Question

For what values of'p' is the points (1,p) ,( 2p,2) , ( 3,3) are collinear

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Answer 1

$\bigstar \large \underbrace \orange{ \pmb{ \frak{ According\:To\:Question }}} : -$

• $\sf x_{1} = 1$

• $\sf y_{1} = p$

• $\sf x_{2} = 2p$

• $\sf y_{2} = 2$

• $\sf x_{3} = 3$

• $\sf y_{3} = 3$

All the three points are colinear.

$\bigstar \large \underbrace \orange{ \pmb{ \frak{Required\: Formula }}} : -$

$\\ \leadsto\underline{ \boxed {\pink{{ \sf{0 = \bigg[ x_{1}( y_{2} - y_{3}) + x_{2}(y_{3} -y_{1} ) + x_{3}(y_{1} - x_{3})\bigg] }}}}} \bigstar$

$\dag \frak{substitute \: the \: given \: values \: in \: the \: formula : - }$

$\\ : \implies \sf{0 = \bigg[ x_{1}( y_{2} - y_{3}) + x_{2}(y_{3} -y_{1} ) + x_{3}(y_{1} - x_{3})\bigg] }$

$\\ : \implies \sf0 = \bigg[ 1( 2 - 3) + 2p(3 -p) + 3(p - 3)\bigg]$

$\\ : \implies \sf0 = \bigg[ - 1 + 6p - 2 {p}^{2} + 3p - 9\bigg]$

$\\ : \implies \sf0 = \bigg[ - 10 + 9p - 2 {p}^{2} \bigg]$

$\\ : \implies \boxed{ \sf 2 {p}^{2} - 9p + 10 = 0} \bigstar$

Now, let's solve this quadratic equation through which we will get 2 values of p and both will be acceptable.

$\\ \rightarrow \sf \ 2 {p}^{2} - 9 + 10 = 0$

$\\ \rightarrow \sf 2 {p}^{2} - (4 + 5)p + 10 = 0$

$\\ \rightarrow \sf 2 {p}^{2} - 4p - 5p + 10 = 0$

$\\ \rightarrow \sf 2p(p - 2) - 5(p - 2) = 0$

$\\ \rightarrow \sf (2p - 5)(p - 2) = 0$

$\\ \rightarrow \underline{ \boxed{ \pink{{ \sf p = 2 , \dfrac{5}{2} }}}} \bigstar$

$\therefore \: \underline{ \frak{2 \: and \: \dfrac{5}{2} \: are \: required \: values \: of \: p \: }}$

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Hope its helpful!