Question

Прямая y=kx-24 проходит через точку (10;6). найдите угловой коэффициент этой прямой и определите,в каких координатных четвертях она расположена

Answer 1

Finding the Slope of the Line

To find the slope (or angular coefficient) of the line passing through the point (10, 6), we can use the slope-intercept form of a linear equation, which is y = kx + b, where k is the slope.

Given that the line passes through the point (10, 6), we can substitute these values into the equation to find the slope:

6 = 10k + b

Since we only need to find the slope, we can ignore the constant term b. Therefore, the equation becomes:

6 = 10k

To solve for k, we divide both sides of the equation by 10:

k = 6/10 = 0.6

Therefore, the slope of the line is 0.6.

Determining the Quadrant of the Line

To determine in which quadrant the line is located, we need to consider the signs of the x and y coordinates.

Since the slope of the line is positive (0.6), it means that as x increases, y also increases. This indicates that the line is located in either the first quadrant or the third quadrant.

To determine the quadrant, we can look at the signs of the x and y coordinates of the given point (10, 6). In the first quadrant, both x and y are positive. In the third quadrant, x is negative, but y is positive.

Since the given point (10, 6) has positive values for both x and y, the line is located in the first quadrant.

Please let me know if there's anything else I can help you with!