1. Class 11
2. Important Question for exams Class 11
3. Chapter 10 Class 11 Straight Lines

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)

Chapter 10 Class 11 Straight Lines

Class 11
Important Question for exams Class 11