Get live Maths 1-on-1 Classs - Class 6 to 12

Examples

Example 1

Example 2 Important

Example 3 Important

Example 4 You are here

Example 5 Important

Example 6

Example 7 Important

Example 8

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 March 22, 2023 by Teachoo

Example 4 Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5). Let the points be A(x, y), B(7, 1), C(3, 5) Given that point A is equidistant from B & C Hence, AB = AC Finding AB & AC using distance formula Finding AB x1 = x, y1 = y x2 = 7, y2 = 1 AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √((7 −𝑥)2+(1 −𝑦)2) Similarly, finding AC x1 = x, y1 = y x2 = 3, y2 = 5 AC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 3 −𝑥)2+(5 −𝑦)2) We know that AB = AC √(( 7 −𝑥)2+(1 −𝑦)2) = √(( 3 −𝑥)2+(5 −𝑦)2) Squaring both sides 〖(√(( 7 −𝑥)2+(1 −𝑦)2) ")2 = (" √(( 3 −𝑥)2+(5 −𝑦)2))〗^2 (7 − x)2 + (1 − y)2 = (3 − x) 2 + (5 − y) 2 72 + x2 – 2 (7) (x) + 12 + y2 – 2y = (3) 2 + x2 − 2 (3)x + 52 + y2 – 2 (5)y 49 + x2 – 14x + 1 + y2 –2y = 9 + x2 − 6x + 25 + y2 – 10y (x2 − x2) – 14x + 6x + y2 − y2 −2y + 10y + 9 – 49 – 1 + 25 = 0 −8x + 8y + 16 = 0 8(−x + y + 2) = 0 −x + y + 2 = 0/8 −x + y + 2 = 0 y + 2 = x x = y + 2 x − y = 2 Since we only have to give relation between x & y So, answer is x − y = 2