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Miscellaneous

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Last updated at Feb. 3, 2020 by Teachoo

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, (π¦_1+ π¦_2)/2) Mid point of PQ joining (3, 8) & (h, k) is = ((3 + β)/2 ", " (8 + π)/2) Coordinate of point R = ((3 + β)/2 ", " (8 + π)/2) Since point R lies on the line AB It will satisfy the equation of line AB Putting x = (3 + β)/2 & y = (8 + π)/2 in equation AB ((3 + β)/2) + 3((8 + π)/2) = 7 (3 + β + 3(8 + π))/2 = 7 3 + h + 24 + 3k = 7(2) h + 3k + 27 = 14 h + 3k = 14 β 27 h + 3k = β13 Also, PQ is perpendicular to AB We know that If two lines are perpendicular then product of their slope is equal to β1 β΄ Slope of AB Γ Slope of PQ = β1 Slope of PQ = (β1)/(πππππ ππ π΄π΅) Finding slope of AB Equation of line AB is x + 3y = 7 3y = 7 β x y = (7 β π₯)/3 y = 7/3 β π₯/3 y = (βπ₯)/3 + 7/3 y = ((β1)/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 = (β1)/3 So, Slope of PQ = (β1)/(πππππ ππ π΄π΅) = (β1)/((β1)/3) = 3 Now, Line PQ is joining point P(3, 8) & Q(h, k) Slope of PQ = (π¦_2 βγ π¦γ_1)/(π₯_2 β π₯_1 ) 3 = (π β 8)/(β β 3) 3(h β 3) = k β 8 3h β 9 = k β 8 3h β k = β 8 + 9 3h β k = 1 Now, our equations are h + 3k = β 13 3h β k = 1 From (1) h + 3k = β13 h = β 13 β 3k Putting value of h in (2) 3h β k = 1 3(β13 β 3k) β k = 1 β39 β 9k β k = 1 β9k β k = 1 + 39 β10k = 40 k = 40/(β10) k = β4 Putting k = β4 in (1) h + 3k = β13 h + 3(β4) = β13 h β 12 = β13 h = β13 + 12 h = β1 Hence Q(h, k) = Q(β1, β4) Hence image is Q(β1, β4)