What is the equation of the line which is parallel to the line 3x +4y =6 and passes through (2, 1)?

Question

Show work and explain please.

Hazel Anglin · Accepted Answer

#y=-3/4x+5/2#

. #3x+4y=6#

Let's solve for #y# so we can have the equation in standard slope-intercept form:

#4y=-3x+6#

#y=-3/4x+3/2#

This is in the form of:

#y=mx+b# where #m# is slope and #b# is the #y#-intercept which is where the line crosses the #y#-axis. Comparing the two we see that:

#m=-3/4# and #b=3/2#

For a line to be parallel to this line, it would have to have the same slope, i.e. its equation would be:

#y=-3/4x+b#

Now, we can use the coordinates of the point the line goes through and plug them into this equation to solve for #b#:

#1=-3/4(2)+b#

#1=-3/2+b#

#b=1+3/2=2/2+3/2=5/2#

Therefore, the equation of the line is:

#y=-3/4x+5/2#

Savannah Anderson · Answer

undefined To find the equation of a line parallel to the line $3x + 4y = 6$, we first need to determine the slope of the given line. The slope-intercept form of a line is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

Rearranging the given line equation into slope-intercept form:
$$4y = -3x + 6$$
$$y = -\frac{3}{4}x + \frac{6}{4}$$
$$y = -\frac{3}{4}x + \frac{3}{2}$$

The slope of the given line is $-\frac{3}{4}$. Since the line we want to find is parallel to this line, it will have the same slope.

Now, we use the point-slope form of a line $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the given point and $m$ is the slope.

Given point: $(2, 1)$
Slope: $-\frac{3}{4}$

Substitute the values into the point-slope form:
$$y - 1 = -\frac{3}{4}(x - 2)$$

Now, we can simplify and rearrange this equation to slope-intercept form:
$$y - 1 = -\frac{3}{4}x + \frac{3}{2}$$
$$y = -\frac{3}{4}x + \frac{3}{2} + 1$$
$$y = -\frac{3}{4}x + \frac{3}{2} + \frac{2}{2}$$
$$y = -\frac{3}{4}x + \frac{5}{2}$$

So, the equation of the line parallel to $3x + 4y = 6$ and passing through $(2, 1)$ is $y = -\frac{3}{4}x + \frac{5}{2}$.