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Misc 18 - Find image of (3, 8) with respect to line x + 3y = 7 - Other Type of questions - Image

  1. Chapter 10 Class 11 Straight Lines
  2. Serial order wise
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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) = ((๐‘ฅ_1+ใ€– ๐‘ฅใ€—_2)/2, (๐‘ฆ_1+ ๐‘ฆ_2)/2) Mid point of PQ joining (3, 8) & (h, k) is = ((3 + โ„Ž)/2, (8 + ๐‘˜)/2) Coordinate of point R = ((3 + โ„Ž)/2, (8 + ๐‘˜)/2) Since point R lies on the line AB It will satisfy the equation of line AB Putting x = (3 + โ„Ž)/2 & y = (8 + ๐‘˜)/2 in equation AB ((3 + โ„Ž)/2) + 3((8 + ๐‘˜)/2) = 7 (3 + โ„Ž + 3(8 + ๐‘˜))/2 = 7 3 + h + 24 + 3k = 7(2) h + 3k + 27 = 14 h + 3k = 14 โˆ’ 27 h + 3k = โˆ’13 Also, PQ is perpendicular to AB We know that If two lines are perpendicular then product of their slope is equal to โ€“1 โˆด Slope of AB ร— Slope of PQ = โ€“1 Slope of PQ = (โˆ’1)/(๐‘†๐‘™๐‘œ๐‘๐‘’ ๐‘œ๐‘“ ๐ด๐ต) Finding slope of AB Equation of line AB is x + 3y = 7 3y = 7 โˆ’ x y = (7 โˆ’ ๐‘ฅ)/3 y = 7/3 โˆ’ ๐‘ฅ/3 y = (โˆ’๐‘ฅ)/3 + 7/3 y = ((โˆ’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 = (โˆ’1)/3 So, Slope of PQ = (โˆ’1)/(๐‘†๐‘™๐‘œ๐‘๐‘’ ๐‘œ๐‘“ ๐ด๐ต) = (โˆ’1)/((โˆ’1)/3) = 3 Now, Line PQ is joining point P(3, 8) & Q(h, k) Slope of PQ = (๐‘ฆ_2 โˆ’ใ€– ๐‘ฆใ€—_1)/(๐‘ฅ_2 โˆ’ ๐‘ฅ_1 ) 3 = (๐‘˜ โˆ’ 8)/(โ„Ž โˆ’ 3) 3(h โˆ’ 3) = k โˆ’ 8 3h โˆ’ 9 = k โˆ’ 8 3h โˆ’ k = โˆ’ 8 + 9 3h โˆ’ k = 1 Now, our equations are h + 3k = โ€“ 13 โ€ฆ(1) 3h โ€“ k = 1 โ€ฆ(2) From (1) h + 3k = โ€“ 13 h = โ€“ 13 โ€“ 3k Putting value of h in (2) 3h โ€“ k = 1 3(โ€“13 โ€“ 3k) โ€“ k = 1 โ€“39 โ€“ 9k โ€“ k = 1 โ€“9k โ€“ k = 1 + 39 โ€“10k = 40 k = 40/(โˆ’10) k = โ€“4 Putting k = โ€“4 in (1) h + 3k = โ€“ 13 h + 3(โ€“4) = โ€“ 13 h โ€“ 12 = โ€“ 13 h = โ€“ 13 + 12 h = โ€“1 Hence Q(h, k) = Q(โˆ’1, โ€“4) Hence image is Q(โˆ’1, โˆ’4)

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