Напишіть рівняння прямої, що проходить через
початок координат і точку м (-2; 5).​

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А) 
∠АMN=90 °; ∠ACN= 90 °.
Сумма противоположных углов четырехугольника СNMA равна 180 °, значит около четырехугольника CNMA можно описать окружность.
∠СMN=∠CAN как вписанные углы, опирающиеся на одну и ту же дугу NC.
б)
Так как точка М– середина гипотенузы является центром окружности, описанной около треугольника АВС, то
ВM=AM=CM 

Треугольник CMB – равнобедренный, так как СM=BM.

Треугольник ANB – равнобедренный, так как NM – серединный перпендикуляр к АВ, поэтому BN=AN.

Угол В в этих треугольниках общий.

По теореме синусов из треугольника АNB
BN/sin∠B=2R1, R1– радиус окружности, описанной около треугольника ANB.
По теореме синусов из треугольника СМВ:
СM/sin ∠B=2R2
R2– радиус окружности, описанной около треугольника СМВ

Значит
R1/R2=BN/CM, так как СМ=ВМ.
R1/R2=BN/BM

Рассмотрим прямоугольный треугольник ВNM:
cos∠B=BM/BN
R1/R2=1/cos∠B

По условию
tg∠A=4/3 ⇒ 1+tg2∠A=1/cos2∠A
значит 
cos2∠A=1/(1+tg2∠A)=1/(1+(4/3)2)=9/25
так как угол А –острый, то cos∠A=3/5
sin∠A=4/5
sin∠A=cos∠B

R1/R2=1/cos∠B=1/(4/5)=5/4
О т в е т. 5/4