Question

Найдите координаты точки графика линейной функции у=х+10, ордината которой в 3 раза больше абциссы

Answer 1

Для решения необходимо просто понять условие. Ордината - по оси Оу, абсцисса - по оси Ох.

Пусть точка M(x;y).

$\begin{matrix} \begin{Bmatrix} y=x+10\\y=3x\end{matrix} &\begin{vmatrix}3x=x+10|-x\\2x=10\Rightarrow x=5\end{matrix} \end{matrix}\\y=3*x=3*5=15\\Otvet:M(5;15).$

Answer 2

Finding the Coordinates of a Linear Function

To find the coordinates of a point on the graph of a linear function, we need to determine the values of the x-coordinate (abscissa) and the y-coordinate (ordinate) for that point.

In this case, we are given that the y-coordinate is three times greater than the x-coordinate. Let's denote the x-coordinate as x. According to the given information, the y-coordinate (y) can be expressed as 3x.

Therefore, the equation of the linear function is y = 3x.

To find the coordinates of a specific point on this graph, we need to substitute a value for x and calculate the corresponding value for y.

Let's choose x = 10 as an example. By substituting this value into the equation, we can find the corresponding value for y:

y = 3 * 10 = 30

Therefore, the coordinates of the point on the graph of the linear function y = 3x, where the y-coordinate is three times greater than the x-coordinate, when x = 10, are (10, 30).

Please note that the given search results did not provide specific information about the coordinates of the point on the graph. However, the process described above is a general method for finding the coordinates of a point on a linear function graph.