# How do you find a standard form equation for the line perpendicular to 3x-2y+5=0 and with the same y-intercept as 3x-y+18=0?

Jul 23, 2016

$2 x + 3 y - 54 = 0$

#### Explanation:

We can find slope and $y$-intercept of a line by converting its equation to point slope form. In a point-slope form of equation $y = m x + c$, $m$ (the coefficient of $x$ variable) is the slope and $c$ is the $y$-intercept of the line.

Now, $3 x - y + 18 = 0$ can be written as $y = 3 x + 18$ and hence its intercept on $y$-axis is $18$.

Further $3 x - 2 y + 5 = 0$ can be written as $2 y = 3 x + 5$ or

$y = \frac{3}{2} x + \frac{5}{2}$ and hence its slope is $\frac{3}{2}$.

Now as product of slopes of two perpendicular lines is $- 1$, slope of line perpendicular to $3 x - 2 y + 5 = 0$ (whose slope is $\frac{3}{2}$) will be $\frac{- 1}{\frac{3}{2}} = - 1 \times \frac{2}{3} = - \frac{2}{3}$.

We now need to find the equation of a line whose slope is $- \frac{2}{3}$ and intercept on $y$-axis is $18$ and using slope intercept form of equation $y = m x + c$, this should be

$y = - \frac{2}{3} x + 18$ or $3 y = - 2 x + 54$ or $2 x + 3 y - 54 = 0$

graph{(3x-y+18)(3x-2y+5)(2x+3y-54)=0 [-32.5, 47.5, -19.36, 20.64]}