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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-amp-ABC-




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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