Equation of the line perpendicular to x - 2y = 1 and passing through (1, 1) is:
y = -2x + 3
The problem asks us to find the equation of a line that satisfies two conditions: it must be perpendicular to the line given by the equation $x - 2y = 1$, and it must pass through the point $(1, 1)$. We can solve this by first determining the slope of the given line, then finding the slope of the perpendicular line, and finally using the point-slope form to find the equation.
The equation of the given line is $x - 2y = 1$. To find its slope, we need to rewrite this equation in the slope-intercept form, which is $y = mx + c$, where '$m$' represents the slope and '$c$' represents the y-intercept.
Rearranging the equation:
From this slope-intercept form, we can see that the slope of the given line ($m_1$) is $\frac{1}{2}$.
Two lines are perpendicular if the product of their slopes is $-1$. Let the slope of the required perpendicular line be $m_2$. Therefore, the relationship between the slopes is:
$$ m_1 \times m_2 = -1 $$
Substitute the slope of the given line ($m_1 = \frac{1}{2}$):
$$ \frac{1}{2} \times m_2 = -1 $$
To find $m_2$, multiply both sides by $2$:
$$ m_2 = -1 \times 2 $$
$$ m_2 = -2 $$
So, the slope of the line perpendicular to the given line is $-2$.
We have the slope of the required line ($m = -2$) and a point it passes through ($(x_1, y_1) = (1, 1)$). We can use the point-slope form of a linear equation, which is $y - y_1 = m(x - x_1)$.
Substitute the values:
$$ y - 1 = -2(x - 1) $$
Now, simplify the equation:
The equation we derived is $y = -2x + 3$. Let's compare this with the given options:
Our calculated equation $y = -2x + 3$ matches Option 4.
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