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## Triangle Centres

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**Given the following triangle, find the:**• centroid • orthocenter • circumcenter**Centroid**• Equation of AD (median) • Strategy…. • Find midpoint D • Find eq’n of AD by • Find slope “m” of AD using A & D • Plug “m” & point A or D into y=mx+b & solve for “b” • Now write eq’n using “m” & “b” Remember – the centroid is useful as the centre of the mass of a triangle – you can balance a triangle on a centroid!**Centroid**Equation of AD (median)**Centroid**• Equation of BE (median) • Strategy…. • Find midpoint E • Find eq’n of BE by • Find slope “m” of BE using B & E • Plug “m” & point B or E into y=mx+b & solve for “b” • Now write eq’n using “m” & “b”**Centroid**Equation of BE (median)**Centroid**Question? Do we have to find the equation of median CF also?**Centroid**No We only need the equations of 2 medians… So, what do we do now?**Centroid**We need to find the Point of Intersection for medians AD & BE using either substitution or elimination**Centroid**Equation of median AD Equation of median BE**Centroid**• Equation of AD (median) • Strategy…. • Find midpoint D • Find eq’n of AD by • Find slope “m” of AD using A & D • Plug “m” & point A or D into y=mx+b & solve for “b” • Now write eq’n using “m” & “b”**Centroid – Intersection of Eq’n AD & BE**AD BE Add AD and BE Simplify and solve for y**Centroid – Intersection of Eq’n AD & BE**Substitute y = 1 into one of the equations Therefore, the point of intersection is (1,1)**Orthocentre**• Equation of altitude AD • Strategy…. • Find “m” of BC • Take –ve reciprocal of “m” of BC to get “m” of AD • Find eq’n of AD by • Plug “m” from 2. & point A into y=mx+b & solve for “b” • Now write eq’n using “m” & “b”**Centroid – Intersection of Eq’n AD & BE**Therefore, the Centroid is (1,1)**Orthocentre**Equation of altitude AD**Orthocentre**• Equation of altitude BE • Strategy…. • Find “m” of AC • Take –ve reciprocal of “m” of AC to get “m” of BE • Find eq’n of BE by • Plug “m” from 2. & point B into y=mx+b & solve for “b” • Now write eq’n using “m” & “b”**Orthocentre**Equation of altitude BE**Orthocentre**Question? Do we have to find the equation of altitude CF also?**Orthocentre**No We only need the equations of 2 altitudes… So, what do we do now?**Orthocentre**We need to find the Point of Intersection for altitudes AD & BE using either substitution or elimination**Orthocentre**Equation of altitude AD Equation of altitude BE**Orthocentre – Intersection of Eq’n AD & BE**AD BE Add AD and BE Simplify and solve for y**Orthocentre – Intersection of Eq’n AD & BE**Substitute y = 1 into one of the equations Therefore, the point of intersection or Orthocentre**Circumcenter**• Equation of ED (perpendicular bisector) • Strategy… (use A (-1, 4), B (-1, -2) & C(5, 1)) • Find midpoint D • Find eq’n of ED by • Find slope “m” of BC using B & E • Take –ve reciprocal to get “m” of ED • Plug “m” ED & point D into y = mx+b & solve for b • Now write eq’n using “m” & “b”**Circumcenter**Equation of ED (perpendicular bisector)**Circumcenter**• Equation of FG (perpendicular bisector) • Strategy… (use A (-1, 4), B (-1, -2) & C(5, 1)) • Find midpoint F • Find eq’n of ED by • Find slope “m” of AC using A & C • Take –ve reciprocal to get “m” of FG • Plug “m” FG & point F into y = mx+b & solve for b • Now write eq’n using “m” & “b”**Circumcenter**Question? Do we have to find the equation of perpendicular bisector HI?**Circumcenter**No We only need the equations of 2 perpendicular bisectors… So, what do we do now?**Circumcenter**We need to find the Point of Intersection for perpendicular bisectors ED & FG using either substitution or elimination**Circumcenter**Equation of perpendicular bisector ED Equation of perpendicular bisector FG**Circumcenter**Equation of FG (perpendicular bisector)