What are the coordinates of the point where the line containing K(-3,5) and L(5,-4) crosses the y-axis?
The slope is -9/8
The equation is y = -9x/8 + b
5 = (-9/8)(-3) + b
b = 5 - 27/8 = 13/8
y = -9x/8 + 13/8
When x = 0, y = b = 13/8
To find the coordinates of the point where the line crosses the y-axis, we need to determine the y-coordinate when the x-coordinate is 0.
The line passing through points K(-3,5) and L(5,-4) can be represented by the equation of a straight line, also known as the slope-intercept form:
y = mx + b
Where m is the slope of the line, and b is the y-intercept (the point where the line crosses the y-axis).
To find the slope (m), we can use the formula:
m = (y2 - y1) / (x2 - x1)
Using the coordinates of the two given points K(-3,5) and L(5,-4):
m = (-4 - 5) / (5 - (-3))
m = (-9) / (8)
m = -9/8
Now that we have the slope, we can substitute it back into the slope-intercept form equation:
y = (-9/8)x + b
To find the value of b (the y-intercept), we can use one of the given points, for example, K(-3,5):
5 = (-9/8)(-3) + b
Now solve for b:
5 = (27/8) + b
b = 5 - (27/8)
b = 40/8 - 27/8
b = 13/8
Therefore, the equation of the line passing through points K(-3,5) and L(5,-4) is:
y = (-9/8)x + 13/8
The point where the line crosses the y-axis occurs when x = 0. Substituting x = 0 into the equation above:
y = (-9/8)(0) + 13/8
y = 0 + 13/8
y = 13/8
So, the coordinates of the point where the line crosses the y-axis are (0, 13/8).