To find the line that is perpendicular to the line given by the equation \( y = 16x + 5 \) and passes through the point \( (1, -3) \), we first need to determine the slope of the original line.
The slope of the line \( y = 16x + 5 \) is \( 16 \). The slope of any line that is perpendicular to this line can be found using the negative reciprocal of the original slope. Therefore, the slope \( m \) of the perpendicular line is:
\[ m = -\frac{1}{16} \]
Next, we can use the point-slope form of the equation of a line, which is:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) \) is the point through which the line passes. Here, \( (x_1, y_1) = (1, -3) \).
Substituting the values into the point-slope form:
\[ y - (-3) = -\frac{1}{16}(x - 1) \]
This simplifies to:
\[ y + 3 = -\frac{1}{16}(x - 1) \]
Distributing the slope:
\[ y + 3 = -\frac{1}{16}x + \frac{1}{16} \]
Now, isolate \( y \):
\[ y = -\frac{1}{16}x + \frac{1}{16} - 3 \]
Converting \( -3 \) to a fraction with a denominator of 16:
\[ -3 = -\frac{48}{16} \]
So now we have:
\[ y = -\frac{1}{16}x + \frac{1}{16} - \frac{48}{16} \] \[ y = -\frac{1}{16}x - \frac{47}{16} \]
None of the provided responses mention a slope of \( -\frac{1}{16} \), but it seems there was a misunderstanding while calculating the perpendicular slope.
Instead, the correct derived slope that best matches what is needed to be perpendicular to \( y = 16x + 5 \) is:
\[ m = -\frac{1}{16} \]
Therefore, the values might seem off, and if another slope choice occurs as the provided options:
- Plugging back into the provided responses again, all measuring \( -6 \), while also needing to check via \( y = mx + c \) choices, then it left without trying any found.
Is there any follow-through or visual provided matching values noted alongside these slopes further?
1 answer