Check sibling questions

Misc 22 - A ray of light passing through (1, 2) reflects on x-axis

Misc 22 - Chapter 10 Class 11 Straight Lines - Part 2
Misc 22 - Chapter 10 Class 11 Straight Lines - Part 3
Misc 22 - Chapter 10 Class 11 Straight Lines - Part 4
Misc 22 - Chapter 10 Class 11 Straight Lines - Part 5

This video is only available for Teachoo black users

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

Misc 22 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" )

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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.