Question

What is the equation of a line that is perpendicular to the line y = 2x + 4 and passes through the point (-2, 3)?

Answer 1

Answer:

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

Step-by-step explanation:

The given equation is written in slope-intercept form:

$y=mx+b$

where:

• m is the slope
• b is the y-intercept
• x and y are the corresponding coordinate points (x,y)

The slope is 2:

When two lines are perpendicular, their slopes will be opposite reciprocals. Find the slope of the line perpendicular to the given equation:

$m_{1}=2$

This can be seen as:

$m_{1}=\frac{2}{1}$

Flip the fraction (reciprocal):

$\frac{2}{1}$ → $\frac{1}{2}$

Now find the opposite:

$\frac{1}{2}$ → $-\frac{1}{2}$

So the slope of the perpendicular line is:

$m_{2}=-\frac{1}{2}$

Now make an equation written in point-slope form:

$y-y_{1}=m(x-x_{1})$

where:

• m is the slope
• $x_{1}$ and $y_{1}$ are the corresponding coordinate points $(x_{1},y_{1})$

Insert information:

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

Simplify to slope-intercept form. Solve for y:

Use the distributive property:

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

Add 3 to both sides to isolate the variable:

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

:Done