Question

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

Answer 1

Answer:

2x + 3y = 5

Step-by-step explanation:

$\bold{METHOD\ 1:}$

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept

Let $k:y=m_1x+b_1,\ l:y=m_2x+b_2$

then

$l\ \parallel\ k\iff m_1=m_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}$

We have the equation of a line:

$2x+3y=6$

Convert to the slope-intercept form:

$2x+3y=6$           subtract 2x from both sides

$3y=-2x+6$              divide both sides by 3

$y=-\dfrac{2}{3}x+2\to m_1=-\dfrac{2}{3}$

therefore the slope is $m_2=-\dfrac{2}{3}$

Put the value of the slope and the coordinates of the point (-2, 3) to the equation of a line:

$3=-\dfrac{2}{3}(-2)+b$

$3=\dfrac{4}{3}+b$            subtract 4/3 from both sides

$\dfrac{9}{3}-\dfrac{4}{3}=b\to b=\dfrac{5}{3}$

Finally:

$y=-\dfrac{2}{3}x+\dfrac{5}{3}$

Convert to the standard form (Ax + By = C):

$y=-\dfrac{2}{3}x+\dfrac{5}{3}$              multiply both sides by 3

$3y=-2x+5$           add 2x to both sides

$2x+3y=5$

$\bold{METHOD\ 2:}$

Let $k:A_1x+B_1y=C_1,\ l:A_2x+B_2y=C_2$.

Lines k and l are parallel iff

$A_1=A_2\ \wedge\ B_1=B_2\to\dfrac{A_2}{A_1}=\dfrac{B_2}{B_1}$

We have the equation:

$2x+3y=6\to A_1=2,\ B_1=3$

then the equation of a line parallel to given lines has the equation:

$2x+3y=C$

Put the coordinates of the point (-2, 3) to the equation:

$C=2(-2)+3(3)\\\\C=-4+9\\\\C=5$

Finally:

$2x+3y=5$