To find the equation of a line passing through the point [tex]\((2,5)\)[/tex] and perpendicular to the line [tex]\(4x - y = 2\)[/tex], we will follow these steps:
### Step 1: Determine the slope of the given line
The given line is [tex]\(4x - y = 2\)[/tex]. We can rewrite this equation in slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] represents the slope:
[tex]\[ 4x - y = 2 \][/tex]
[tex]\[ - y = -4x + 2 \][/tex]
[tex]\[ y = 4x - 2 \][/tex]
Thus, the slope of the given line [tex]\(4x - y = 2\)[/tex] is [tex]\(4\)[/tex].
### Step 2: Find the slope of the perpendicular line
The slope of a line perpendicular to another line is the negative reciprocal of the original line's slope. Since the slope of the given line is [tex]\(4\)[/tex], the negative reciprocal is:
[tex]\[ m_{\text{perpendicular}} = -\frac{1}{4} \][/tex]
### Step 3: Use the point-slope form to write the equation of the perpendicular line
The perpendicular line passes through the point [tex]\((2, 5)\)[/tex] and has a slope of [tex]\(-\frac{1}{4}\)[/tex]. Using the point-slope form [tex]\(y - y_1 = m(x - x_1)\)[/tex]:
[tex]\[ y - 5 = -\frac{1}{4}(x - 2) \][/tex]
### Step 4: Simplify the equation
Distribute [tex]\(-\frac{1}{4}\)[/tex] on the right side:
[tex]\[ y - 5 = -\frac{1}{4}x + \frac{1}{2} \][/tex]
Add 5 to both sides to isolate [tex]\(y\)[/tex]:
[tex]\[ y = -\frac{1}{4}x + \frac{1}{2} + 5 \][/tex]
[tex]\[ y = -\frac{1}{4}x + \frac{1}{2} + \frac{10}{2} \][/tex]
[tex]\[ y = -\frac{1}{4}x + \frac{11}{2} \][/tex]
### Step 5: Convert to standard form
To convert this equation to standard form [tex]\(Ax + By = C\)[/tex], we'll eliminate the fraction by multiplying every term by 4:
[tex]\[ 4y = -x + 22 \][/tex]
Rearrange to get all terms on one side:
[tex]\[ x + 4y = 22 \][/tex]
Thus, the standard form of the equation of the line passing through [tex]\((2,5)\)[/tex] and perpendicular to [tex]\(4x - y = 2\)[/tex] is:
[tex]\[ \boxed{x + 4y = 22} \][/tex]
