Question

Line perpendicular to g(x)=2x-8 and passing through the point (6,3)

Answer 1

The line perpendicular to
`\(g(x) = 2x - 8\)`and passing through the point
`\((6,3)\) is \(y = -(1)/(2)x + 6\)`.

Step-by-step explanation:

To find the line perpendicular to
`\(g(x)\)`, we need to determine the negative reciprocal of the slope of
`\(g(x)\)`. The given function
`\(g(x) = 2x - 8\)` has a slope of 2. The negative reciprocal of
`2 is \(-(1)/(2)\)`. Therefore, the slope of the perpendicular line is
`\(-(1)/(2)\)`.

Next, we use the point-slope form of a linear equation,
`\(y - y₁ = m(x - x₁)\)` where
`\((x₁, y₁)\)` is the given point. Plugging in \((6,3)\)
`\((x₁, y₁)\)` and
`\(-(1)/(2)\) for \(m\), we get \(y - 3 = -(1)/(2)(x - 6)\).`

Now, we simplify the equation to slope-intercept form
`\(y = mx + b\), where \(m\)` is the slope and \(b\) is the y-intercept. Distributing the
`\(-(1)/(2)\) and rearranging terms, we get \(y = -(1)/(2)x + 6\).`

In conclusion, the line perpendicular to
`\(g(x) = 2x - 8\)` and passing through the point
`\((6,3)\)` is represented by the equation
`\(y = -(1)/(2)x + 6\).` This line has a slope of
`\(-(1)/(2)\)`, ensuring its perpendicularity to the original line, and it passes through the specified point
`\((6,3)\)`.