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.?

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  • Indica
    Lv 7
    2 years ago
    Favorite Answer

    Take position vectors 𝙖,𝙗,π™˜,π™₯ for A,B,C,P and the centroid of Ξ”ABC as origin so 𝙖+𝙗+π™˜=𝙀

    PAΒ² = |π™₯βˆ’π™–|Β² = (π™₯βˆ’π™–)β€’(π™₯βˆ’π™–) = |π™₯|Β²+π™–β€’π™–βˆ’2π™₯‒𝙖 with similar results for PBΒ² and PCΒ²

    PAΒ²+PBΒ²+PCΒ² = 3|π™₯|Β²+𝙖‒𝙖+𝙗‒𝙗 +π™˜β€’π™˜βˆ’2π™₯β€’(𝙖+𝙗+π™˜) = 3|π™₯|Β²+|𝙖|Β²+|𝙗|Β²+|π™˜|Β²

    Using result at end, PAΒ²+PBΒ²+PCΒ² = 3|π™₯|Β²+245/3 = 90 ⟹ |π™₯|=5/3

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

    ………………………………………………..

    𝙖+𝙗+π™˜=𝙀 ⟹ (𝙖+𝙗+π™˜)β€’(𝙖+𝙗+π™˜)=0 ⟹ |𝙖|Β²+|𝙗|Β²+|π™˜|Β²+2(𝙖‒𝙗+π™—β€’π™˜+π™˜β€’π™–) = 0

    ABΒ²+BCΒ²+CAΒ² = |π™–βˆ’π™—|Β²+|π™—βˆ’π™˜|Β²+|π™˜βˆ’π™–|Β² = 2(|𝙖|Β²+|𝙗|Β²+|π™˜|Β²)βˆ’2(𝙖‒𝙗+π™—β€’π™˜+π™˜β€’π™–)

    ABΒ²+BCΒ²+CAΒ² = 3(|𝙖|Β²+|𝙗|Β²+|π™˜|Β²) ⟹ |𝙖|Β²+|𝙗|Β²+|π™˜|Β² = (8Β²+9Β²+10Β²)/3 = 245/3

    ……………………………………………….

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