Question

if the line segment joining the points A(2,5),B(-5,-2) is divided by the point C such that AC:AB=3:7 find the coordinates of C​

Answer 1

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${ \bold { \underline { \underline { \maltese \:\red{Given} \:\maltese }}}} \:$

$\:\:\:\:\:\:\bullet\:\:\sf{Line \:segment\: AB \:is\: divided\: by\: the \:point \:C}$

$\:\:\:\:\:\:\bullet\:\:\sf{Coordinates\: of\: A\: and\: B\: are\: (2\: , \:5) \:,\: (-5\: ,\: -2)}$

$\:\:\:\:\:\:\bullet\:\:\sf{AC\:\ratio\:AB\:=\:3\:\ratio\:7}$

${ \bold { \underline { \underline { \maltese\:\red{ To\:Find }\:\maltese }}}} \:$

• $\sf{Coordinates \: of \: point \:C}$

${ \bold { \underline { \underline { \maltese\: \red{Solution} \:\maltese }}}} \:$

• $\sf{AC\:\ratio\:AB\:=\:3\:\ratio\:7\:(Given)}$

$\:\:\:\:\:\:\therefore\:\underline{\sf{ (AC\: =\: 3) \:and\: (AB\:= \:7)}}$

$\underline\bold\red{Now,}$

$\:\leadsto\:\tt{BC \:= \:AB \:-\:AC }$

$\:\leadsto\:\tt{BC \:= \:7 \:-\:3 }$

$\:\leadsto\:\underline{\underline{\boxed{\tt {\therefore\:BC \:= \:4}}} }\:\orange\bigstar$

$\star\:\underline{\sf{[So, \:AC \:\ratio\:BC\: =\: 3 \:\ratio\: 4]}}\:\dag$

$\dag\:\underline{\frak\red{To\:find\:the\:coordinates\:of\:point\:C\:,\:formula\:that\:we\:will\:use\:(Section\:formula)\::-}}$

${\pink\bigstar\:\underline{\underline{\boxed {\sf \blue {Section\:formula\:=\:\bigg(\dfrac{m_{1}x_{2}\:+\:m_{2}x_{1}}{m_{1}\:+\:m_{2}}\:,\:\dfrac{m_{1}y_{2}\:+\:m_{2}y_{1}}{m_{1}\:+\:m_{2}}\bigg)}}}}}$

$\dag\:\underline{\frak\red{The\: values\: we\: have \::-}}$

• $\sf{(m_{1}\:=\:3)\:and\:(m_{2}\:=\:4)}$

• $\sf{(x_{1}\:=\:2)\:and\:(x_{2}\:=\:-5)}$

• $\sf{(y_{1}\:=\:2)\:and\:(y_{2}\:=\:-2)}$

$\dag\:\underline{\frak\red{Putting \:all \:values\: in\: the\: formula\: :-}}$

$\ratio\dashrightarrow\:\sf{\bigg(\dfrac{m_{1}x_{2}\:+\:m_{2}x_{1}}{m_{1}\:+\:m_{2}}\:,\:\dfrac{m_{1}y_{2}\:+\:m_{2}y_{1}}{m_{1}\:+\:m_{2}}}\bigg)$

$\ratio\dashrightarrow\:\sf{\bigg(\dfrac{[3\:\times\:(-5)]\:+\:(4\:\times\:2)}{3\:+\:4}\:,\:\dfrac{[3\:\times\:(-2)]\:+\:(4\:\times\:5)}{3\:+\:4}}\bigg)$

$\ratio\dashrightarrow\:\sf{\bigg(\dfrac{(-15\:+\:8)}{7}\:,\:\dfrac{(-6\:+\:20)}{7}}\bigg)$

$\ratio\dashrightarrow\:\sf{\bigg(\dfrac{-7}{7}\:,\:\dfrac{14}{7}}\bigg)$

$\ratio\dashrightarrow\:\sf{\bigg(\dfrac{\cancel{-7}}{\cancel{7}}\:,\:\dfrac{\cancel{14}}{\cancel{7}}}\bigg)$

$\ratio\dashrightarrow\:\sf{(-1\:,\:2)}$

$\underline{\underline{\boxed {\frak {\therefore \green {Coordinates\:of\:point\:C\:\leadsto\:(-1\:,\:2)}}}}}\:\red\bigstar$

${ \bold { \underline { \underline { \maltese \:\red{Note} \:\maltese }}}} \:$

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Answer 2

$\huge\fcolorbox{black}{lime}{Answer}$

Given points are A(−2,5) and B(3,2) 

BCAC=12 , C divides line segment AB externally.

If A(x1,y1) and B(x2,y2) be two end points of a line segment, then the coordinates of the point P(x,y) that divides the line segment externally in the ratio m:n is (m−nmx2−nx1,m−nmy2−ny1).

Thus, the coordinates of C are (2−12(3)−1(−2),2−12(2)−1(5)).

=(16+2,14−5)

=(8,−1)