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Question Number 140555 by liberty last updated on 09/May/21

The coordinates of A,B ,C are (6,3),(−3,5) and (4,−2) respectively and P is any point (x,y). Find ratio of area △PBC & △ABC.

$$\mathrm{The}\:\mathrm{coordinates}\:\mathrm{of}\:\mathrm{A},\mathrm{B}\:,\mathrm{C}\:\mathrm{are}\: \\ $$ $$\left(\mathrm{6},\mathrm{3}\right),\left(−\mathrm{3},\mathrm{5}\right)\:\mathrm{and}\:\left(\mathrm{4},−\mathrm{2}\right)\:\mathrm{respectively} \\ $$ $$\mathrm{and}\:\mathrm{P}\:\mathrm{is}\:\mathrm{any}\:\mathrm{point}\:\left(\mathrm{x},\mathrm{y}\right).\:\mathrm{Find}\:\mathrm{ratio} \\ $$ $$\mathrm{of}\:\mathrm{area}\:\bigtriangleup\mathrm{PBC}\:\&\:\bigtriangleup\mathrm{ABC}. \\ $$

Answered by EDWIN88 last updated on 09/May/21

area △PBC = (1/2)∣ determinant ((( x y 1)),((−3 5 1)),(( 4 −2 1)))∣ =(7/2)∣x+y−7∣ area △ABC = (1/2)∣ determinant ((( 6 3 1)),((−3 5 1)),(( 4 −2 1)))∣= ((49)/2) ((area △PBC)/(area △ABC)) = ((∣x+y−7∣)/7) .

$$\mathrm{area}\:\bigtriangleup\mathrm{PBC}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mid\begin{vmatrix}{\:\:\:\mathrm{x}\:\:\:\:\:\mathrm{y}\:\:\:\:\:\mathrm{1}}\\{−\mathrm{3}\:\:\:\:\mathrm{5}\:\:\:\:\:\mathrm{1}}\\{\:\:\:\mathrm{4}\:\:\:−\mathrm{2}\:\:\:\mathrm{1}}\end{vmatrix}\mid\:=\frac{\mathrm{7}}{\mathrm{2}}\mid\mathrm{x}+\mathrm{y}−\mathrm{7}\mid \\ $$ $$\mathrm{area}\:\bigtriangleup\mathrm{ABC}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mid\:\begin{vmatrix}{\:\:\:\mathrm{6}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{3}\:\:\:\mathrm{5}\:\:\:\:\:\:\mathrm{1}}\\{\:\:\mathrm{4}\:\:\:−\mathrm{2}\:\:\:\:\mathrm{1}}\end{vmatrix}\mid=\:\frac{\mathrm{49}}{\mathrm{2}} \\ $$ $$\frac{\mathrm{area}\:\bigtriangleup\mathrm{PBC}}{\mathrm{area}\:\bigtriangleup\mathrm{ABC}}\:=\:\frac{\mid\mathrm{x}+\mathrm{y}−\mathrm{7}\mid}{\mathrm{7}}\:. \\ $$

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