





Get live Maths 1-on-1 Classs - Class 6 to 12
Miscellaneous
Misc 2 Important Deleted for CBSE Board 2023 Exams
Misc 3 Important
Misc 4
Misc 5
Misc 6 Important
Misc 7
Misc 8 Important
Misc 9 Important
Misc 10
Misc 11 Important
Misc 12 Important
Misc 13
Misc 14
Misc 15 Important
Misc 16 Important You are here
Misc 17
Misc 18 Important
Misc 19 Important
Misc 20 Important
Misc 21 Important
Misc 22 Important
Misc 23
Misc 24 Important
Last updated at March 22, 2023 by Teachoo
Misc 16 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