Which of the following equations is parallel to y = (2/3)x + 6 and contains the point (4, -2)?

Question

Avery Alexander · Accepted Answer

#y=2/3x-14/3#

As we are aware,

#(1)# If slop line #l_1# is #m_1# and slop of # l_2# is #m_2#, then

#l_1////l_2<=>m_1=m_2#

Here,

#l_1 :y=(2/3)x+6,and l_1////l_2#

Comparing with #y=mx+c#

#=>#Slop of the line #l_1# is #m_1=2/3#

#=>#Slop of the line #l_2# is #m_2=2/3...to[as,m_1=m_2]#

Currently, the line's "point-slop" form is:

#y-y_1=m(x-x_1)#

For line #l_2#,#m=2/3and #point #(4,-2)#

Thus, the line's equation is:

#y-(-2)=2/3(x-4)#

#=>3(y+2)=2(x-4)#

#=>3y+6=2x-8#

#=>3y=2x-14#

#=>y=2/3x-14/3#

There isn't a single equation to contrast.

Jameson Allen · Answer

undefined An equation that is parallel to $ y = \frac{2}{3}x + 6 $ has the same slope. Therefore, the equation we seek will also have a slope of $ \frac{2}{3} $. To find the equation that passes through the point $ (4, -2) $ with this slope, we can use the point-slope form of a linear equation:

$$ y - y_1 = m(x - x_1) $$

Substituting $ m = \frac{2}{3} $ and $ (x_1, y_1) = (4, -2) $:

$$ y - (-2) = \frac{2}{3}(x - 4) $$

Simplify:

$$ y + 2 = \frac{2}{3}(x - 4) $$

$$ y + 2 = \frac{2}{3}x - \frac{8}{3} $$

$$ y = \frac{2}{3}x - \frac{8}{3} - 2 $$

$$ y = \frac{2}{3}x - \frac{8}{3} - \frac{6}{3} $$

$$ y = \frac{2}{3}x - \frac{14}{3} $$

So, the equation parallel to $ y = \frac{2}{3}x + 6 $ and passing through the point $ (4, -2) $ is $ y = \frac{2}{3}x - \frac{14}{3} $.