Ex 9.1

Chapter 9 Class 11 Straight Lines
Serial order wise

### Transcript

Ex 9.1, 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" )

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.