Question

Find an equation for the line perpendicular to the line 3x+2y=−1 3x+2y=−1 having the same y-intercept as 2x+7y=2

Answer 1

The general form for the equation of a line is:

`y = mx + c`

Where:

m is the gradient of the line

c is the y intercept of the line (y - intercept is where the graph crosses the y-axis)

So if you had the following equation:

`y = 3x + 2`

Then:

m = 3

c = 2

So gradient = 3, and y intercept = 2

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Rearranging

So first rearrange both of the equations in the form y = mx + c :

`3x + 2y = -1` becomes
`y = -(3)/(2) x-(1)/(2)`
`(where: \ m = -(3)/(2) \ and, \ c = -(1)/(2) )`

and:

`2x+7y=2` becomes
`y=-(2)/(7)x +(2)/(7)`
`(where: \ m = -(2)/(7) \ and, \ c = (2)/(7))`

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The question tells us that the equation of the line we are looking for has the same y-intercept as:

`2x+7y=2`

So the line we are trying to work out will also have a y intercept of
`(2)/(7)`

(refer to rearranging)

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The question also tells us that the line is perpendicular to
`3x + 2y = -1`

Perpendicular gradient = negative reciprocal of the gradient of the line it is perpendicular to.

So the gradient of the new line will be the negative reciprocal of the gradient of
`3x + 2y = -1`

Gradient of
`3x + 2y = -1` is:
`-(3)/(2)`

(refer to rearranging)

Gradient of new line: = negative reciprocal of
`-(3)/(2)` , which is
`(2)/(3)`

(just flip fraction and change the sign)

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So for the new line:
`m = (2)/(3) \ and, \ c = (2)/(7)`

So just substitute in the values for m and c into: y = mx + c

`y = mx + c\\y = (2)/(3) x + (2)/(7)`

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Answer:

So equation of the new line is:

`y = (2)/(3)x + (2)/(7)`

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Any questions, just ask.