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fiasKO [112]
3 years ago
12

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

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
5 0

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.

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