B(0, 0).
therefore space area space of space increment ABP space equals space 0
therefore space space space 1 half open vertical bar table row 1 cell space space 3 end cell cell space space 1 end cell row 0 cell space 0 end cell cell space space 1 end cell row straight x cell space space straight y end cell cell space space 1 end cell end table close vertical bar space equals space 0 space space space space space rightwards double arrow space space space space space open vertical bar table row 1 cell space space 3 end cell cell space space 1 end cell row 0 cell space space 0 end cell cell space space 1 end cell row straight x cell space space straight y end cell cell space space 1 end cell end table close vertical bar space equals space 0
therefore space space space space minus 1 space open vertical bar table row 1 cell space space 3 end cell row straight x cell space space space straight y end cell end table close vertical bar space equals space 0 space space space space space space space space space space rightwards double arrow space space space space space open vertical bar table row 1 cell space space space 3 end cell row straight x cell space space space straight y end cell end table close vertical bar space equals space 0
therefore space space space space space space space space straight y minus 3 straight x space equals space 0 space space space or space space straight y space equals space 3 straight x

which is equation of line AB.
Now D is (k, 0).
From given condition,
area of ∆ABD = 3
therefore space space space space space space space 1 half open vertical bar table row 1 cell space space 3 end cell cell space 1 end cell row 0 cell space space 0 end cell cell space 1 end cell row straight k cell space space 0 end cell cell space 1 end cell end table close vertical bar space equals space plus-or-minus 3
therefore space space space space space space space space space open vertical bar table row 1 cell space space space space 3 end cell cell space space space 1 end cell row 0 cell space space space 0 end cell cell space space space 1 end cell row straight k cell space space space 0 end cell cell space space space space 1 end cell end table close vertical bar space equals space plus-or-minus 6 space space space space space rightwards double arrow space space minus space open vertical bar table row 1 cell space space space 3 end cell row straight k cell space space space 0 end cell end table close vertical bar space equals space plus-or-minus 6
therefore space space space space space minus left parenthesis 0 minus 3 space straight k right parenthesis space equals space plus-or-minus 6 space space space space space space rightwards double arrow space space space 3 space straight k space equals space plus-or-minus 6
therefore space space space space space straight k space equals space plus-or-minus 2

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