Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Miscellaneous

Misc 1
Important

Misc 2 Important

Misc 3

Misc 4

Misc 5 Important

Misc 6

Misc 7 Important

Misc 8 Important

Misc 9

Misc 10 Important

Misc 11 Important

Misc 12

Misc 13

Misc 14 Important

Misc 15 Important You are here

Misc 16

Misc 17 Important

Misc 18 Important

Misc 19 Important

Misc 20 Important

Misc 21 Important

Misc 22

Misc 23 Important

Question 1 Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

Misc 15 Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point. Let line AB be x + y = 4 & point P be (–1, 2) Q is the point of intersection of line through P(-1, 2) to AB & PQ = 3 units We need to find the equation of line PQ First we will find the co-ordinates of point Q Since point Q(h,k) lies on line AB It will satisfy its equation x + y = 4 h + k = 4 h = 4 – k Also, Given PQ = 3 √((𝑥_2 − 𝑥_1 )^2 + (𝑦_2 − 𝑦_1 )^2 ) = 3 Putting values √((ℎ−(−1))^2 + (𝑘−2)^2 ) = 3 √((ℎ+1)^2 + (𝑘−2)^2 ) = 3 Squaring both sides (√((ℎ+1)^2 + (𝑘−2)^2 ))^2 = 32 (ℎ+1)^2 + (𝑘−2)^2 = 9 h2 + 12 + 2h + k2 + 22 – 4k = 9 h2 + 1 + 2h + k2 + 4 – 4k = 9 h2 + 2h + k2 – 4k = 9 – 1 – 4 h2 + 2h + k2 – 4k = 4 Putting h = 4 – k from (1) (4 – k)2 + 2(4 – k) + k2 – 4k = 4 42 + k2 – 8k + 8 – 2k + k2 – 4k = 4 16 + k2 – 8k + 8 – 2k + k2 – 4k = 4 k2 + k2 – 8k – 2k – 4k + 16 + 8 – 4 = 0 2k2 – 14k + 20 = 0 2(k2 – 7k + 10) = 0 k2 – 7k + 10 = 0 k2 – 5k – 2k + 10 = 0 k(k – 5) – 2(k – 5) = 0 (k – 2)(k – 5)= 0 So, k = 2 & k = 5 Now, finding equation of line PQ We know that equation of line joining two point (x1, y1) & (x2, y2) is (y – y1) = (𝑦2 − 𝑦1)/(𝑥2 − 𝑥1) (x – x1) So, line is parallel to x axis or y -axis