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Составьте уравнение прямой, проходящей через начало координат и точку (2, 3) кто нибудь , не просто с ответов , а с объяснением , а то в ответах из (-1.5bx+by=0) из ax+by=0, до 2х-3у=0

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dnikolay365
Раз проходит через начало координат, значит с=0. остаётся ах+ву=0. подставив значения точки д получим 3а=2в. в простейшем случае а=2; в=3 уравнение прямой 2х+3у=0
kuziv31
Посчитаем расстояния меж точками cd =  sqrt((2-6)^2+(2-5)^2)  = sqrt(4^2+3^2) =  sqrt(16+9) = sqrt(25) =  5 de  = sqrt((6-5)^2+())^2)  = sqrt(1^2+7^2)  = sqrt(50) = 5sqrt(2) ec =  sqrt((5-2)^2+(-2-2)^2)  = sqrt(3^2+4^2)  = sqrt(9+16) = sqrt(25) =  5 длины двух сторон совпали, и это хорошо, треугольник действительно равнобедренный. просят найти биссектрису, проведённую из вершины равнобедренного треугольника. а биссектриса эта совпадает с высотой и медианой. медиана делит основание пополам в точке м м = (d+e)/2 = ((6+5)/2; (5-2)/2)  = (11/2; 3/2) = (5,5; 1,5) cm =    sqrt((2-5,5)^2+(2-1,5)^2)  = sqrt(3,5^2+0,5^2)  = 5/sqrt(2)
Golovinskii_Sirotkina1944

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Составьте уравнение прямой, проходящей через начало координат и точку (2, 3) кто нибудь , не просто с ответов , а с объяснением , а то в ответах из (-1.5bx+by=0) из ax+by=0, до 2х-3у=0
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