Question

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

Answer 1

`\displaystyle y = (1)/(2) x - 7`

Explanation:

We want to determine the equation of the line that passes through the point (6, -4) and is perpendicular to the line:

`\displaystyle y = -2x - 3`

Recall that the slopes of perpendicular lines are negative reciprocals of each other.

The slope of the given line is -2.

Hence, the slope of the perpendicular line is 1/2.

Therefore, the slope of our new line is 1/2. We also know that it passes through the point (6, -4). Since we are given the slope and a point, we can consider using point-slope form:

`\displaystyle y - y_1 = m(x - x_1)`

Where m is the slope and (x₁, y₁) is a point.

Let (6, -4) be (x₁, y₁). Substitute:

`\displaystyle y - (-4) = (1)/(2)(x - (6))`

Simplify. Distribute:

`\displaystyle y + 4 = (1)/(2) x - 3`

And subtract 4 from both sides:

`\displaystyle y = (1)/(2) x - 7`

In conclusion, the equation of our line is:

`\displaystyle y = (1)/(2) x - 7`