Question

the points (-3,-3) ,(-3,2) , and (-3,5) are collinear ? give reasons​

Answer 1

$\underbrace{\underline{\bigstar\:\bf Required\:Answer:}}$

» We are given with three points and we are supposed to tell whether they are collinear or not.

» Collinear Points : Three or more points that lie on a straight line are know as collinear points.

» To solve this question, we would use the coordinate formula of finding the area of a triangle . If the area comes zero, then the points would be collinear.

◑ Let these points be $(x_1,y_1)\:;\:(x_2,y_2)\: and \:(x_3,y_3)$

It means :

• $\sf x_1=-3\: ,\: y_1=-3$
• $\sf x_2=-3\:,\: y_2=2$
• $\sf x_3 = -3\:,\: y_3= 5$

◑ Area of the triangle (a) :

$\dashrightarrow \: \boxed{ \sf a = \frac{1}{2} |x _{1} (y _{2} - y_{3}) + x _{2}(y _{3}- y _{1}) + x _{3}(y _{1} - y _{2}) | }$

$\colon \longrightarrow \sf \: a = \dfrac{1}{2} |( - 3)(2 - 5) + ( - 3)(5 - ( - 3)) + ( - 3)( - 3 - 2)|$

$\colon \longrightarrow \sf \: a = \frac{1}{2} |( - 3)( - 3) + ( - 3)(8) + ( - 3)( - 5)|$

$\colon \longrightarrow \sf \: a = \dfrac{1}{2} |9 + ( - 24) + 15|$

$\colon \longrightarrow \sf \: a = \dfrac{1}{2} |24 - 24|$

$\colon \longrightarrow \sf \: a = \dfrac{1}{2} \times 0$

$\colon \implies \: \boxed{ \bf{ \pink{a = 0}}} \: \: \bigstar$

$\red\therefore\underline{\sf The \: given\:points\:are\:\bf{collinear.}}$

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Answer 2

Answer:

$\underbrace{\underline{\bigstar\:\bf Required\:Answer:}}</p><p>\begin{gathered} \\ \end{gathered}</p><p></p><p>» We are given with three points and we are supposed to tell whether they are collinear or not.</p><p></p><p>» Collinear Points : Three or more points that lie on a straight line are know as collinear points.</p><p></p><p>» To solve this question, we would use the coordinate formula of finding the area of a triangle . If the area comes zero, then the points would be collinear.</p><p></p><p>\begin{gathered} \\ \end{gathered}</p><p></p><p>◑ Let these points be (x_1,y_1)\:;\:(x_2,y_2)\: and \:(x_3,y_3)(x1,y1);(x2,y2)and(x3,y3)</p><p></p><p>It means :</p><p></p><p>\sf x_1=-3\: ,\: y_1=-3x1=−3,y1=−3</p><p></p><p>\sf x_2=-3\:,\: y_2=2x2=−3,y2=2</p><p></p><p>\sf x_3 = -3\:,\: y_3= 5x3=−3,y3=5</p><p></p><p>\begin{gathered} \\ \end{gathered}</p><p></p><p>◑ Area of the triangle (a) :</p><p></p><p>\begin{gathered} \\ \end{gathered}</p><p></p><p>\begin{gathered} \dashrightarrow \: \boxed{ \sf a = \frac{1}{2} |x _{1} (y _{2} - y_{3}) + x _{2}(y _{3}- y _{1}) + x _{3}(y _{1} - y _{2}) | } \\ \end{gathered}⇢a=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣</p><p></p><p>\begin{gathered} \\ \end{gathered}</p><p></p><p>\begin{gathered} \colon \longrightarrow \sf \: a = \dfrac{1}{2} |( - 3)(2 - 5) + ( - 3)(5 - ( - 3)) + ( - 3)( - 3 - 2)| \\ \\ \end{gathered}:⟶a=21∣(−3)(2−5)+(−3)(5−(−3))+(−3)(−3−2)∣</p><p></p><p>\begin{gathered} \colon \longrightarrow \sf \: a = \frac{1}{2} |( - 3)( - 3) + ( - 3)(8) + ( - 3)( - 5)| \\ \\ \end{gathered}:⟶a=21∣(−3)(−3)+(−3)(8)+(−3)(−5)∣</p><p></p><p>\begin{gathered} \colon \longrightarrow \sf \: a = \dfrac{1}{2} |9 + ( - 24) + 15| \\ \\ \end{gathered}:⟶a=21∣9+(−24)+15∣</p><p></p><p>\begin{gathered} \colon \longrightarrow \sf \: a = \dfrac{1}{2} |24 - 24| \\ \\ \end{gathered}:⟶a=21∣24−24∣</p><p></p><p>\begin{gathered} \colon \longrightarrow \sf \: a = \dfrac{1}{2} \times 0 \\ \\ \end{gathered}:⟶a=21×0</p><p></p><p>\colon \implies \: \boxed{ \bf{ \pink{a = 0}}} \: \: \bigstar:⟹a=0★</p><p></p><p>\begin{gathered} \\ \end{gathered}</p><p></p><p>\begin{gathered}\red\therefore\underline{\sf The \: given\:points\:are\:\bf{collinear.}}\\\end{gathered}∴Thegivenpointsarecollinear.</p><p></p><p>\begin{gathered} \\ \end{gathered}</p><p></p><p>_____________________________</p><p></p><p>$