Last updated at May 29, 2018 by Teachoo

Transcript

Misc 18 Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror. Let line AB be x + 3y = 7 & point P be (3, 8) Let Q (h, k) be the image of point P (3, 8) in the line AB x + 3y = 7 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 (3, 8) & (h, k) is = ((3 + )/2, (8 + )/2) Coordinate of point R = ((3 + )/2, (8 + )/2) Since point R lies on the line AB It will satisfy the equation of line AB Putting x = (3 + )/2 & y = (8 + )/2 in equation AB ((3 + )/2) + 3((8 + )/2) = 7 (3 + + 3(8 + ))/2 = 7 3 + h + 24 + 3k = 7(2) h + 3k + 27 = 14 h + 3k = 14 27 h + 3k = 13 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 = 7 3y = 7 x y = (7 )/3 y = 7/3 /3 y = ( )/3 + 7/3 y = (( 1)/3)x + 7/3 Equation of line is of the form y = mx + c Where m is slope of line Hence ,Slope of line AB = ( 1)/3 So, Slope of PQ = ( 1)/( ) = ( 1)/(( 1)/3) = 3 Now, Line PQ is joining point P(3, 8) & Q(h, k) Slope of PQ = ( _2 _1)/( _2 _1 ) 3 = ( 8)/( 3) 3(h 3) = k 8 3h 9 = k 8 3h k = 8 + 9 3h k = 1 Now, our equations are h + 3k = 13 (1) 3h k = 1 (2) From (1) h + 3k = 13 h = 13 3k Putting value of h in (2) 3h k = 1 3( 13 3k) k = 1 39 9k k = 1 9k k = 1 + 39 10k = 40 k = 40/( 10) k = 4 Putting k = 4 in (1) h + 3k = 13 h + 3( 4) = 13 h 12 = 13 h = 13 + 12 h = 1 Hence Q(h, k) = Q( 1, 4) Hence image is Q( 1, 4)

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Misc 6 Important

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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.