1 Answer

T D Nguyen · Answer 1 · 2023-07-23T01:28:06+0000

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

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.

$\begin{gathered} m=-(1)/(m_1)_{}_{} \\ m=-(1)/(2) \end{gathered}$

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

$(y-y_1)=m(x-x_1)_{}$

Let us substitute the slope and the given point into the above equation and simplify the equation.

$\begin{gathered} (y-4)=-(1)/(2)(x-6)_{} \\ y-4=-(1)/(2)x+3 \\ y=-(1)/(2)x+3+4 \\ y=-(1)/(2)x+7 \end{gathered}$

Therefore, the equation of the line is

Please log in or register to add a comment.