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Question Number 43944 by Tawa1 last updated on 18/Sep/18

If the points (−3, 5) , (4, −2) and (6, 2) are the vertices of a triangle. (i) Find the equation of the perpendicular bisector of the sides (ii) Find the coordinate of the circumcenter. (The circumcenter of a triangle is the point of intersection of the perpendicular bisector of the side (iii) Find the radius of the circumcircle.

$$\mathrm{If}\:\mathrm{the}\:\mathrm{points}\:\left(−\mathrm{3},\:\mathrm{5}\right)\:,\:\left(\mathrm{4},\:−\mathrm{2}\right)\:\mathrm{and}\:\left(\mathrm{6},\:\mathrm{2}\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}. \\ $$$$\left(\mathrm{i}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{perpendicular}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{coordinate}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circumcenter}.\:\left(\mathrm{The}\:\mathrm{circumcenter}\:\mathrm{of}\:\mathrm{a}\right. \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{perpendicular}\:\mathrm{bisector}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{side} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circumcircle}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Sep/18

A(−3,5) B(4,−2) C(6,2) AB= (y−5)=((−2−5)/(4+3))(x+3) 7y−35=−7x−21 7y+7x=14 mid of AB ((−3+4)/2),((5−2)/2) ((1/2),(3/2)) perpendicular to AB and passing through((1/2),(3/2)) is perpdndiculR bisector of AB now slope of AB calculation 7y+7x=14 y=−x+2 m=−1 m_1 ×m_2 =−1 (−1)×m_2 =−1 m_2 =1 so ⊥ bisector of AB is (y−(3/2))=1(x−(1/2)) 2y−3=2x−1 2y−2x=2 y−x=1 similar way calculate other ⊥ bisector of BC B(4,−2) C(6,2) BC eqn is y+2=((2+2)/(6−4))(x−4) 2y+4=4x−16 2y−4x+20=0 y−2x+10=0 slope=2 m_1 ×m_2 =−1 2×m_2 =−1 m_2 =((−1)/2) mid point of BC ((4+6)/2)′((−2+2)/2)=(5,0) eqn of ⊥bisector of BC is y−0=((−1)/2)(x−5) 2y+x−5=0 intersecting point between y−x=1 and 2y+x−5=0 is the centre of circle y−x=1 2y+x=5 3y=6 y=2 x=1 so centre if cir cle is (1,2) eqn (x−1)^2 +(y−2)^2 =r^2 r=distance between (1,2) and(−3,5)←A point r=(√((−3−1)^2 +(5−2)^2 )) =(√(16+9)) =5 so eqn circle (x−1)^2 +(y−2)^2 =5^2 another way ii)eqn of circle is x^2 +y^2 +2gx+2fy+c=0 passes through(−3,5) (4,−2) and(6,2) 9+25−6g+10f+c=0 ←eqn 1 16+4+8g−4f+c=0 ←eqn 2 36+4+12g+4f+c=0 ←eqn 3f circle 34−6g+10f+c=0 20+8g−4f+c=0 40+12g+4f+c=0 eqn1 −eqn2 14−14g+14f=0 1−g+f=0 eqn2 −eqn 3 −20−4g−8f=0 5+g+2f=0 solve 5+g+2f=0 1−g+f=0 6+3f=0 f=−2 g=−1 centre is (−g,−f)=(1,2) put g=−1 anf f=−2 in eqn 1 to[find c 34−6g+10f+c=0 34−6(−1)+10(−2)+c=0 34+6−20+c=0 c=−20 radius r=(√(g^2 +f^2 −c)) r=(√(1+4+20)) =5 eqn circle (x−1)^2 +(y−2)^2 =5^2

$${A}\left(−\mathrm{3},\mathrm{5}\right)\:\:{B}\left(\mathrm{4},−\mathrm{2}\right)\:\:{C}\left(\mathrm{6},\mathrm{2}\right) \\ $$$${AB}=\:\:\left({y}−\mathrm{5}\right)=\frac{−\mathrm{2}−\mathrm{5}}{\mathrm{4}+\mathrm{3}}\left({x}+\mathrm{3}\right) \\ $$$$\mathrm{7}{y}−\mathrm{35}=−\mathrm{7}{x}−\mathrm{21} \\ $$$$\mathrm{7}{y}+\mathrm{7}{x}=\mathrm{14} \\ $$$${mid}\:{of}\:{AB}\:\:\frac{−\mathrm{3}+\mathrm{4}}{\mathrm{2}},\frac{\mathrm{5}−\mathrm{2}}{\mathrm{2}}\:\:\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${perpendicular}\:{to}\:{AB}\:{and}\:{passing}\:{through}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right)\:{is} \\ $$$${perpdndiculR}\:{bisector}\:{of}\:{AB} \\ $$$${now}\:{slope}\:{of}\:{AB}\:\:{calculation} \\ $$$$\mathrm{7}{y}+\mathrm{7}{x}=\mathrm{14} \\ $$$${y}=−{x}+\mathrm{2}\:\:\:{m}=−\mathrm{1} \\ $$$${m}_{\mathrm{1}} ×{m}_{\mathrm{2}} =−\mathrm{1} \\ $$$$\left(−\mathrm{1}\right)×{m}_{\mathrm{2}} =−\mathrm{1}\:\:\:{m}_{\mathrm{2}} =\mathrm{1} \\ $$$${so}\:\bot\:{bisector}\:{of}\:{AB}\:{is}\:\left({y}−\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{1}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{2}{y}−\mathrm{3}=\mathrm{2}{x}−\mathrm{1} \\ $$$$\mathrm{2}{y}−\mathrm{2}{x}=\mathrm{2} \\ $$$${y}−{x}=\mathrm{1} \\ $$$$\boldsymbol{{similar}}\:\boldsymbol{{way}}\:\boldsymbol{{calculate}}\:\boldsymbol{{other}}\:\bot\:\boldsymbol{{bisector}} \\ $$$${of}\:{BC}\:\:\:{B}\left(\mathrm{4},−\mathrm{2}\right)\:{C}\left(\mathrm{6},\mathrm{2}\right) \\ $$$${BC}\:{eqn}\:{is}\:{y}+\mathrm{2}=\frac{\mathrm{2}+\mathrm{2}}{\mathrm{6}−\mathrm{4}}\left({x}−\mathrm{4}\right) \\ $$$$\mathrm{2}{y}+\mathrm{4}=\mathrm{4}{x}−\mathrm{16} \\ $$$$\mathrm{2}{y}−\mathrm{4}{x}+\mathrm{20}=\mathrm{0}\:\:\:\:\:{y}−\mathrm{2}{x}+\mathrm{10}=\mathrm{0} \\ $$$${slope}=\mathrm{2} \\ $$$${m}_{\mathrm{1}} ×{m}_{\mathrm{2}} =−\mathrm{1}\:\:\:\:\mathrm{2}×{m}_{\mathrm{2}} =−\mathrm{1}\:\:\:\:\:\:{m}_{\mathrm{2}} =\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${mid}\:{point}\:{of}\:{BC}\:\:\frac{\mathrm{4}+\mathrm{6}}{\mathrm{2}}'\frac{−\mathrm{2}+\mathrm{2}}{\mathrm{2}}=\left(\mathrm{5},\mathrm{0}\right) \\ $$$${eqn}\:{of}\:\bot{bisector}\:{of}\:{BC}\:{is} \\ $$$${y}−\mathrm{0}=\frac{−\mathrm{1}}{\mathrm{2}}\left({x}−\mathrm{5}\right) \\ $$$$\mathrm{2}{y}+{x}−\mathrm{5}=\mathrm{0} \\ $$$${intersecting}\:\:{point}\:{between}\:{y}−{x}=\mathrm{1} \\ $$$${and}\:\mathrm{2}{y}+{x}−\mathrm{5}=\mathrm{0} \\ $$$${is}\:{the}\:{centre}\:{of}\:{circle} \\ $$$${y}−{x}=\mathrm{1} \\ $$$$\mathrm{2}{y}+{x}=\mathrm{5} \\ $$$$\mathrm{3}{y}=\mathrm{6}\:\:\:\:{y}=\mathrm{2}\:\:\:\:\:\:{x}=\mathrm{1} \\ $$$${so}\:{centre}\:{if}\:{cir}\:{cle}\:{is}\:\left(\mathrm{1},\mathrm{2}\right) \\ $$$${eqn}\: \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${r}={distance}\:{between}\:\left(\mathrm{1},\mathrm{2}\right)\:{and}\left(−\mathrm{3},\mathrm{5}\right)\leftarrow{A}\:{point} \\ $$$${r}=\sqrt{\left(−\mathrm{3}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{5}−\mathrm{2}\right)^{\mathrm{2}} }\:=\sqrt{\mathrm{16}+\mathrm{9}}\:=\mathrm{5} \\ $$$${so}\:{eqn}\:{circle}\:\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${another}\:{way} \\ $$$$\left.\boldsymbol{{ii}}\right)\boldsymbol{{eqn}}\:{of}\:{circle}\:{is}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{gx}+\mathrm{2}{fy}+{c}=\mathrm{0} \\ $$$${passes}\:{through}\left(−\mathrm{3},\mathrm{5}\right)\:\left(\mathrm{4},−\mathrm{2}\right)\:{and}\left(\mathrm{6},\mathrm{2}\right) \\ $$$$\mathrm{9}+\mathrm{25}−\mathrm{6}{g}+\mathrm{10}{f}+{c}=\mathrm{0}\:\:\leftarrow{eqn}\:\mathrm{1} \\ $$$$\mathrm{16}+\mathrm{4}+\mathrm{8}{g}−\mathrm{4}{f}+{c}=\mathrm{0}\:\leftarrow{eqn}\:\mathrm{2} \\ $$$$\mathrm{36}+\mathrm{4}+\mathrm{12}{g}+\mathrm{4}{f}+{c}=\mathrm{0}\:\leftarrow{eqn}\:\mathrm{3}{f}\:{circle} \\ $$$$\mathrm{34}−\mathrm{6}{g}+\mathrm{10}{f}+{c}=\mathrm{0} \\ $$$$\mathrm{20}+\mathrm{8}{g}−\mathrm{4}{f}+{c}=\mathrm{0} \\ $$$$\mathrm{40}+\mathrm{12}{g}+\mathrm{4}{f}+{c}=\mathrm{0} \\ $$$${eqn}\mathrm{1}\:\:−{eqn}\mathrm{2} \\ $$$$\mathrm{14}−\mathrm{14}{g}+\mathrm{14}{f}=\mathrm{0}\:\:\:\mathrm{1}−{g}+{f}=\mathrm{0} \\ $$$${eqn}\mathrm{2}\:\:\:−{eqn}\:\mathrm{3} \\ $$$$−\mathrm{20}−\mathrm{4}{g}−\mathrm{8}{f}=\mathrm{0} \\ $$$$\mathrm{5}+{g}+\mathrm{2}{f}=\mathrm{0} \\ $$$${solve} \\ $$$$\mathrm{5}+{g}+\mathrm{2}{f}=\mathrm{0} \\ $$$$\mathrm{1}−{g}+{f}=\mathrm{0} \\ $$$$\mathrm{6}+\mathrm{3}{f}=\mathrm{0}\:\:\:\:{f}=−\mathrm{2}\:\:\:\:\:{g}=−\mathrm{1} \\ $$$${centre}\:{is}\:\left(−{g},−{f}\right)=\left(\mathrm{1},\mathrm{2}\right) \\ $$$${put}\:{g}=−\mathrm{1}\:{anf}\:{f}=−\mathrm{2}\:{in}\:{eqn}\:\mathrm{1}\:{to}\left[{find}\:{c}\right. \\ $$$$\mathrm{34}−\mathrm{6}{g}+\mathrm{10}{f}+{c}=\mathrm{0} \\ $$$$\mathrm{34}−\mathrm{6}\left(−\mathrm{1}\right)+\mathrm{10}\left(−\mathrm{2}\right)+{c}=\mathrm{0} \\ $$$$\mathrm{34}+\mathrm{6}−\mathrm{20}+{c}=\mathrm{0} \\ $$$${c}=−\mathrm{20} \\ $$$${radius}\:{r}=\sqrt{{g}^{\mathrm{2}} +{f}^{\mathrm{2}} −{c}}\: \\ $$$${r}=\sqrt{\mathrm{1}+\mathrm{4}+\mathrm{20}}\:=\mathrm{5} \\ $$$${eqn}\:{circle}\: \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Tawa1 last updated on 18/Sep/18

God bless you sir. I really appreciate

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate} \\ $$

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