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

Misc 18 - Chapter 10 Class 11 Straight Lines - Part 2
Misc 18 - Chapter 10 Class 11 Straight Lines - Part 3
Misc 18 - Chapter 10 Class 11 Straight Lines - Part 4
Misc 18 - Chapter 10 Class 11 Straight Lines - Part 5
Misc 18 - Chapter 10 Class 11 Straight Lines - Part 6
Misc 18 - Chapter 10 Class 11 Straight Lines - Part 7
Misc 18 - Chapter 10 Class 11 Straight Lines - Part 8

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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) = ((π‘₯_1+γ€– π‘₯γ€—_2)/2, (𝑦_1+ 𝑦_2)/2) Mid point of PQ joining (3, 8) & (h, k) is = ((3 + β„Ž)/2 ", " (8 + π‘˜)/2) Coordinate of point R = ((3 + β„Ž)/2 ", " (8 + π‘˜)/2) Since point R lies on the line AB It will satisfy the equation of line AB Putting x = (3 + β„Ž)/2 & y = (8 + π‘˜)/2 in equation AB ((3 + β„Ž)/2) + 3((8 + π‘˜)/2) = 7 (3 + β„Ž + 3(8 + π‘˜))/2 = 7 3 + h + 24 + 3k = 7(2) h + 3k + 27 = 14 h + 3k = 14 βˆ’ 27 h + 3k = βˆ’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 Γ— Slope of PQ = –1 Slope of PQ = (βˆ’1)/(π‘†π‘™π‘œπ‘π‘’ π‘œπ‘“ 𝐴𝐡) Finding slope of AB Equation of line AB is x + 3y = 7 3y = 7 βˆ’ x y = (7 βˆ’ π‘₯)/3 y = 7/3 βˆ’ π‘₯/3 y = (βˆ’π‘₯)/3 + 7/3 y = ((βˆ’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 = (βˆ’1)/3 So, Slope of PQ = (βˆ’1)/(π‘†π‘™π‘œπ‘π‘’ π‘œπ‘“ 𝐴𝐡) = (βˆ’1)/((βˆ’1)/3) = 3 Now, Line PQ is joining point P(3, 8) & Q(h, k) Slope of PQ = (𝑦_2 βˆ’γ€– 𝑦〗_1)/(π‘₯_2 βˆ’ π‘₯_1 ) 3 = (π‘˜ βˆ’ 8)/(β„Ž βˆ’ 3) 3(h βˆ’ 3) = k βˆ’ 8 3h βˆ’ 9 = k βˆ’ 8 3h βˆ’ k = βˆ’ 8 + 9 3h βˆ’ k = 1 Now, our equations are h + 3k = – 13 3h – k = 1 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/(βˆ’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(βˆ’1, –4) Hence image is Q(βˆ’1, βˆ’4)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.