# Misc 18

Last updated at March 9, 2017 by Teachoo

Last updated at March 9, 2017 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 โฆ(1) 3h โ k = 1 โฆ(2) From (1) h + 3k = โ 13 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 = โ 13 h + 3(โ4) = โ 13 h โ 12 = โ 13 h = โ 13 + 12 h = โ1 Hence Q(h, k) = Q(โ1, โ4) Hence image is Q(โ1, โ4)

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .