Check sibling questions

Example 22 - Straight lines work as plane mirror for a point

Example 22 - Chapter 10 Class 11 Straight Lines - Part 2
Example 22 - Chapter 10 Class 11 Straight Lines - Part 3
Example 22 - Chapter 10 Class 11 Straight Lines - Part 4
Example 22 - Chapter 10 Class 11 Straight Lines - Part 5
Example 22 - Chapter 10 Class 11 Straight Lines - Part 6
Example 22 - Chapter 10 Class 11 Straight Lines - Part 7
Example 22 - Chapter 10 Class 11 Straight Lines - Part 8
Example 22 - Chapter 10 Class 11 Straight Lines - Part 9

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Example 22 Assuming that straight lines work as the plane mirror for a point, find the image of the point (1, 2) in the line x − 3y + 4 = 0 . Let line AB be x – 3y + 4 = 0 & point P be (1, 2) Let Q (h, k) be the image of point P (1, 2) in line AB Since line AB is mirror Point P & Q are at equal distance from line AB, i.e. PR = QR, i.e. R is the mid point of PQ Image is formed perpendicular to mirror i.e. line PQ is perpendicular to line AB Since R is the mid point of PQ We know that mid point of a line joining (x1, y1) & (x2, y2) = ((𝑥_1+〖 𝑥〗_2)/2, (𝑦_1+ 𝑦_2)/2) Mid point of PQ joining (1, 2) & (h, k) is = ((1 + ℎ)/2 ", " (2 + 𝑘)/2) Coordinate of point R = ((1 + ℎ)/2 ", " (2 + 𝑘)/2) Since point R lies on the line AB It will satisfy the equation of line AB Putting x = (ℎ + 1)/2 & y = (𝑘 + 2)/2 in equation of AB x – 3y + 4 = 0 ((ℎ + 1)/2) – 3((𝑘 + 2)/2) + 4 = 0 (ℎ + 1 − 3(𝑘 + 2) + 4 × 2)/2 = 0 h + 1 – 3k – 6 + 8 = 0 h – 3k + 1 – 6 + 8 = 0 h – 3k + 3 = 0 h – 3k = –3 Also, PQ is perpendicular to AB We know that If two lines are perpendicular then product of their slope is equal to – 1 ∴ Slope of AB × Slope of PQ = –1 Slope of PQ = (−1)/(𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝐴𝐵) Finding slope of AB Equation of line AB is x – 3y + 4 = 0 x + 4 = 3y 3y = x + 4 y = (𝑥 + 4)/3 y = (1/3)x + 4/3 The above equation is of the form y = mx + c where m = Slope of line So, Slope of line AB = 1/3 Now, Slope of line PQ = (−1)/(𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝐴𝐵) = (−1)/(1/3) = –3 Now, Line PQ is formed by joining points P(1, 2) & Q(h, k) Slope of PQ = (𝑦_2 − 𝑦_1)/(𝑥_2− 𝑥_1 ) –3 = (𝑘 − 2)/(ℎ − 1) –3(h – 1) = k – 2 –3h + 3 = k – 2 –3h – k = –2 – 3 –3h – k = –5 –(3h + k) = –5 3h + k = 5 Now, our equations are h – 3k = –3 & 3h + k = 5 From (1) h – 3k = –3 h = 3k – 3 Putting value of h in (2) 3h + k = 5 3(3k – 3) + k = 5 9k – 9 + k = 5 9k + k = 5 + 9 10k = 14 k = 14/10 k = 7/5 Putting k = 7/5 in (1) 3h + k = 5 3h + 7/5 = 5 3h = 5 – 7/5 3h = (5(5) − 7)/5 3h = (25 − 7)/5 3h = 18/5 h = 18/(5 × 3) h = 6/5 Hence Q = (6/5, 7/5) Hence, image is (𝟔/𝟓, 𝟕/𝟓)

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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.