{"id":1366,"date":"2023-05-05T10:00:53","date_gmt":"2023-05-05T04:30:53","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=1366"},"modified":"2023-05-06T09:18:52","modified_gmt":"2023-05-06T03:48:52","slug":"division-of-a-line-segment","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/division-of-a-line-segment\/","title":{"rendered":"Division Of A Line Segment Into A Given Ratio"},"content":{"rendered":"

Division Of A Line Segment<\/a> Into A Given Ratio<\/strong><\/h2>\n

\"\"\"\"<\/p>\n

Given a line segment AB, we want to divide it in the ratio m : n, where both m and n are positive integers. To help you to understand it, we shall take m = 3 and n = 2.
\nSteps of Construction:<\/strong>
\n1.<\/strong>\u00a0 Draw any ray AX, making an acute angle with AB.
\n2.\u00a0<\/strong> Locate 5(= m + n) points A1<\/sub>, A2<\/sub>,\u00a0 A3<\/sub>, A4<\/sub> and A5<\/sub> on AX so that AA1<\/sub> = A1<\/sub>A2<\/sub> = A2<\/sub>A3<\/sub> = A3<\/sub>A4<\/sub> = A4<\/sub>A5<\/sub>.
\n3.\u00a0<\/strong> Join BA5<\/sub>.
\n4.<\/strong>\u00a0 Through the point A3<\/sub> (m = 3), draw a line parallel to A5<\/sub>B (by making an angle equal to \u2220AA5<\/sub>B) at A3<\/sub> intersecting AB at the point C (see figure). Then, AC : CB = 3 : 2.
\n\"Division
\nLet use see how this method gives us the required division.
\nSince A3<\/sub>C is parallel to A5<\/sub>B, therefore,
\n\\( \\frac{A{{A}_{3}}}{{{A}_{3}}{{A}_{5}}}=\\frac{AC}{CB}\\text{ }\\left( \\text{By the Basic Proportionality Theorem} \\right) \\)
\n\\( \\frac{A{{A}_{3}}}{{{A}_{3}}{{A}_{5}}}=\\frac{3}{2}\\text{ (By construction) } \\)
\n\\( \\text{ }\\frac{AC}{CB}=\\frac{3}{2}\\text{ } \\)
\nThis shows that C divides AB in the ratio 3 : 2.<\/p>\n