Question

Find an equation of the line perpendicular to the line $3x+6y=5$ and passing through the point $(1,3)$. Write the equation in the standard form.

Answer 1

We must must transform the standard form equation $3x+6y=5$ into a slope-intercept form equation ($y=mx+b$) to find its slope.

$3x+6y=5$ (Subtract $3x$ on both sides.)

$6y=-3x+5$ (Divide both sides by $6$.)

$y=-\frac{3}{6}x+\frac{5}{6}$

$y=-\frac{1}{2}x+\frac{5}{6}$

The slope of our first line is equal to $-\frac{1}{2}$ . Perpendicular lines have negative reciprocal slopes, so if the slope of one is $x$, the slope of the other is $-\frac{1}{x}$ .

The negative reciprocal of $-\frac{1}{2}$ is equal to $2$, therefore $2$ is the slope of our line.

Since the equation of line passing through the point $(1,3)$, therefore substitute the given point in the equation $y=2x+b$:

$3=(2\times 1)+b$

$3=2+b$

$b=3-2=1$

Substitute this value for $b$ in the equation $y=2x+b$:

$y=2x+1$

Hence, the equation of the line is $y=2x+1$.