Question

Write the equation of the line that passes through (–2, 1) and is perpendicular to the line 3x – 2y = 5.

Answer 1

So first you will want to put the equation into slope-intercept form (y=mx+b). 

-3x+2y=-6  Add 3x to both sides.

2y=3x-6   Then divide by 2.

y=3/2x-6  Now it is in slope-intercept form.

 

The slope for this particular equation is 3/2 (m).

The slope of a line that is perpendicular to this would be -2/3, the negative reciprocal of 3/2.

 

Next you need to plug (3,3) and the slope into the slope-intercept form to find the y-intercept (b).

3=-2/3(3)+b

3=-2+b  Add 2 to both sides.

5=b  Now you know your y-intercept.

 

Now you have your final equation: y=-2/3x+5  :)

Answer 2

Answer:

$y-+2x+1=0$

Step-by-step explanation:

Given:

Line =  3x – 2y = 5

Point (-2,1)

To find:

Equation of the line

Solution:

If x = k, where k is a constant, then write the general form of a straight line that is parallel to the y axis. The line's separation from the y axis represents the absolute value of the constant k. To obtain the equation of the vertical line, substitute the x coordinate in the presumptive equation.

The equation of a line must be written in point slope form using the knowledge that the sum of the slopes of any two perpendicular lines is -1. Without using these facts, we won't be able to write the equation for a line. Any line equation depicts the relationship between points along the line. Any general point lying on the line and the slope of the line are related by the point slope form of the line.

The given line $3x-2y=5$

for the form,

ax+by=c

Thus here slope = $\frac{-a}{b} =\frac{-3}{-2}$

Slope of the line parallel to and extending through,

(-2,1) is $\frac{-2}{1}$

The line's equation through (-2, 1) is

$(y-3)=\frac{-2}{1}(x+2)\\ y-3=-2x-4\\y-+2x+1=0$

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