Example 22 - Straight lines work as plane mirror for a point - Examples

  1. Class 11
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Example 22 Assuming that straight lines work as the plane mirror for a point, find the image of the point (1, 2) in the line x โˆ’ 3y + 4 = 0 . Let line AB be x โ€“ 3y + 4 = 0 & point P be (1, 2) Let Q (h, k) be the image of point P (1, 2) in line AB 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 (1, 2) & (h, k) is = ((1 + โ„Ž)/2, (2 + ๐‘˜)/2) Coordinate of point R = ((1 + โ„Ž)/2, (2 + ๐‘˜)/2) Since point R lies on the line AB It will satisfy the equation of line AB Putting x = (โ„Ž + 1)/2 & y = (๐‘˜ + 2)/2 in equation of AB x โ€“ 3y + 4 = 0 ((โ„Ž + 1)/2) โ€“ 3((๐‘˜ + 2)/2) + 4 = 0 (โ„Ž + 1 โˆ’ 3(๐‘˜ + 2) + 4 ร— 2)/2 = 0 h + 1 โ€“ 3k โ€“ 6 + 8 = 0 h โ€“ 3k + 1 โ€“ 6 + 8 = 0 h โ€“ 3k + 3 = 0 h โ€“ 3k = โ€“3 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 โ€“ 3y + 4 = 0 x + 4 = 3y 3y = x + 4 y = (๐‘ฅ + 4)/3 y = (1/3)x + 4/3 The above equation is of the form y = mx + c where m = Slope of line So, Slope of line AB = 1/3 Now, Slope of line PQ = (โˆ’1)/(๐‘†๐‘™๐‘œ๐‘๐‘’ ๐‘œ๐‘“ ๐ด๐ต) = (โˆ’1)/(1/3) = โ€“3 Now, Line PQ is formed by joining points P(1, 2) & Q(h, k) Slope of PQ = (๐‘ฆ_2 โˆ’ ๐‘ฆ_1)/(๐‘ฅ_2โˆ’ ๐‘ฅ_1 ) โ€“3 = (๐‘˜ โˆ’ 2)/(โ„Ž โˆ’ 1) โ€“3(h โ€“ 1) = k โ€“ 2 โ€“3h + 3 = k โ€“ 2 โ€“3h โ€“ k = โ€“2 โ€“ 3 โ€“3h โ€“ k = โ€“5 โ€“(3h + k) = โ€“5 3h + k = 5 Now, our equations are h โ€“ 3k = โ€“3 โ€ฆ(1) & 3h + k = 5 โ€ฆ(2) From (1) h โ€“ 3k = โ€“3 h = 3k โ€“ 3 Putting value of h in (2) 3h + k = 5 3(3k โ€“ 3) + k = 5 9k โ€“ 9 + k = 5 9k + k = 5 + 9 10k = 14 k = 14/10 k = 7/5 Putting k = 7/5 in (1) 3h + k = 5 3h + 7/5 = 5 3h = 5 โ€“ 7/5 3h = (5(5) โˆ’ 7)/5 3h = (25 โˆ’ 7)/5 3h = 18/5 h = 18/(5 ร— 3) h = 6/5 Hence Q = (6/5, 7/5) Hence, image is (6/5, 7/5)

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