Question

Nico is trying to write an equation in slope-intercept form for a line that passes 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 write the equation of a line in slope-intercept form (which is \( y = mx + b \)), we need to know the slope \( m \) and the y-intercept \( b \).

Given:
- Slope \( m = -\frac{1}{2} \)
- Point \( (1, 2) \)

We can use the point-slope formula to find the y-intercept. The point-slope formula is given by:

\[
y - y_1 = m(x - x_1)
\]

where \( (x_1, y_1) \) is the point the line passes through. Substituting in the values we have:

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

Substituting these into the formula:

\[
y - 2 = -\frac{1}{2}(x - 1)
\]

Now, we simplify this equation:

\[
y - 2 = -\frac{1}{2}x + \frac{1}{2}
\]

Now, add 2 to both sides:

\[
y = -\frac{1}{2}x + \frac{1}{2} + 2
\]

To combine the constants, convert 2 into a fraction with a denominator of 2:

\[
2 = \frac{4}{2}
\]

So now we have:

\[
y = -\frac{1}{2}x + \frac{1}{2} + \frac{4}{2}
\]

This simplifies to:

\[
y = -\frac{1}{2}x + \frac{5}{2}
\]

Now we need to determine the y-intercept \( b \):

\[
b = \frac{5}{2} = 2.5
\]

Thus, rounding to the nearest tenth, the y-intercept is:

\[
\boxed{2.5}
\]