Menu Close

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-circumcen

Tinku Tara 0 Comments
Pin




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.
Ifthepoints(−3,5),(4,−2)and(6,2)aretheverticesofatriangle.(i)Findtheequationoftheperpendicularbisectorofthesides(ii)Findthecoordinateofthecircumcenter.(Thecircumcenterofatriangleisthepointofintersectionoftheperpendicularbisectoroftheside(iii)Findtheradiusofthecircumcircle.
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(−3,5)B(4,−2)C(6,2)AB=(y−5)=−2−54+3(x+3)7y−35=−7x−217y+7x=14midofAB−3+42,5−22(12,32)perpendiculartoABandpassingthrough(12,32)isperpdndiculRbisectorofABnowslopeofABcalculation7y+7x=14y=−x+2m=−1m1×m2=−1(−1)×m2=−1m2=1so⊥bisectorofABis(y−32)=1(x−12)2y−3=2x−12y−2x=2y−x=1similarwaycalculateother⊥bisectorofBCB(4,−2)C(6,2)BCeqnisy+2=2+26−4(x−4)2y+4=4x−162y−4x+20=0y−2x+10=0slope=2m1×m2=−12×m2=−1m2=−12midpointofBC4+62′−2+22=(5,0)eqnof⊥bisectorofBCisy−0=−12(x−5)2y+x−5=0intersectingpointbetweeny−x=1and2y+x−5=0isthecentreofcircley−x=12y+x=53y=6y=2x=1socentreifcircleis(1,2)eqn(x−1)2+(y−2)2=r2r=distancebetween(1,2)and(−3,5)←Apointr=(−3−1)2+(5−2)2=16+9=5soeqncircle(x−1)2+(y−2)2=52anotherwayii)eqnofcircleisx2+y2+2gx+2fy+c=0passesthrough(−3,5)(4,−2)and(6,2)9+25−6g+10f+c=0←eqn116+4+8g−4f+c=0←eqn236+4+12g+4f+c=0←eqn3fcircle34−6g+10f+c=020+8g−4f+c=040+12g+4f+c=0eqn1−eqn214−14g+14f=01−g+f=0eqn2−eqn3−20−4g−8f=05+g+2f=0solve5+g+2f=01−g+f=06+3f=0f=−2g=−1centreis(−g,−f)=(1,2)putg=−1anff=−2ineqn1to[findc34−6g+10f+c=034−6(−1)+10(−2)+c=034+6−20+c=0c=−20radiusr=g2+f2−cr=1+4+20=5eqncircle(x−1)2+(y−2)2=52
Commented by Tawa1 last updated on 18/Sep/18
God bless you sir. I really appreciate
Godblessyousir.Ireallyappreciate

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *