Miscellaneous

Misc 1
Important

Misc 2 Important

Misc 3 Important

Misc 4

Misc 5

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13

Misc 14

Misc 15 Important

Misc 16 Important

Misc 17

Misc 18 Important You are here

Misc 19 Important

Misc 20 Important

Misc 21 Important

Misc 22 Important

Misc 23

Misc 24 Important

Chapter 10 Class 11 Straight Lines (Term 1)

Serial order wise

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)