Last updated at Feb. 3, 2020 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) = ((๐ฅ_1+ใ ๐ฅใ_2)/2, (๐ฆ_1+ ๐ฆ_2)/2) Mid point of PQ joining (3, 8) & (h, k) is = ((3 + โ)/2 ", " (8 + ๐)/2) Coordinate of point R = ((3 + โ)/2 ", " (8 + ๐)/2) Since point R lies on the line AB It will satisfy the equation of line AB Putting x = (3 + โ)/2 & y = (8 + ๐)/2 in equation AB ((3 + โ)/2) + 3((8 + ๐)/2) = 7 (3 + โ + 3(8 + ๐))/2 = 7 3 + h + 24 + 3k = 7(2) h + 3k + 27 = 14 h + 3k = 14 โ 27 h + 3k = โ13 Also, PQ is perpendicular to AB We know that If two lines are perpendicular then product of their slope is equal to โ1 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 + 3y = 7 3y = 7 โ x y = (7 โ ๐ฅ)/3 y = 7/3 โ ๐ฅ/3 y = (โ๐ฅ)/3 + 7/3 y = ((โ1)/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 = (โ1)/3 So, Slope of PQ = (โ1)/(๐๐๐๐๐ ๐๐ ๐ด๐ต) = (โ1)/((โ1)/3) = 3 Now, Line PQ is joining point P(3, 8) & Q(h, k) Slope of PQ = (๐ฆ_2 โใ ๐ฆใ_1)/(๐ฅ_2 โ ๐ฅ_1 ) 3 = (๐ โ 8)/(โ โ 3) 3(h โ 3) = k โ 8 3h โ 9 = k โ 8 3h โ k = โ 8 + 9 3h โ k = 1 Now, our equations are h + 3k = โ 13 3h โ k = 1 From (1) h + 3k = โ13 h = โ 13 โ 3k Putting value of h in (2) 3h โ k = 1 3(โ13 โ 3k) โ k = 1 โ39 โ 9k โ k = 1 โ9k โ k = 1 + 39 โ10k = 40 k = 40/(โ10) k = โ4 Putting k = โ4 in (1) h + 3k = โ13 h + 3(โ4) = โ13 h โ 12 = โ13 h = โ13 + 12 h = โ1 Hence Q(h, k) = Q(โ1, โ4) Hence image is Q(โ1, โ4)

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Chapter 10 Class 11 Straight Lines (Term 1)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.