Question

The equation for line k can be written as y - 8 = -8/3 (x + 6). Line L includes the point (-8, -5) and is perpendicular to line k. What is the equation of line L?

Answer 1

To find the equation of line L, take the negative reciprocal of the slope from line k, which is 3/8, and use the point-slope form with the given point (-8, -5). The final equation of line L is y = 3/8 x + 2.

Step-by-step explanation:

The equation of line k is given as y - 8 = -8/3 (x + 6). To find the equation of line L, which is perpendicular to line k and passes through point (-8, -5), we must first identify the slope of line k. The slope of line k is -8/3, so the slope of line L, which is perpendicular to line k, will be the negative reciprocal, which is 3/8.

Now that we have the slope of line L, we can use the point-slope form of a line to find its equation. Since line L passes through the point (-8, -5), we can express its equation as y - (-5) = 3/8 (x - (-8)).

Simplifying, we get y + 5 = 3/8 (x + 8). To express this in slope-intercept form, we distribute the 3/8 and move the 5 to the other side of the equation to get the final equation of line L: y = 3/8 x + 2.

Answer 2

So, the equation of line
`\(L\) is \(8y - 3x = -16\)`.

The given equation for line
`\(k\)` is in the point-slope form
`\(y - y_1 = m(x - x_1)\)`, where
`\((x_1, y_1)\)` is a point on the line, and \(m\) is the slope.

For line k:

`\[ y - 8 = -(8)/(3)(x + 6) \]`

Now, let's find the slope
`(\(m_k\))` of line
`\(k\)` by comparing it with the standard point-slope form
`\(y - y_1 = m(x - x_1)\)`:

`\[ m_k = -(8)/(3) \]`

Since line
`\(L\)` is perpendicular to line k, the slope
`(\(m_L\))` of line
`\(L\)` is the negative reciprocal of the slope of line
`\(k\)`. The negative reciprocal of
`\(-(8)/(3)\) is \((3)/(8)\)`.

Now, we know the slope
`(\(m_L = (3)/(8)\))` and a point on line
`\(L\) \((-8, -5)\)`. We can use the point-slope form to write the equation of line \(L\):

`\[ y - y_1 = m_L(x - x_1) \]`

`\[ y - (-5) = (3)/(8)(x - (-8)) \]`

Simplify the equation:

`\[ y + 5 = (3)/(8)(x + 8) \]`

Multiply through by 8 to get rid of the fraction:

`\[ 8y + 40 = 3(x + 8) \]`

`\[ 8y + 40 = 3x + 24 \]`

Subtract 3x and 24 from both sides:

`\[ 8y - 3x = -16 \]`