Question

Please help me!! Find the equation of the line that is perpendicular to y=1/6x+3 and contains the point (-3,23).

Answer 1

`\displaystyle y = -6x + 5`

Where ? = -6.

Explanation:

We want to find the equation of a line that is perpendicular to:

`\displaystyle y = (1)/(6) x + 3`

And contains the point (-3, 23).

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

The negative reciprocal of 1/6 is -6. Hence, the slope of the perpendicular line is -6.

We are also given that it passes through the point (-3, 23). Since we know the slope and a point, we can consider using the point-slope form:

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

Substitute:

`\displaystyle y - (23) = -6(x - (-3))`

Simplify:

`\displaystyle y - 23 = -6 (x+3)`

Distribute:

`\displaystyle y - 23 = -6x - 18`

And add. Hence:

`\displaystyle y = -6x + 5`

In conclusion, our equation is:

`\displaystyle y = -6x + 5`

Where ? = -6.