Equation of a line (point/parallel line)

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Find the equation of the line passsing through the point P $\left(0,\frac{1}{2}\right)$ and perpendicular to the line $8x+3y-5= 0$

**Answer**

The slope of the line $8x+3y-5= 0$ is given by $m = -\frac{8}{3}$

The second line is perpendicular to this line.

So, the product of their slopes must be $-1$. Thus, the slope of the second line must me $m=\frac{3}{8}$

The second line passes through the point $\left(0,\frac{1}{2}\right)$ and has slope $m=\frac{3}{8}$

The equation of the line with slope $m$ and passing through the point $\left(x_0,y_0\right)$ is given by $$y-y_0=m\left(x-x_0\right)$$

Here $m=\frac{3}{8}$, $x_0=0$ and $y_0=\frac{1}{2}$

Substituting these values in the equation, we get

$ y-\left(\frac{1}{2}\right)=\frac{3}{8}\left(x-\left(0\right)\right)$

or, $ y-\frac{1}{2}=\frac{3}{8}\left(x\right)$

or, $y-\frac{1}{2}=\frac{3}{8}x$

or, $y=\frac{3}{8}x+\frac{1}{2}$