# Example 8 - Chapter 7 Class 10 Coordinate Geometry (Term 1)

Last updated at Aug. 16, 2021 by Teachoo

Examples

Example 1

Example 2 Important

Example 3 Important

Example 4

Example 5 Important

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Example 7 Important

Example 8 You are here

Example 9 Important

Example 10 Important

Example 11 Deleted for CBSE Board 2023 Exams

Example 12 Deleted for CBSE Board 2023 Exams

Example 13 Important Deleted for CBSE Board 2023 Exams

Example 14 Important Deleted for CBSE Board 2023 Exams

Example 15 Deleted for CBSE Board 2023 Exams

Chapter 7 Class 10 Coordinate Geometry

Serial order wise

Last updated at Aug. 16, 2021 by Teachoo

Example 8 Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4). Let the given points be A(2, −2) & B(−7, 4) P & Q are two points on AB such that AP = PQ = QB Let k = AP = PQ = QB Hence comparing AP & PB AP = m PB = PQ + QB = k + k Hence, Ratio between AP & PB = AP/PB = 𝑚/2𝑚 = 1/2 Thus P divides AB in the ratio 1:2 Finding P Let P(x, y) Hence, m1 = 1 , m2 = 2 And for AB x1 = 2 , x2 = −2 y1 = −7 , y2 = 4 x = (𝑚_1 𝑥_2 + 𝑚_2 𝑥_1)/(𝑚_1 + 𝑚_2 ) = (1 × −7 + 2 × 2)/(1 + 2) = (−7 + 4)/3 = (−3)/3 = −1 y = (𝑚_1 𝑦_2 + 𝑚_2 𝑦_1)/(𝑚_1 + 𝑚_2 ) = (1 × 4 + 2 × −2)/(1 + 2) = (4 − 4)/3 = 0/3 = 0 Hence, point P is P(x, y) = P(−1, 0) 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 = 2, x2 = −2 y1 = −7, y2 = 4 x = (𝑚_1 𝑥_2 + 𝑚_2 𝑥_1)/(𝑚_1 + 𝑚_2 ) = (2 ×−7 + 1 × 2)/(2 + 1) = (−14 + 2)/3 = (−12)/3 = −4 y = (𝑚_1 𝑦_2+ 𝑚_2 𝑦_1)/(𝑚_1+ 𝑚_2 ) = (2 × 4 + 1 × −2)/(1 + 2) = (8 − 2)/3 = 6/3 = 2 Hence, Point Q is Q (x, y) = Q (−4, 2)