# Ex 7.1, 8 - Chapter 7 Class 10 Coordinate Geometry

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 7.1 , 8 Find the values of y for which the distance between the points P(2, โ 3) and Q(10, y) is 10 units. Let the points be P(โ2, 3) & Q(10, y) Given that PQ = 10 units By distance formula PQ = โ((๐ฅ2 โ๐ฅ1)2+(๐ฆ2 โ๐ฆ1)2) x1 = 2 , y1 = โ3 x2 = 10 , y2 = y PQ = โ(( 10 โ2)2+(๐ฆโ(โ3))2) 10 = โ((8)2+(๐ฆ+3)2) Squaring both sides (10)2 = (โ((8)2+(๐ฆ+3)2))2 (10)2 = (8)2 + (y+ 3)2 100 = 64 + (y+ 3)2 100 = 64 + y2 + 32 + 2 ร 3 ร y 100 = 64 + y2 + 9 + 6y 0 = y2 + 6y + 64 + 9 โ 100 0 = y2 + 6y โ 27 y2 + 6y โ 27 = 0 y2 + 9y โ 3y โ 27 = 0 y(y + 9) โ 3(y + 9) = 0 (y โ 3) (y + 9) = 0 Hence y = 3 or y = โ9 is the solution

About the Author

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.