My question is, “ A line with slope 3/2 that passes through the point (-2,0)” I have to find an equation. Can someone give me a rough explanation of how to do this, so i can figure this out? Thanks!
y = m x + b
is a straight line that has slope m and when x = 0, y = b
now we already know m = 3/2
so
y = (3/2) x + b
and we have y = 0 when x = -2
so
0 = (3/2)(-2) + b
0 = -3 + b
b = 3
so
y = (3/2) x + 3
or
2 y = 3 x + 6
THANK YOU!!!!!!!
You are welcome.
use the point-slope form of a line. The line with slope m that passes through (h,k) is
y-k = m(x-h)
So, using your numbers,
y-0 = 3/2 (x+2)
you can massage that as you will
To find the equation of a line with a given slope and that passes through a given point, you can follow these steps:
1. Write down the given slope: In this case, the slope is 3/2.
2. Write down the given point: The point is (-2, 0), where (-2) is the x-coordinate and 0 is the y-coordinate.
3. Use the point-slope form of a linear equation: The point-slope form of a linear equation is y - y1 = m(x - x1), where (x1, y1) represents the given point, and m represents the slope.
4. Plug in the values into the point-slope form equation: In this case, plugging in the values gives us: y - 0 = (3/2)(x - (-2)).
5. Simplify the equation: Simplify the equation further by multiplying 3/2 with x - (-2). This gives us: y = (3/2)(x + 2).
6. Distribute the slope: Distribute the slope 3/2 to both x and 2: y = (3/2)x + (3/2)(2).
7. Simplify the equation further: Simplify the equation by multiplying (3/2) with 2: y = (3/2)x + 3.
Therefore, the equation of the line with a slope of 3/2 that passes through the point (-2, 0) is y = (3/2)x + 3.