Miscellaneous
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
Misc 16
Misc 17 Important
Misc 18 Important
Misc 19 Important
Misc 20 Important
Misc 21 Important You are here
Misc 22
Misc 23 Important
Question 1 Important
Last updated at April 16, 2024 by Teachoo
You saved atleast 2 minutes by viewing the ad-free version of this page. Thank you for being a part of Teachoo Black.
Misc 21 A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A. There is a point A on x-axis on which ray reflects A ray passing through P(1, 2) reflects on point A On reflection, the ray passes through point Q(5, 3) We need to find coordinate of A θ + ∠ MAQ = 90° ∠ MAQ = 90° – θ Also, ∠ MAP = ∠ MAQ = 90 – θ Now, ∠ PAX = ∠ MAP + ∠ MAQ + ∠ QAX = (90 – θ) + (90 – θ) + θ = 180 – θ Now, we find slope of line PA & QA We know that slope of line that passes through points (x1, y1) & (x2, y2) is m = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) Line PA Slope of line PA passing through points (1, 2) & (k, 0) is Slope of PA = (0 − 2)/(𝑘 − 1) = ( −2)/(𝑘 − 1) But PA makes angle 180 – θ with positive x-axis Slope of PA = tan (180 – θ) = – tan θ So, – tan θ = ( −2)/(𝑘 − 1) tan θ = ( 2)/(𝑘 − 1) Line QA Slope of line QA passing through points (5, 3) & (k, 0) is Slope of QA = (0 − 3)/(𝑘 − 5) = ( −3)/(𝑘 − 5) But QA makes angle θ with positive x-axis Slope of QA = tan θ So, tan θ = ( −3)/(𝑘 − 5) From (1) & (2) 2/(𝑘 − 1) = ( −3)/(𝑘 − 5) 2(k − 5) = −3(k − 1) 2k − 10 = −3k + 3 2k + 3k = 3 + 10 5k = 13 k = 13/5 Hence point A (𝟏𝟑/𝟓 ", 0" )