Last updated at May 29, 2018 by Teachoo

Transcript

Ex 10.2, 19 (Method 1) Point R (h, k) divides a line segment between the axes in the ratio 1 : 2. Find equation of the line. Let the AB be a line between axis & point R(h, k) divides AB in the ratio 1: 2 Let AB make x-intercept a & y-intercept b Point A = (a, 0) & B = (0, b) So, equation of line AB by intercept form is 𝑥𝑎 + 𝑦 𝑏 = 1 Coordinate of point which divide line segment joining two points (x1, y1) & (x2, y2) in the ratio of m1 : m2 are = 𝑚2 𝑥2 + 𝑚1 𝑥1 𝑚1 + 𝑚2, 𝑚2 𝑦2 + 𝑚1 𝑦1 𝑚1 + 𝑚2 Point R(h, k) divide AB joining two points (a, 0) & (0, b) in the ratio of 1 : 2 (h, k) = 1 × 0 + 2 × 𝑎1 + 2, 1 × 𝑏 + 2 × 01 + 2 (h, k) = 0 + 2𝑎3, 𝑏3 (h, k) = 2𝑎3, 𝑏3 Putting value of a & b in (1) 𝑥𝑎 + 𝑦 𝑏 = 1 𝑥 3ℎ2 + 𝑦 3𝑘 = 1 2𝑥3ℎ + 𝑦 3𝑘 = 1 13 2𝑥ℎ + 𝑦 𝑘 = 1 2𝑥 𝑘 + 𝑦(ℎ)ℎ𝑘 = 1 × 3 2𝑘𝑥 + ℎ𝑦ℎ𝑘 = 3 2kx + hy = 3hk Which is the required equation Ex 10.2, 19 (Method 2) Point R (h, k) divides a line segment between the axes in the ratio 1:2. Find equation of the line. Let the AB be a line between axis & point R(h, k) divides AB in the ratio 1: 2 Let Point A = (a, 0) B = (0, b) We have to find equation of line AB We know that by two point form equation of line passing through (x1, y1) & (x2, y2) is (y – y1) = 𝑦2 − 𝑦1 𝑥2 − 𝑥1 (x – x1) So, equation of line AB passing through (a, 0) & (0, b) (y − 0) = b − 00 − a(x − a) y = 𝑏 − 𝑎(x − a) We need to find value of a & b Given that point R(h, k) divide AB in the ratio of 1 : 2 Coordinate of point which divide line segment joining two points (x1, y1) & (x2, y2) in the ratio of m1 : m2 are = 𝑚2 𝑥2 + 𝑚1 𝑥1 𝑚1 + 𝑚2, 𝑚2 𝑦2 + 𝑚1 𝑦1 𝑚1 + 𝑚2 Point R(h, k) divide AB joining two points (a, 0) & (0, b) in the ratio of 1 : 2 (h, k) = 1 × 0 + 2 × 𝑎1 + 2, 1 × 𝑏 + 2 × 01 + 2 (h, k) = 0 + 2𝑎3, 𝑏3 (h, k) = 2𝑎3, 𝑏3 Putting value of a & b in (1) y = − 𝑏𝑎 (x − a) y = − 3𝑘 3ℎ2 (x − 3ℎ2) y = − 3 × 2𝑘3ℎ(x – 3ℎ2) y = − 2𝑘ℎ (x – 3ℎ2) yh = − 2k ( 2𝑥 − 3ℎ2) yh = − 2𝑘2 (2x − 3h) yh = − k(2x − 3h) yh = − 2kx + 3hk hy + 2kx = 3hk 2kx + hy = 3hk Which is the required equation

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.