# Misc 22 - Chapter 9 Class 11 Straight Lines

Last updated at April 16, 2024 by Teachoo

Miscellaneous

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Misc 22 You are here

Misc 23 Important

Question 1 Important Deleted for CBSE Board 2024 Exams

Last updated at April 16, 2024 by Teachoo

Misc 22 Prove that the product of the lengths of the perpendiculars drawn from the points (β(π^2 β π^2 ), 0) and ( β β(π^2 β π^2 ), 0) to the line π₯/π cos π + π¦/π sin π = 1 is b2 . Let point A be (β(π^2βπ^2 ), 0) & point B be (ββ(π^2βπ^2 ), 0) The given line is π₯/π cos ΞΈ + π¦/π sinΞΈ = 1 We need to show that Product of the length of perpendiculars from point A & point B to the line π₯/π cos ΞΈ + π¦/π sin ΞΈ = 1 is b2 Let p1 be the perpendicular distance from point A(β(π^2 β π^2 ), 0) to the line π₯/π cos ΞΈ + π¦/π sin ΞΈ = 1 & p2 be the perpendicular distance from point B( β β(π^2 β π^2 ), 0) to the line π₯/π cos ΞΈ + π¦/π sin ΞΈ = 1 We need to show p1 Γ p2 = b2 Calculating p1 & p2 Given line is (π₯ )/π cos ΞΈ + π¦/πsin ΞΈ = 1 (cosβ‘π/π) x + (sinβ‘π/π)y β 1 = 0 The above equation is of the form Ax + By + C = 0 Here A = cosβ‘π/π, B = sinβ‘π/π & C = β 1 We know that Distance of a point (x1, y1) from line Ax + By + C = 0 is d = |γπ΄π₯γ_1 + γπ΅π¦γ_1 + π|/β(π΄^2 + π΅^2 ) Perpendicular distance p1 of point A(β(π^2 β π^2 ), 0) from the line π₯/π cos ΞΈ + π¦/π sin ΞΈ = 1 is d = |γπ΄π₯γ_1 + γπ΅π¦γ_1 + π|/β(π΄^2 + π΅^2 ) Putting values p1 = |cosβ‘π/π β(π^2 + π^2 ) + sinβ‘π/π (0) + ( β 1)|/β((cosβ‘π/π)^2 + (sinβ‘π/π)^2 ) p1 = (β(π^2 β π^2 )/π cosβ‘π + 0 β 1)/β((γπππ γ^2 π)/π^2 + (γπ ππγ^2 π)/π^2 ) p1 = (β(π^2 β π^2 )/π cosβ‘π β 1)/β((γπππ γ^2 π)/π^2 + (γπ ππγ^2 π)/π^2 ) Perpendicular distance p2 of point B = ( β β(π^2 β π^2 ), 0) from the line π₯/π cos ΞΈ + π¦/π sin ΞΈ = 1 is d = |π΄π₯1 + π΅π¦1 + π|/β(π^2 + π^2 ) Putting values p2 = |cosβ‘π/π ( β β(π^2 β π^2 )) + sinβ‘π/π (0) + ( β 1)|/β((cosβ‘π/π)^2 + (sinβ‘π/π)^2 ) p2 = ( β β(π^2 β π^2 )/π cosβ‘π + 0 β 1)/β((cosβ‘π/π)^2 + (sinβ‘π/π)^2 ) p2 = ( β β(π^2 β π^2 )/π cosβ‘π β 1)/β(γπππ γ^2/π^2 + (γπ ππγ^2 π)/π^2 ) Finding p1p2