Question

State true or false. (2, 3), (4, 0) and (6, -3) are collinear points.

Answer 1

$Let \: A(2,3), B(4,0) \:and \: C(6,-3) \: are$

$Vertices \: of \: a \: triangle \: ABC$

/* By using Area of the Triangle Formula: */

$\triangle = \frac{1}{2}|x_{1}(y_{2}-y_{3}) +x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|$

$= \frac{1}{2}|2[0-(-3)] + 4(-3+3)+6(3-0)|$

$= \frac{1}{2}|2\times 3 + 4 \times 0 + 6 \times 3 |$

$= \frac{1}{2}| 6 + 0 + 18|$

$= \frac{1}{2} \times 24$

$= 12 \: square \:units$

$\neq 0$

$\green{ \therefore Given \:points \: are \: not \: collinear}.$

$\red{ Given \: statement \:is \:false }$

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