John asked in Science & MathematicsMathematics Β· 2 years ago

# A point P is situated inside a triangle ABC with sides 8, 9, 10. It is given that PA^2+PB^2+PC^2=90. Find the area of the locus of point P.?

Relevance
• Indica
Lv 7
2 years ago

Take position vectors π,π,π,π₯ for A,B,C,P and the centroid of ΞABC as origin so π+π+π=π€

PAΒ² = |π₯βπ|Β² = (π₯βπ)β’(π₯βπ) = |π₯|Β²+πβ’πβ2π₯β’π with similar results for PBΒ² and PCΒ²

PAΒ²+PBΒ²+PCΒ² = 3|π₯|Β²+πβ’π+πβ’π +πβ’πβ2π₯β’(π+π+π) = 3|π₯|Β²+|π|Β²+|π|Β²+|π|Β²

Using result at end, PAΒ²+PBΒ²+PCΒ² = 3|π₯|Β²+245/3 = 90 βΉ |π₯|=5/3

So locus is circle centre at centroid and radius 5/3

β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..

π+π+π=π€ βΉ (π+π+π)β’(π+π+π)=0 βΉ |π|Β²+|π|Β²+|π|Β²+2(πβ’π+πβ’π+πβ’π) = 0

ABΒ²+BCΒ²+CAΒ² = |πβπ|Β²+|πβπ|Β²+|πβπ|Β² = 2(|π|Β²+|π|Β²+|π|Β²)β2(πβ’π+πβ’π+πβ’π)

ABΒ²+BCΒ²+CAΒ² = 3(|π|Β²+|π|Β²+|π|Β²) βΉ |π|Β²+|π|Β²+|π|Β² = (8Β²+9Β²+10Β²)/3 = 245/3

β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.