Last updated at May 29, 2018 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 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) = (( _1+ _2)/2, ( _1+ _2)/2) Mid point of PQ joining (1, 2) & (h, k) is = ((1 + )/2, (2 + )/2) Coordinate of point R = ((1 + )/2, (2 + )/2) Since point R lies on the line AB It will satisfy the equation of line AB Putting x = ( + 1)/2 & y = ( + 2)/2 in equation of AB x 3y + 4 = 0 (( + 1)/2) 3(( + 2)/2) + 4 = 0 ( + 1 3( + 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)/( ) Finding slope of AB Equation of line AB is x 3y + 4 = 0 x + 4 = 3y 3y = x + 4 y = ( + 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)/( ) = ( 1)/(1/3) = 3 Now, Line PQ is formed by joining points P(1, 2) & Q(h, k) Slope of PQ = ( _2 _1)/( _2 _1 ) 3 = ( 2)/( 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 (1) & 3h + k = 5 (2) 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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Example 22 Important You are here

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Class 11

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 has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.