Learn all Concepts of Chapter 7 Class 10 (with VIDEOS). Check - Coordinate Geometry - Class 10     1. Chapter 7 Class 10 Coordinate Geometry
2. Serial order wise
3. Ex 7.1

Transcript

Ex 7.1 , 9 If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR. Since Q is equidistant from P & R QP = QR Finding QP x1 = 0 , y1 = 1 x2 = 5 , y2 = −3 QP= √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 5 −0)2+(−3 −1)2) = √((5)2+(−4)2) = √(25+16) = √41 Similarly, Finding QR x1 = 0 , y1 = 1 x2 = x , y2 = 6 QR= √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 𝑥 −0)2+(6 −1)2) = √((𝑥)2+(5)2) = √(𝑥2+ 25) Since , QP = QR √41 = √(𝑥2+ 25) Squaring both sides (√41)2 = (√(𝑥2+ 25)) 2 41 = x 2 + 25 0 = x 2 + 25 − 41 0 = x 2 − 16 x 2 − 16 = 0 x 2 = 0 + 16 x 2 = 16 x = ± √16 x = ± 4 So, x = 4 or x = −4 Therefore, point R(x, 6) is (4, 6) or (−4, 6) Now we need to find the distances PR & QR Finding QR QR = √(𝑥2+ 25) Hence QR = √41 Finding PR x1 = 5 , y1 = −3 x2 = x , y2 = 6 PR =√((𝑥 −5)2+(6 −(−3))2) = √((𝑥 −5)2+(6+3)2) = √((𝑥 −5)2+(9)2) Hence PR = √82 or √162

Ex 7.1 