Show that points A(a, b +  c),  B(b, c + a),  C(c, a + b) are

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 Multiple Choice QuestionsMultiple Choice Questions

131. If a, b, c are in A.P, then the determinant
open vertical bar table row cell straight x plus 2 end cell cell space space space straight x plus 3 end cell cell space space space straight x plus 2 straight a end cell row cell straight x plus 3 end cell cell space space straight x plus 4 end cell cell space space straight x plus 2 straight b end cell row cell straight x plus 4 end cell cell space space straight x plus 5 end cell cell space space space straight x plus 2 straight c end cell end table close vertical bar


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 Multiple Choice QuestionsShort Answer Type

132. Find the area of the triangle with vertices (2, 7), (1, 1), (10, 8).
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133.

Find the area of the triangle, whose vertices are (3, 1), (4, 3) and (-5, 4).

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134.

Find the area of the triangle with vertices at the points given in each of the following:
(1, 0)  (6,0), (4, 3) 

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135.

Find the area of the triangle with vertices at the points given in each of the following:
(2, 7), (1, 1), (10, 8)

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136.

Find the area of the triangle with vertices at the points given in each of the following:
(–2,–3), (3, 2), (–1,–8)

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137.

Show that points A(a, b +  c),  B(b, c + a),  C(c, a + b) are collinear.


Let ∆ be the area of the triangle formed by the points
A (a, b + c), B (b, c + a), C(c, a + b)
therefore space space space space space increment space equals space 1 half open vertical bar table row straight a cell space space straight b plus straight c end cell cell space space 1 end cell row straight b cell space straight c plus straight a end cell cell space space 1 end cell row straight c cell space straight a plus straight b end cell cell space space 1 end cell end table close vertical bar space equals space 1 half open vertical bar table row cell straight a plus straight b plus straight c end cell cell space space straight b plus straight c end cell cell space space 1 end cell row cell straight b plus straight c plus straight a end cell cell space space straight c plus straight a end cell cell space space 1 end cell row cell straight c plus straight a plus straight b end cell cell space space straight a plus straight b end cell cell space space 1 end cell end table close vertical bar comma space by space straight C subscript 1 space rightwards arrow space straight C subscript 1 plus straight C subscript 2
space space space space space space space space space space space space space space equals space 1 half left parenthesis straight a plus straight b plus straight c right parenthesis space open vertical bar table row 1 cell space space straight b plus straight c end cell cell space space 1 end cell row 1 cell space straight c plus straight a end cell cell space space 1 end cell row 1 cell space space straight a plus straight b end cell cell space space 1 end cell end table close vertical bar
space space space space space space space space space space space space space space equals space 1 half left parenthesis straight a plus straight b plus straight c right parenthesis thin space left parenthesis 0 right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space two space columns space are space identical close square brackets
space space space space space space space space space space space space space space space equals space 0
therefore space space space space given space points space straight A left parenthesis straight a comma space straight b space plus space straight c right parenthesis comma space straight B left parenthesis straight b comma space straight c space plus space straight a right parenthesis comma space straight C left parenthesis straight c comma space straight a space plus space straight b right parenthesis space are space collinear.

space space

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138.

Using determinants, show that the points (11, 7), (5, 5) and (– 1, 3) are collinear.

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139.

Determine x so that the points (3, 2), (x, 2) and (8, 8) lie on a line. 

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140. Find the value of x if the area of triangle is 35 square cm. with vertices (x, 4), (2, – 6) and (5, 4).
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