Check sibling questions

Ex 10.1, 4 - Find a point on x-axis, which is equidistant

Ex 10.1, 4 - Chapter 10 Class 11 Straight Lines - Part 2
Ex 10.1, 4 - Chapter 10 Class 11 Straight Lines - Part 3
Ex 10.1, 4 - Chapter 10 Class 11 Straight Lines - Part 4
Ex 10.1, 4 - Chapter 10 Class 11 Straight Lines - Part 5

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Transcript

Ex 10.1, 4 Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4). Let the given points be A(7, 6) & B (3, 4) Let C be a point on the x-axis Coordinates of C = C(x, 0) Given that point C is equidistant from the points A & B (As it is on the x-axis, y = 0) Hence, Distance AC = Distance BC We know that distance D between two points (x1 , y1)&(x2 , y2)is D = √((𝑥2−𝑥1)2+(𝑦2 − 𝑦1)2) Distance between A(7, 6) & C(x, 0) AC = √((𝑥−7)^2+(0−6)2) = √((𝑥−7)^2+36) Distance between B(3, 4) & C(x, 0) BC = √((𝑥−3)^2+(0−4)2) = √((𝑥−3)^2+16) Since, AC = BC √((𝑥−7)^2+36) = √((𝑥−3)^2+16) Squaring both sides (√((𝑥−7)^2+36))^2 = (√((𝑥−3)^2+16))^2 (x – 7)2 + 36 = (x – 3)2 + 16 (x – 7)2 – (x – 3)2 = 16 – 36 x2 + 49 – 14x – (x2 + 9 – 6x) = –20 x2 + 49 – 14x – x2 – 9 + 6x = –20 0 – 8x + 40 = −20 –8x = – 20 – 40 –8x = −60 x = (−60)/(−8) x = 15/2 Thus, Required point = C(x, 0) = (𝟏𝟓/𝟐 ", 0" )

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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.