Respuesta :
We are asked to find the equation of a line that is perpendicular to the line y = 2x - 3 and passes through the point (6, 4)
Recall that the equation of a line in slope-intercept form is given by
[tex]y=mx+b[/tex]Where m is the slope and b is the y-intercept.
Comparing the general form with the given equation we see that the slope is 2
Since we are given that the lines are perpendicular so the slope of the other line must be negative reciprocal of the given line.
[tex]\begin{gathered} m=-\frac{1}{m_1}_{}_{} \\ m=-\frac{1}{2} \end{gathered}[/tex]So, the slope of the required equation is m = -1/2
Since we are also given that the line passes through the point (6, 4)
The point-slope form of the equation of a line is given by
[tex](y-y_1)=m(x-x_1)_{}[/tex]Let us substitute the slope and the given point into the above equation and simplify the equation.
[tex]\begin{gathered} (y-4)=-\frac{1}{2}(x-6)_{} \\ y-4=-\frac{1}{2}x+3 \\ y=-\frac{1}{2}x+3+4 \\ y=-\frac{1}{2}x+7 \end{gathered}[/tex]Therefore, the equation of the line is
[tex]y=-\frac{1}{2}x+7[/tex]