Question Number 190075 by Spillover last updated on 26/Mar/23
Answered by PowerMaths last updated on 27/Mar/23
![(a) p_1 = ⊥^r from A(1,3) to line =((∣3×1+4×3−9∣)/( (√(3^2 +4^2 )))) = (6/5) = 1.2 p_2 = ⊥^r from B(2,7) to line =((∣3×2+4×7−9∣)/( (√(3^2 +4^2 )))) = ((25)/5) = 5 ratio of division = ((1.2)/5) = ((12)/(50)) = (6/(25)) from A [notic: the ratio of division of segment = the ratio between perpendiculars due to similarity] (b) equation of line: (x/a) + (y/b) = 1 ⇒ bx + ay = ab p = ((∣b×0+a×0−ab∣)/( (√(a^2 +b^2 )))) ⇒ p = ((ab)/( (√(a^2 +b^2 )))) squaring p^2 = ((a^2 b^2 )/( a^2 +b^2 )) ⇒ (1/p^2 ) = ((a^2 +b^2 )/(a^2 b^2 )) ⇒ (1/p^2 ) = (1/b^2 ) + (1/a^2 )](https://www.tinkutara.com/question/Q190086.png)
$$\left({a}\right)\:{p}_{\mathrm{1}} =\:\bot^{{r}} \:{from}\:{A}\left(\mathrm{1},\mathrm{3}\right)\:{to}\:{line}\:=\frac{\mid\mathrm{3}×\mathrm{1}+\mathrm{4}×\mathrm{3}−\mathrm{9}\mid}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }}\:=\:\frac{\mathrm{6}}{\mathrm{5}}\:=\:\mathrm{1}.\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:{p}_{\mathrm{2}} \:=\:\bot^{{r}} \:{from}\:{B}\left(\mathrm{2},\mathrm{7}\right)\:{to}\:{line}\:=\frac{\mid\mathrm{3}×\mathrm{2}+\mathrm{4}×\mathrm{7}−\mathrm{9}\mid}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }}\:=\:\frac{\mathrm{25}}{\mathrm{5}}\:=\:\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:{ratio}\:{of}\:{division}\:=\:\frac{\mathrm{1}.\mathrm{2}}{\mathrm{5}}\:=\:\frac{\mathrm{12}}{\mathrm{50}}\:=\:\frac{\mathrm{6}}{\mathrm{25}}\:\:{from}\:{A} \\ $$$$\:\:\:\:\:\:\:\:\:\left[{notic}:\:{the}\:{ratio}\:{of}\:{division}\:{of}\:{segment}\:=\:{the}\:{ratio}\:{between}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:{perpendiculars}\:{due}\:{to}\:{similarity}\right] \\ $$$$\left({b}\right)\:{equation}\:{of}\:{line}: \\ $$$$\:\:\:\:\:\:\:\:\frac{{x}}{{a}}\:+\:\frac{{y}}{{b}}\:=\:\mathrm{1}\:\Rightarrow\:{bx}\:+\:{ay}\:=\:{ab} \\ $$$$\:\:\:\:\:\:\:\:{p}\:=\:\frac{\mid{b}×\mathrm{0}+{a}×\mathrm{0}−{ab}\mid}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:\Rightarrow\:{p}\:=\:\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:\:\:\:\:\:\:{squaring} \\ $$$$\:\:\:\:\:\:\:\:{p}^{\mathrm{2}} \:=\:\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\Rightarrow\:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:=\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\:\Rightarrow\:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\: \\ $$
Commented by Spillover last updated on 27/Mar/23
$$\mathrm{good}.\mathrm{thanks} \\ $$