Last updated at March 9, 2017 by Teachoo

Transcript

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 โ 3y + 4 = 0 . Let line AB be x โ 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) = ((๐ฅ_1+ใ ๐ฅใ_2)/2, (๐ฆ_1+ ๐ฆ_2)/2) Mid point of PQ joining (1, 2) & (h, k) is = ((1 + โ)/2, (2 + ๐)/2) Coordinate of point R = ((1 + โ)/2, (2 + ๐)/2) Since point R lies on the line AB It will satisfy the equation of line AB Putting x = (โ + 1)/2 & y = (๐ + 2)/2 in equation of AB x โ 3y + 4 = 0 ((โ + 1)/2) โ 3((๐ + 2)/2) + 4 = 0 (โ + 1 โ 3(๐ + 2) + 4 ร 2)/2 = 0 h + 1 โ 3k โ 6 + 8 = 0 h โ 3k + 1 โ 6 + 8 = 0 h โ 3k + 3 = 0 h โ 3k = โ3 Also, PQ is perpendicular to AB We know that If two lines are perpendicular then product of their slope is equal to โ 1 โด Slope of AB ร Slope of PQ = โ1 Slope of PQ = (โ1)/(๐๐๐๐๐ ๐๐ ๐ด๐ต) Finding slope of AB Equation of line AB is x โ 3y + 4 = 0 x + 4 = 3y 3y = x + 4 y = (๐ฅ + 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 = (โ1)/(๐๐๐๐๐ ๐๐ ๐ด๐ต) = (โ1)/(1/3) = โ3 Now, Line PQ is formed by joining points P(1, 2) & Q(h, k) Slope of PQ = (๐ฆ_2 โ ๐ฆ_1)/(๐ฅ_2โ ๐ฅ_1 ) โ3 = (๐ โ 2)/(โ โ 1) โ3(h โ 1) = k โ 2 โ3h + 3 = k โ 2 โ3h โ k = โ2 โ 3 โ3h โ k = โ5 โ(3h + k) = โ5 3h + k = 5 Now, our equations are h โ 3k = โ3 โฆ(1) & 3h + k = 5 โฆ(2) From (1) h โ 3k = โ3 h = 3k โ 3 Putting value of h in (2) 3h + k = 5 3(3k โ 3) + k = 5 9k โ 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 โ 7/5 3h = (5(5) โ 7)/5 3h = (25 โ 7)/5 3h = 18/5 h = 18/(5 ร 3) h = 6/5 Hence Q = (6/5, 7/5) Hence, image is (6/5, 7/5)

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Example 15 Important

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Example 22 Important You are here

Misc 6 Important

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Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.