Question

Find an equation of a perpendicular line that contains given point. Write the equation in slope-intercept. Line 2x-3y=8 point (4, -1)

Answer 1

To find a perpendicular line to the line 2x -3y = 8, and that passes through the point (4, -1), we can proceed as follows:

1. We need to find the slope of the perpendicular line using the next property:

$\begin{gathered} m_a\cdot m_b=-1 \\ m_b=-\frac{1}{m_a} \end{gathered}$

That is, the slope of the perpendicular line is the negative reciprocal of the slope of the original line, and the product of both slopes is equal to -1.

2. We need to know the slope of the original line 2x - 3y = 8, and we have to rewrite the equation in slope-intercept form since we can identify its slope easily in this way. Then we have:

The slope-intercept form of a line is given by:

$y=mx+b$

Where

• m is the slope of the line

,

• b is the y-intercept of the line, and at this point, the value of x = 0. It is the point where the line passes through the y-axis.

Then we have:

$\begin{gathered} 2x-3y=8 \\ 2x-2x-3y=8-2x \\ -3y=8-2x \\ \end{gathered}$

We subtracted 2x from both sides of the equation. Now, we have to divide by -3 to both sides of the equation:

$\begin{gathered} \frac{-3y}{-3}=\frac{8-2x}{-3} \\ y=\frac{8}{-3}+\frac{-2x}{-3} \\ y=-\frac{8}{3}+\frac{2}{3}x \\ y=\frac{2}{3}x-\frac{8}{3} \end{gathered}$

3. We can see that the slope of the line is m = 2/3. Now, to find a perpendicular line to this one, we have to apply the property of the slopes we wrote above:

$\begin{gathered} m_b=-\frac{1}{m_a} \\ m_a=\frac{2}{3} \\ m_b=-\frac{1}{\frac{2}{3}} \\ m_b=-\frac{3}{2} \end{gathered}$

Therefore, the slope of the perpendicular line is m = -3/2.

4. Now, we can use this slope, and the point (4, -1), to find the equation of the perpendicular line. To do this, we can use the point-slope form of the line:

$\begin{gathered} y-y_1=m(x-x_1) \\ (x_1,y_1)=(4,-1)\Rightarrow x_1=4,y_1=-1 \end{gathered}$

Then we have:

$\begin{gathered} y-(-1)=-\frac{3}{2}(x-4) \\ y+1=(-\frac{3}{2})(x)+(-\frac{3}{2})(-4)_{} \\ y+1=-\frac{3}{2}x+(-3)(-\frac{4}{2}) \\ y+1=-\frac{3}{2}x+(-3)(-2) \\ y+1=-\frac{3}{2}x+6 \\ y+1-1=-\frac{3}{2}x+6-1 \\ y=-\frac{3}{2}x+5 \end{gathered}$

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