(iii)What are the coordinates of the point on y axis equidistant from A and G?

 

Slide139.JPG

Slide140.JPG
Slide141.JPG
'


Transcript

Question 36 (iii) – Choice 1 What are the coordinates of the point on y axis equidistant from A and G? Here, A(-2,2) and G(-4,7) Since the required point is in y-axis, its x –coordinate will be zero Let Required point = Z (0, a) As per question, point Z is equidistant from A & B Hence, AZ = BZ Finding AC x1 = βˆ’2 , y1 = 2 x2 = 0 , y2 = a AC = √((π‘₯2 βˆ’π‘₯1)2+(𝑦2 βˆ’π‘¦1)2) = √(( 0 βˆ’(βˆ’2))2+(π‘Žβˆ’2)2) = √((2)2+(π‘Žβˆ’2)2) = √((2)2+ π‘Ž2+22βˆ’2(2)(π‘Ž) ) = √(4+ π‘Ž2+4βˆ’4π‘Ž) = √(π’‚πŸβˆ’πŸ’π’‚+πŸ–) Finding BC x1 = βˆ’4 , y1 = 7 x2 = 0 , y2 = a BC = √((π‘₯2 βˆ’π‘₯1)2+(𝑦2 βˆ’π‘¦1)2) = √(( 0 βˆ’(βˆ’4))2+(7 βˆ’π‘Ž)2) = √((0+4)2+(7 βˆ’π‘Ž)2) = √((4)2+(7 βˆ’π‘Ž)2) = √((4)2+ π‘Ž2+72βˆ’2(7)(π‘Ž) ) = √(16+ π‘Ž2+49βˆ’14π‘Ž) = √(π’‚πŸβˆ’πŸπŸ’π’‚+πŸ”πŸ“) Now, AC = BC √(π‘Ž2βˆ’4π‘Ž+8) = √(π‘Ž2βˆ’14π‘Ž+65) Squaring both sides (√(π‘Ž2βˆ’4π‘Ž+8) " )2 = (" √(π‘Ž2βˆ’14π‘Ž+65))^2 π’‚πŸβˆ’πŸ’π’‚+πŸ– = π’‚πŸβˆ’πŸπŸ’π’‚+πŸ”πŸ“ βˆ’4a + 8 = βˆ’14a + 65 βˆ’4a + 14a = 65 βˆ’ 8 10a = 57 a = 57/10 a = 5.7 Hence the required point is C(0, a) = (0, 9)

Go Ad-free
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.