© examsiri.com
Question : 5 of 81
Marks:
+1 ,
-0
Solution:
‌ 4 x + 3 y − 1 = 0 ‌ ‌ Normal vector ‌ = ⟨ 4 , 3 ⟩ Direction vectors are perpendicular to the normal vector
d = ⟨ 3 , − 4 ⟩ ‌ or ‌ ⟨ − 3 , 4 ⟩ Unit vector in direction vector
‌ = ‌ ⟨ 3 , − 4 ⟩ ‌ = ‌ ⟨ 3 , − 4 ⟩ So, two points 10 units away from
A = ( − 2 , 3 ) along the line are
‌ ( x 1 , y 1 ) = ( − 2 , 3 ) + 10 ⋅ ‌ < 3 , − 4 > ‌ = ( − 2 + 6 , 3 − 8 ) = ( 4 , − 5 ) ‌ ‌ and ‌ ( x 2 , y 2 ) = ( − 2 , 3 ) − 10 ⋅ ‌ < 3 , − 4 ) ‌ = ( − 2 − 6 , 3 + 8 ) ‌ = ( − 8 , 11 ) Therefore,
( x 1 + y 1 ) 2 + ( x 2 + y 2 ) 2 = ( 4 − 5 ) 2 + ( − 8 + 11 ) 2 = 1 + 9 = 10
© examsiri.com
Go to Question:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81