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Ex 7.2, 2 Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3). Let the given points be A (4, −1) & B (−2, 3) P & Q are two points on AB such that AP = PQ = QB Let AP = PQ = QB = m Point P divides AP & PB in the ratio AP = m PB = PQ + QB = k + k = 2k Hence, Ratio between AP & PB = AP/PB = 𝑘/2𝑘 = 1/2 Thus P divides AB in the ratio 1:2 Finding P Let P(x, y) m1 = 1, m2 = 2 And for AB x1 = 4, x2 = −2 y1 = −1, y2 = −3 x = (𝑚_1 𝑥_2 + 𝑚_2 𝑥_1)/(𝑚_1 + 𝑚_2 ) = (1 ×−2 + 2 × 4)/(1 + 2) = (−2 + 8)/3 = 6/3 = 2 y = (𝑚_1 𝑦_(2 )+ 𝑚_2 𝑦_1)/(𝑚_1 + 𝑚_2 ) = (1 × −3 + 2 × −1)/(1 + 2) = (−3 − 2)/3 = (−5)/3 Hence, point P is P(x, y) = P ("2, " (−5)/3) Similarly, Point Q divides AB in the ratio AQ & QB 𝐴𝑄/𝑄𝐵 = (𝐴𝑃 + 𝑃𝑄)/𝑄𝐵 = (𝑘 + 𝑘 )/𝑘 = 2𝑘/𝑘 = 2/1 = 2 : 1 Finding Q Let Q be Q(x , y) m1 = 2, m2 = 1 x1 = 4, x2 = −2 y1 = −1, y2 = −3 x = (𝑚_1 𝑥_2 + 𝑚_2 𝑥_1)/(𝑚_1 + 𝑚_2 ) = (2 ×−2 + 1 × 4)/(2 + 1) = (−4 + 4)/3 = 0/3 = 0 y = (𝑚_1 𝑦_2 + 𝑚_2 𝑦_1)/(𝑚_1 + 𝑚_2 ) = (2 × −3 + 1 × −1)/(1 + 2) = (−6 − 1)/3 = (−7)/3 Hence, point Q is Q(x, y) = Q ("0, " (−𝟕)/𝟑)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.