Question

Дана прямая уровнение которой -3х-3y+48=0 найдите координаты точек в которых эта прямая пересекает оси координат

Answer 1

Ответ:

А(16;0) - точка пересечения прямой с осью абсцисс

В(0;16) - точка пересечения прямой с осью ординат

Объяснение:

-3х-3у+48=0

-3у=3х-48 | : (-3)

у=-х+16

1). у=0, -х+16=0, х=16

А(16;0)

2). х=0, у=16

В(0;16)

Answer 2

Finding the Intersection Points of a Line with the Coordinate Axes

The given equation of the line is -3x - 3y + 48 = 0. To find the coordinates of the points where this line intersects the coordinate axes, we can solve for the x and y intercepts separately.

X-Intercept: To find the x-intercept, we set y = 0 and solve for x: -3x - 3(0) + 48 = 0 -3x + 48 = 0 -3x = -48 x = -48 / -3 x = 16

Y-Intercept: To find the y-intercept, we set x = 0 and solve for y: -3(0) - 3y + 48 = 0 -3y + 48 = 0 -3y = -48 y = -48 / -3 y = 16

Therefore, the coordinates of the x-intercept are (16, 0) and the coordinates of the y-intercept are (0, 16).

So, the line intersects the x-axis at the point (16, 0) and the y-axis at the point (0, 16).