Find the equation of the line with the given information.
I know how to do this when given (x1,y1) & (x2,y2) but the given information is:
Through (2,-5) and parallel to y=3.
Can someone please explain the process to me?
And also for this given information:
x-intercept=3
y-intercept=-2
You know the slope is zero (from parallel to y=3). So it must be of the form
y= 0x + c
putting in (2.-5) yields c -5
y= -5 is the line.
help with find the domain of the function f(x)=log(5x+1)
To find the equation of a line parallel to y=3 and passing through a given point, you need to use the fact that parallel lines have the same slope.
The equation y=3 is in the form y=mx+b, where m is the slope of the line. In this case, since the line is horizontal (y=3), the slope is equal to 0. Therefore, any line parallel to y=3 will also have a slope of 0.
Now, for a line with a slope of 0 passing through the point (2, -5), the equation can be written as y = 0x + c, where c represents the y-coordinate of the point.
Substituting the values of (2, -5) into the equation, you get -5 = 0(2) + c. Simplifying the equation, you have -5 = c. Therefore, the equation of the line passing through (2, -5) and parallel to y=3 is y = -5.
For the second given information, the x-intercept is 3, which means the point (3, 0) lies on the line. Similarly, the y-intercept is -2, so the point (0, -2) lies on the line.
To find the equation, you can use the slope-intercept form of a line, y=mx+b, where m represents the slope and b represents the y-intercept.
Since the line passes through (3, 0), you can find the slope using the formula:
m = (y2 - y1) / (x2 - x1) = (0 - (-2)) / (3 - 0) = 2/3.
Now you have the slope as 2/3 and the y-intercept as -2. Plugging these values into the slope-intercept form, you get:
y = (2/3)x - 2
Therefore, the equation of the line with an x-intercept at 3 and a y-intercept at -2 is y = (2/3)x - 2.