Помогите срочно Общее уравнение прямой имеет вид x-y-3=0 .Найти угол наклона прямой к
положительному направлению оси абсцисс . Записать уравнение прямой на отрезках .Ответы на вопрос
y=kx+b, где k=tgα и α - угол наклона к положит. направлению оси Ох
y=x-3 => k=1 => tgα=1 => α=45°
Запишем уравнение прямой в отрезках:
Полученное уравнение и является уравнением прямой в отрезках.
Finding the Slope of the Line:
The general equation of a line is given by Ax + By + C = 0. In this case, the equation of the line is x - y - 3 = 0. To find the slope of the line, we can rewrite the equation in the slope-intercept form, which is y = mx + c, where m is the slope.By rearranging the given equation, we get: y = x - 3
Comparing this with the slope-intercept form, we can see that the slope of the line is 1.
Calculating the Angle of Inclination:
The angle of inclination of a line to the positive direction of the x-axis is given by the arctan of the slope. Therefore, the angle of inclination (θ) is given by: θ = arctan(m)Substituting the value of the slope, we get: θ = arctan(1) = 45 degrees
So, the angle of inclination of the line to the positive direction of the x-axis is 45 degrees.
Writing the Equation of the Line in Segments:
To write the equation of the line in segments, we can consider two points on the line and use the point-slope form of the equation of a line, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.Let's consider two points on the line: - Point 1: (0, -3) - Point 2: (3, 0)
Using the point-slope form with Point 1, we get: y + 3 = x
Using the point-slope form with Point 2, we get: y - 0 = x - 3
Therefore, the equation of the line in segments is: y + 3 = x for the segment between the y-intercept and the x-intercept, and y = x - 3 for the segment between the x-intercept and the y-intercept.
I hope this helps! If you have further questions or need additional assistance, feel free to ask.
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