Question

find the equation of the straight line passing through P -2,1and parallel to the line with equation 2x - 3y=1​

Answer 1

Answer:

$y=\frac{2}{3} x+\frac{7}{3}$

Step-by-step explanation:

Parallel lines have the same slope. Therefore, the line we're trying to find the equation for and the given line 2x-3y=1​ must have the same slope.

1) Find the slope of 2x-3y=1​

To do this, rewrite this equation in slope-intercept form: $y=mx+b$ where $m$ is the slope and $b$ is the y-intercept (the value of y when the line crosses the y-axis)

$2x-3y=1$

Subtract 2x from both sides

$2x-3y-2x=1-2x\\-3y=-2x+1$

Divide both sides by -3 to isolate y

$\frac{-3y}{-3}=\frac{-2}{-3}x+(\frac{1}{-3} )\\y=\frac{2}{3}x-\frac{1}{3}$

Now, we can easily identify that $\frac{2}{3}$ is in the position of m, the slope. Plug this into $y=mx+b$:

$y=\frac{2}{3} x+b$

2) Find the y-intercept

$y=\frac{2}{3} x+b$

To find the y-intercept, plug the given point P(-2,1) into the equation and solve for b

$1=\frac{2}{3} (-2)+b\\1=\frac{-4}{3}+b$

Add $\frac{4}{3}$ to both sides of the equation

$1+\frac{4}{3}= \frac{-4}{3} +b+\frac{4}{3} \\\frac{3}{3} +\frac{4}{3}= b\\\frac{7}{3}$

Therefore, the y-intercept of the line is $\frac{7}{3}$. Plug this into our original equation:

$y=\frac{2}{3} x+b\\y=\frac{2}{3} x+\frac{7}{3}$

I hope this helps!