from point A2, draw A2B’ || A3B, meeting AB at B’. Take P as a centre and draw an arc on PX and mark that point as A. AB is a line which doesn’t have an ending. Now, So take 6 parts  out of 7 equal parts on BX i.e. This means that you have to find a point \(C\) on \(AB\) such that \(AC:CB = 3:4\) , as shown below: How will you geometrically locate point \(C\)? Solution: Let D be the point \(\left( {h,\;k} \right)\). = (–4 + 8,1 + 4) = (4,5) The following figure shows the graph of this line segment and the points that divide it into three equal parts. Thus, C divides AB in the ratio 3 : 2. In the figure, we have shown one arc (to construct the first interval). Since \(C\) divides \(AB\) externally in the ratio \(m:n\), we have: \[\begin{align}&\frac{{AC}}{{CB}} = \frac{m}{n}\\& \Rightarrow  \;\;\; \frac{{CB}}{{AC}} = \frac{n}{m}\\&\Rightarrow  \;\;\; 1 - \frac{{CB}}{{AC}} = 1 - \frac{n}{m}\\ &\Rightarrow  \;\;\; \frac{{AC - CB}}{{AC}} = \frac{{m - n}}{m}\\& \Rightarrow  \;\;\; \frac{{AB}}{{AC}} = \frac{{m - n}}{m}\end{align}\], \[\begin{align}&  \frac{{AP}}{{AQ}} = \frac{{BP}}{{CQ}} = \frac{{m - n}}{m}\\& \Rightarrow  \quad \frac{{{x_2} - {x_1}}}{{h - {x_1}}} = \frac{{{y_2} - {y_1}}}{{k - {y_1}}} = \frac{{m - n}}{m}\\ &\Rightarrow  \quad\left\{ \begin{gathered}\frac{{{x_2} - {x_1}}}{{h - {x_1}}} = \frac{{m - n}}{m}\\\frac{{{y_2} - {y_1}}}{{k - {y_1}}} = \frac{{m - n}}{m}\end{gathered} \right.\\ &\Rightarrow  \quad \left\{ \begin{gathered}h - {x_1} = \left( {\frac{m}{{m - n}}} \right)\left( {{x_2} - {x_1}} \right)\\k - {y_1} = \left( {\frac{m}{{m - n}}} \right)\left( {{y_2} - {y_1}} \right)\end{gathered} \right.\\& \Rightarrow \quad \left\{ \begin{gathered}h = {x_1} = \left( {\frac{m}{{m - n}}} \right)\left( {{x_2} - {x_1}} \right)\\k = {y_1} + \left( {\frac{m}{{m - n}}} \right)\left( {{y_2} - {y_1}} \right)\end{gathered} \right. He has been teaching from the past 9 years. Let us divide a line segment AB into 3:2 ratio. Because(m + n = 3 + 1) . To watch interesting videos on the topic, download BYJU’S – The Learning App from Google Play Store. Locate 5(= m + n) points A1, A2,  A3, A4 and A5 on AX so that AA1 = A1A2 = A2A3 = A3A4 = A4A5. 2. Since the given ratio is 3:1, the number of points to be located on PX should be 4. Applying BPT in \(\Delta AB{A_7}\) , we have: \[\frac{{A{A_3}}}{{{A_3}{A_7}}} = \frac{{AC}}{{CB}}\], \[ \Rightarrow \boxed{\frac{{AC}}{{CB}} = \frac{3}{4}}\]. In the latter case, \(C\) would be a point on the extended line \(AB\), outside of the segment \(AB\), such that \({\rm{BC:CA  =  3:1}}\), as shown in the figure below: Now, how do we geometrically locate \(C\) if it divides \(AB\) externally in the ratio 3:1. Consider a line segment \(AB\): We want to find out a point lying on the extended line \(AB\), outside of the segment \(AB\), such that \({\rm{AC:CB  =  3:1}}\) , as shown in the figure below: We will say that \(C\) externally divides \(AB\) in the ratio 3:1. 1. 3. 〖〗_3/(_3 _5 )=/ There is a better way to mark point while dividing a line in a given ratio, which explained as follows. Draw ray BX making an acute ∠ABX. ... By geometrical construction, it is possible to divide a line segment in the ratio . on the opposite side of  the vertex C. Step IV :   Along AX, mark-off m (large of m and n) points A1, A2,…,Am on AX such that AA1 = A1A2 = ….. = Am–1Am. Then, we can equivalently say that \(C\) divides \(AB\) internally in the ratio (1/3):1. Let us divide a line segment AB into 3:2 ratio. Since ∠ AA5B = ∠ AA3C, Step III : Draw a line GH || AB at a distance of 3 cm, intersecting BP at C. Step V : Extend AB to D such that AD = 3/2 AB =cm = 6 cm. Step II : With A as centre andradius = AC = 6 cm, draw an arc. Now, as in the case of internal division, let us revisit this problem of external division from the perspective of coordinates. Suppose that the coordinates of \(A\) and \(B\) are: \[\begin{array}{l}A \equiv \left( {{x_1},\;{y_1}} \right)\\B \equiv \left( {{x_2},\;{y_2}} \right)\end{array}\].

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