# A is the point with coordinates (2,8)and B is the point with coordinates (14,4). A straight line L is perpendicular to line AB and passes through A.Find the equation of the straight line L.

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In order to find the equation of a line, we need the slope and a point. We already know a point on the line, it is A(2,8), so we just need to calculate the slope.

We are given the line AB, and it has slope

`m={y_2-y_1}/{x_2-x_1}`

`={4-8}/{14-2}`

`=-4/12`

`=-1/3`

But, the line L is perpendicular to AB, so its slope is the negative reciprocal of the slope of AB. This means that the line L has slope of 3.

Now we can use find the equation of the line using y=mx+b.

`y=mx+b` sub in the point and the slope, solve for b

`8=3(2)+b`

`8-6=b`

`b=2`

**The line is `y=3x+2` . In standard form, the line is `3x-y+2=0` .**

This is my solution:

First find the gradient of AB

A (2,8) B (14,4)

gradient = (4 - 8) / (14 - 2)

= -4 / 12

= -1/ 3

Gradient of perpendicular AB = 3

Since it passes through A

( y - 8)/(x - 2) = 3

y - 8 = 3x - 6

y = 3x + 2