Question

Write an equation of the line that contains the point (3.-2) and is perpendicular to the line 3x - 6y = 7.

Answer 1

y = -2x + 4

Explanation:

Perpendicular lines intersect at a point that form right angles; these lines also have negative reciprocal slopes, in which the product of their slopes result in a negative one (-1 ).

Given the linear equation in standard form, 3x - 6y = 7, and the point (3, -2):

Transform the standard form into slope-intercept form: y = mx + b

3x - 6y = 7

3x - 3x - 6y = -3x + 7

-6y = -3x + 7

Divide both sides by -6 to isolate y:

`(-6y)/(-6) = (-3x + 7)/(-6)`

y = ½x - 7/6 (This is the slope-intercept form of the given equation, 3x - 6y = 7).

Now that we have the slope of the given equation,
`m_1` = ½, then we can determine that the slope of the other line must be
`m_2` = -2:

`m_1` ×
`m_2` = ½ × -2 = -1

Next, using the slope of the other line,
`m_2` = -2, and the given point, (3, -2), substitute these values into the slope-intercept form to solve for the value of the y-intercept, b:

y = mx + b

-2 = -2(3) + b

-2 = -6 + b

Add 6 to both sides to isolate b:

-2 + 6 = -6 + 6 + b

4 = b (This is the y-intercept of the other line).

Therefore, the linear equation of the line that is perpendicular to 3x - 6y = 7 is:

y = -2x + 4