To determine which equation represents the line that passes through the points [tex]\((-8, 11)\)[/tex] and [tex]\(\left(4, \frac{7}{2}\right)\)[/tex], let's follow these steps:
1. Calculate the slope (m) of the line:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
Here, [tex]\( (x_1, y_1) = (-8, 11) \)[/tex] and [tex]\( (x_2, y_2) = \left(4, \frac{7}{2}\right) \)[/tex].
Plugging in the values, we get:
[tex]\[
m = \frac{\frac{7}{2} - 11}{4 - (-8)}
\][/tex]
2. Simplify the numerator and the denominator:
- The numerator is:
[tex]\[
\frac{7}{2} - 11 = \frac{7}{2} - \frac{22}{2} = \frac{7 - 22}{2} = \frac{-15}{2}
\][/tex]
- The denominator is:
[tex]\[
4 - (-8) = 4 + 8 = 12
\][/tex]
So the slope [tex]\(m\)[/tex] is:
[tex]\[
m = \frac{\frac{-15}{2}}{12} = \frac{-15}{2} \times \frac{1}{12} = \frac{-15}{24} = \frac{-5}{8}
\][/tex]
Hence, the slope of the line is:
[tex]\[
m = -\frac{5}{8}
\][/tex]
3. Calculate the y-intercept (b) using the slope-intercept form [tex]\( y = mx + b \)[/tex]:
Take one of the given points [tex]\((-8, 11)\)[/tex]. Substitute [tex]\(y = 11\)[/tex], [tex]\(x = -8\)[/tex], and [tex]\(m = -\frac{5}{8} \)[/tex] into the equation [tex]\( y = mx + b \)[/tex]:
[tex]\[
11 = -\frac{5}{8}(-8) + b
\][/tex]
4. Solve for [tex]\( b \)[/tex]:
[tex]\[
11 = \frac{5 \times 8}{8} + b = 5 + b
\][/tex]
[tex]\[
11 - 5 = b
\][/tex]
[tex]\[
b = 6
\][/tex]
Therefore, the equation of the line is:
[tex]\[
y = -\frac{5}{8}x + 6
\][/tex]
5. Compare with the given equations:
\begin{align}
(i) & \quad y = -\frac{5}{8}x + 6 \\
(ii) & \quad y = -\frac{5}{8}x + 16 \\
(iii) & \quad y = -\frac{15}{2}x - 49 \\
(iv) & \quad y = -\frac{15}{2}x + 71 \\
\end{align}
The correct equation that matches our derived equation is:
[tex]\[
y = -\frac{5}{8}x + 6
\][/tex]
Thus, the equation that represents the line passing through the points [tex]\((-8, 11)\)[/tex] and [tex]\(\left(4, \frac{7}{2}\right)\)[/tex] is:
[tex]\[
\boxed{1}
\][/tex]
