Question

Write an equation of the line containing the point (6,-4) and perpendicular to.the line y=-2x-3

Answer 1

Answer:

$y=\frac{1}{2}x-7$

Step-by-step explanation:

So we need to find an equation of a line that crosses the point (6,-4) and is perpendicular to y = -2x -3.

First, let's find the slope of the line we want to write. The line we want is perpendicular to y = -2x -3. Recall that if two lines are perpendicular to each other, their slopes are negative reciprocals of each other. What this means is that:

$m_1\cdot m_2=-1$

Plug -2 for one of the slopes.

$-2\cdot m_2=-1$

Divide by -2 to find the slope of our line.

$m_2=1/2$

Thus, our line needs to have a slope of 1/2.

Now, let's use the point-slope form. The point-slope form is given by:

$y-y_1=m(x-x_1)$

Plug in 1/2 for the slope m and let's let our point (6,-4) be x₁ and y₁. Thus:

$y-(-4)=\frac{1}{2}(x-6)$

Simplify and distribute:

$y+4=\frac{1}{2} x-3$

Subtract 4 from both sides:

$y=\frac{1}{2}x-7$

The above is the equation that passes the point (6,-4) and is perpendicular to y = -2x -3.