Question

nico is trying to write an equation in slope-intercept form for a line through the point (1,2) and has a slope of -1/2 what is the y-intercept of this line rounded to the nearest tenth?

Answer 1

To find the equation of a line in slope-intercept form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept, we can use the slope (-1/2) and the point (1, 2).

We have:

- \( m = -\frac{1}{2} \)
- The point \( (x_1, y_1) = (1, 2) \)

First, we can substitute the slope and the coordinates of the point into the equation \( y = mx + b \):

1. Start with \( y = mx + b \):
\[
y = -\frac{1}{2}x + b
\]

2. Substitute the coordinates of the point \( (1, 2) \):
\[
2 = -\frac{1}{2}(1) + b
\]
\[
2 = -\frac{1}{2} + b
\]

3. To isolate \( b \), add \( \frac{1}{2} \) to both sides:
\[
2 + \frac{1}{2} = b
\]
\[
b = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}
\]

Now convert \( \frac{5}{2} \) to a decimal:
\[
\frac{5}{2} = 2.5
\]

Thus, the y-intercept \( b \) is \( 2.5 \).

So the y-intercept of this line, rounded to the nearest tenth, is \( \boxed{2.5} \).