rpendicular to \( y=2 x+4 \) and passing through \( (0,3) \)

To find the equation of a line that is perpendicular to \( y=2x+4 \) and passes through the point \( (0,3) \), we start by identifying the slope of the original line. The slope of the line \( y=2x+4 \) is \( 2 \). The slope of a line perpendicular to this one will be the negative reciprocal, which is \( -\frac{1}{2} \). Now, we can use the point-slope form of the equation of a line which is given by \( y - y_1 = m(x - x_1) \). Substituting the point \( (0,3) \) and the perpendicular slope \( -\frac{1}{2} \): \[ y - 3 = -\frac{1}{2}(x - 0) \] Simplifying this gives: \[ y - 3 = -\frac{1}{2}x \] \[ y = -\frac{1}{2}x + 3 \] So, the equation of the line that is perpendicular to \( y=2x+4 \) and passes through \( (0,3) \) is: \[ y = -\frac{1}{2}x + 3 \] Now, get ready to graph some slopes! The negative slope means this line descends as it moves to the right, crossing the y-axis at 3. Keep an eye on how it interacts with the original line—such fun geometry!