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Question Number 78453 by mathocean1 last updated on 17/Jan/20

ABC is a triangle with points A(−5;−5) B(−5;10) C(15;−5). the cartesian equtions of (AB); (AC) and (BC) are respectively x=−5 y=−5 x+y=5 1) Please help me to determinate the cartesian equations of the interior bisectors of A^� ; B^� ; C^� . 2) Demonstrate that these bisectors meet in some point H(25;25) 3) Give a cartesian equation of inscrited circle in ABC triangle.

$$\mathrm{ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{points} \\ $$$$\mathrm{A}\left(−\mathrm{5};−\mathrm{5}\right)\:\mathrm{B}\left(−\mathrm{5};\mathrm{10}\right)\:\mathrm{C}\left(\mathrm{15};−\mathrm{5}\right). \\ $$$$\mathrm{the}\:\mathrm{cartesian}\:\mathrm{equtions}\:\mathrm{of}\:\left(\mathrm{AB}\right);\:\left(\mathrm{AC}\right) \\ $$$$\mathrm{and}\:\left(\mathrm{BC}\right)\:\mathrm{are}\:\mathrm{respectively} \\ $$$$\mathrm{x}=−\mathrm{5} \\ $$$$\mathrm{y}=−\mathrm{5} \\ $$$$\mathrm{x}+\mathrm{y}=\mathrm{5} \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{determinate}\: \\ $$$$\mathrm{the}\:\mathrm{cartesian}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{interior}\:\mathrm{bisectors}\:\mathrm{of}\:\hat {\mathrm{A}}\:;\:\hat {\mathrm{B}}\:;\:\hat {\mathrm{C}}. \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Demonstrate}\:\mathrm{that}\:\mathrm{these}\:\mathrm{bisectors} \\ $$$$\:\mathrm{meet}\:\mathrm{in}\:\mathrm{some}\:\mathrm{point}\:\mathrm{H}\left(\mathrm{25};\mathrm{25}\right) \\ $$$$\left.\mathrm{3}\right)\:\mathrm{Give}\:\mathrm{a}\:\mathrm{cartesian}\:\mathrm{equation}\:\mathrm{of}\: \\ $$$$\mathrm{inscrited}\:\mathrm{circle}\:\:\mathrm{in}\:\mathrm{ABC}\:\mathrm{triangle}. \\ $$$$ \\ $$

Commented by mathocean1 last updated on 17/Jan/20

Hello Please sirs can you help me...

$$\mathrm{Hello}\:\mathrm{Please}\:\mathrm{sirs}\:\mathrm{can}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me}... \\ $$$$ \\ $$

Commented by john santu last updated on 18/Jan/20

cartesian equation of interior bisector A^� : slope m = tan (π/4)=1 y= 1(x+5)+(−5) ⇒y=x

$${cartesian}\:{equation}\:{of}\:{interior}\:{bisector}\: \\ $$$$\hat {{A}}\::\:{slope}\:{m}\:=\:\mathrm{tan}\:\frac{\pi}{\mathrm{4}}=\mathrm{1} \\ $$$${y}=\:\mathrm{1}\left({x}+\mathrm{5}\right)+\left(−\mathrm{5}\right)\:\Rightarrow{y}={x} \\ $$$$ \\ $$

Commented by john santu last updated on 18/Jan/20

cartesian equation of the interior bisector of B^� with slope = −cot α = −2 and passing througt B(−5,10) ⇒y= −2(x+5)+10 y = −2x

$${cartesian}\:{equation}\:{of} \\ $$$${the}\:{interior}\:{bisector}\:{of}\:\hat {{B}}\: \\ $$$${with}\:{slope}\:=\:−\mathrm{cot}\:\alpha\:=\:−\mathrm{2}\:{and} \\ $$$${passing}\:{througt}\:{B}\left(−\mathrm{5},\mathrm{10}\right)\: \\ $$$$\Rightarrow{y}=\:−\mathrm{2}\left({x}+\mathrm{5}\right)+\mathrm{10} \\ $$$${y}\:=\:−\mathrm{2}{x} \\ $$

Commented by john santu last updated on 18/Jan/20

cartesian equation of the interior bisector C^� with slope = −tan β = −(1/3) y= −(1/3)(x−15)+(−5) y=−(1/3)x

$${cartesian}\:{equation}\:{of}\:{the} \\ $$$${interior}\:{bisector}\:\hat {{C}}\: \\ $$$${with}\:{slope}\:=\:−\mathrm{tan}\:\beta\:=\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${y}=\:−\frac{\mathrm{1}}{\mathrm{3}}\left({x}−\mathrm{15}\right)+\left(−\mathrm{5}\right) \\ $$$${y}=−\frac{\mathrm{1}}{\mathrm{3}}{x}\: \\ $$

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