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?

1 answer

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:

\[ \boxed{2.5} \]