A triangle has its three sides equal to a, b and c. If the coord

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 Multiple Choice QuestionsShort Answer Type

141.

Find values of k if area of triangle is 4 sq. units and vertices are (k, 0), (4, 0), (0, 2).

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142.

Find values of k if area of triangle is 4 sq. units and vertices are (– 2, 0), (0, 4), (0, k)

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143. 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.
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144.

Find equation of line joining (1, 2) and (3, 6) using determinants.

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145.

Find equation of line joining (3, 1) and (9, 3) using determinants.

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 Multiple Choice QuestionsLong Answer Type

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146. A triangle has its three sides equal to a, b and c. If the coordinates of its vertices are A(x1 y1), B(x2, y2) and C(x3 y3), show that
space open vertical bar table row cell straight x subscript 1 end cell cell space space straight y subscript 1 end cell cell space 2 end cell row cell straight x subscript 2 end cell cell space straight y subscript 2 end cell cell space 2 end cell row cell straight x subscript 3 end cell cell space straight y subscript 3 end cell cell space 2 end cell end table close vertical bar squared space equals space left parenthesis straight a plus straight b plus straight c right parenthesis thin space left parenthesis straight b plus straight c minus straight a right parenthesis thin space left parenthesis straight c plus straight a minus straight b right parenthesis thin space left parenthesis straight a plus straight b minus straight c right parenthesis.


We know that
      increment space equals space open vertical bar table row cell straight x subscript 1 end cell cell space straight y subscript 1 end cell 1 row cell straight x subscript 2 end cell cell straight y subscript 2 end cell 1 row cell straight x subscript 3 end cell cell straight y subscript 3 end cell 1 end table close vertical bar space space or space space space 2 increment space equals space open vertical bar table row cell straight x subscript 1 end cell cell space straight y subscript 1 end cell cell space 1 end cell row cell straight x blank presubscript 2 end cell cell space straight y subscript 2 end cell cell space 1 end cell row cell straight x subscript 3 end cell cell space straight y subscript 3 end cell cell space 1 end cell end table close vertical bar
rightwards double arrow space space 4 space increment space equals space space open vertical bar table row cell straight x subscript 1 end cell cell straight y subscript 1 end cell 2 row cell straight x subscript 2 end cell cell straight y subscript 2 end cell 2 row cell straight x subscript 3 end cell cell straight y subscript 3 end cell 2 end table close vertical bar space space or space space space space 16 increment squared space equals space open vertical bar table row cell straight x subscript 1 end cell cell space straight y subscript 1 end cell cell space 2 end cell row cell straight x subscript 2 end cell cell space straight y subscript 2 end cell cell space 2 end cell row cell straight x subscript 3 end cell cell space straight y subscript 3 end cell cell space 2 end cell end table close vertical bar squared space space space space space space space space space space space space... left parenthesis 1 right parenthesis

Again,    increment space equals space square root of straight s left parenthesis straight s minus straight a right parenthesis thin space left parenthesis straight s minus straight b right parenthesis thin space left parenthesis straight s minus straight c right parenthesis end root space space where space straight s space equals space 1 half left parenthesis straight a plus straight b plus straight c right parenthesis
therefore space space space space space space space straight s minus straight a space equals space 1 half left parenthesis straight a plus straight b plus straight c right parenthesis minus straight a space equals space 1 half left parenthesis straight b plus straight c minus straight a right parenthesis
Similarly,   straight s minus straight b space equals space 1 half left parenthesis straight c plus straight a minus straight b right parenthesis comma space space space straight s minus straight c space equals space 1 half left parenthesis straight a plus straight b minus straight c right parenthesis
therefore space space space space space space space space increment squared space equals space straight s left parenthesis straight s minus straight a right parenthesis thin space left parenthesis straight s minus straight b right parenthesis thin space left parenthesis straight s minus straight c right parenthesis
rightwards double arrow space space space space space space space space increment squared space equals space 1 half left parenthesis straight a plus straight b plus straight c right parenthesis. space 1 half left parenthesis straight b plus straight c minus straight a right parenthesis thin space. space 1 half left parenthesis straight c plus straight a minus straight b right parenthesis. space 1 half left parenthesis straight a plus straight b minus straight c right parenthesis
rightwards double arrow space space space space space 16 increment squared space equals space left parenthesis straight a plus straight b plus straight c right parenthesis space left parenthesis straight b plus straight c minus straight a right parenthesis thin space left parenthesis straight c plus straight a minus straight b right parenthesis thin space left parenthesis straight a plus straight b minus straight c right parenthesis
rightwards double arrow space space space space space open vertical bar table row cell straight x subscript 1 end cell cell space space space straight y subscript 1 end cell cell space space 2 end cell row cell straight x subscript 2 end cell cell space space straight y subscript 2 end cell cell space space 2 end cell row cell straight x subscript 3 end cell cell space space straight y subscript 3 end cell cell space space 2 end cell end table close vertical bar squared space equals space left parenthesis straight a plus straight b plus straight c right parenthesis thin space left parenthesis straight b plus straight c minus straight a right parenthesis thin space left parenthesis straight c plus straight a minus straight b right parenthesis thin space left parenthesis straight a plus straight b minus straight c right parenthesis thin space open square brackets because space of space left parenthesis 1 right parenthesis close square brackets

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 Multiple Choice QuestionsMultiple Choice Questions

147. If area of triangle is 35 sq. units with vertices (2, – 6), (5, 4) and (k, 4). Then k is
  • 12

  • -2

  • -12, -2

  • -12, -2

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 Multiple Choice QuestionsShort Answer Type

148.

Find the minor of element 6 in the determinant increment space equals space open vertical bar table row 1 cell space 2 end cell cell space 3 end cell row 4 cell space 5 end cell cell space 6 end cell row 7 cell space 8 end cell cell space 9 end cell end table close vertical bar

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149.

Find the minors and co-factors of the elements a11, a21 in the determinant.

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150.

Find minors and cofactors of all the elements of the determinant:
open vertical bar table row 1 cell space space minus 2 end cell row 4 cell space space space 3 end cell end table close vertical bar.

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