Write an equation of the line passing through point $P\left(4,\ 0\right)$ that is parallel to the line $-x+2y\ =\ 12$
Why did the parallel line refuse to pass through the point $P\left(4,\ 0\right)$? Because it found out that $P$ was in a parallel universe! 🌌
But don't worry, finding the equation of a line parallel to another one is actually quite simple. We can start by rewriting the given equation $-x+2y=12$ in slope-intercept form:
$2y = x + 12$
$y = \frac{1}{2}x + 6$
Since our new line is parallel, it will have the same slope as the given line, which is $\frac{1}{2}$. Now we can use the point-slope form $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the given point $(4, 0)$, and $m$ is the slope:
$y - 0 = \frac{1}{2}(x - 4)$
Simplifying, we find:
$y = \frac{1}{2}x - 2$
So the equation of the line parallel to $-x+2y=12$ and passing through $P(4, 0)$ is $y = \frac{1}{2}x - 2$.
To find the equation of a line parallel to the line $-x+2y=12$, we need to determine the slope of the line $-x+2y=12$.
First, we can rewrite the equation in slope-intercept form, $y=mx+b$, by solving for $y$.
Rearranging the equation $-x+2y=12$, we have:
$2y=x+12$
Dividing both sides of the equation by 2, we get:
$y=\frac{1}{2}x+6$
Comparing this equation with the general form $y=mx+b$, we can see that the slope $m$ of the line $-x+2y=12$ is $\frac{1}{2}$.
Since the line we want to find is parallel to this line, it will also have a slope of $\frac{1}{2}$.
Now, we can use the point-slope form of a linear equation to find the equation of the line passing through point $P(4, 0)$:
$y-y_1=m(x-x_1)$
Substituting the slope, $m=\frac{1}{2}$, and the coordinates of point $P$, $x_1=4$ and $y_1=0$, we get:
$y-0=\frac{1}{2}(x-4)$
Simplifying the equation, we have:
$y=\frac{1}{2}(x-4)$
Therefore, the equation of the line passing through point $P(4, 0)$ that is parallel to the line $-x+2y=12$ is $y=\frac{1}{2}(x-4)$.
To find the equation of a line that is parallel to another line, we can use the fact that parallel lines have the same slope.
The given equation of the line is $-x+2y=12$. To find its slope, let's rewrite this equation in slope-intercept form $y=mx+b$, where $m$ is the slope and $b$ is the y-intercept.
Start by isolating $y$ in the equation $-x+2y=12$:
\begin{align*}
-1x+2y &= 12 \\
2y &= 1x + 12 \\
y &= \frac{1}{2}x + 6
\end{align*}
Now we can see that the slope of the given line is $\frac{1}{2}$. Since we want a line that is parallel, the slope of our new line will also be $\frac{1}{2}$.
We know that the new line passes through point $P(4,0)$. To find its equation, we can use the point-slope form of a linear equation: $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the point on the line and $m$ is the slope.
Substituting $x_1 = 4$, $y_1 = 0$, and $m = \frac{1}{2}$ into the point-slope form, we get:
\begin{align*}
y - 0 &= \frac{1}{2}(x - 4) \\
y &= \frac{1}{2}x - 2
\end{align*}
Thus, the equation of the line passing through point $P(4,0)$ that is parallel to the line $-x+2y=12$ is $y=\frac{1}{2}x-2$.
Good grief!!
Just type it like this:
Write an equation of the line passing through point P(4,0) that is parallel to the line -x + 2y = 12
The standard way would be to have the leading term as positive, so your
given equation is
x - 2y = -12
Your new equation will differ only in the constant, so
x - 2y = c
given: (4,0) lies on it, so
4 - 0 = c
c = 4
x - 2y = 4 is your new equation.