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=-(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} (-3y)/(-3)=(8-2x)/(-3) \\ y=(8)/(-3)+(-2x)/(-3) \\ y=-(8)/(3)+(2)/(3)x \\ y=(2)/(3)x-(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=-(1)/(m_a) \\ m_a=(2)/(3) \\ m_b=-(1)/((2)/(3)) \\ m_b=-(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)=-(3)/(2)(x-4) \\ y+1=(-(3)/(2))(x)+(-(3)/(2))(-4)_{} \\ y+1=-(3)/(2)x+(-3)(-(4)/(2)) \\ y+1=-(3)/(2)x+(-3)(-2) \\ y+1=-(3)/(2)x+6 \\ y+1-1=-(3)/(2)x+6-1 \\ y=-(3)/(2)x+5 \end{gathered}`

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