Here is a nice little problem from the Irish Leaving Certificate maths paper (2016, paper 2, question 1, part a). It is asking us to find the equation of the line for the line between point B and |AC|, where the line is ⊥ to |AC|.

We know the equation of the line to be ** y=mx+c**, but as we don’t know

*c*of the |B

*c*| ⊥ to |AC| we find ourselves a little stuck. We know |B

*c*| is ⊥ to |AC| ∴

*m*pertaining to |B

*c*| is the negative reciprocal (

*m*

^{-1}) of the

*m*pertaining to |AC|. This helps. So to stop us getting confused over which

*m*belongs to which line we will rename them

*m*

_{1}and

*m*

_{2}, where

*m*

_{1}belongs to |AC| and

*m*

_{2}belongs to |B

*c*|. Also note the difference between C and

*c*. C is the point (-3,4) on the |AC| and

*c*is where the |B

*c*| ⊥ to |AC| intersects the

*y*axis.

### Finding the slop of |AC|

This is a simple matter of calculating the differential of the two given points (A and C) on the |AC|, and we know this to be ** m=Δy/Δx**. Thus:

**m _{1}=Δy/Δx**

*m*

_{1 }= (y_{2}-y_{1})/(x_{2}-x_{1})*m*

_{1 }= (4- -2)/(-3 – 6)*∴ m*

_{1 }= -2/3 ∨ -0.666### Finding the negative reciprocal

Now that we have the value of *m*_{1} as -2/3 and we know that *m*_{2} is is the negative reciprocal of *m*_{1} (or the *m*_{1}^{-1}), then all we have to do is the following:

**m _{2}=m_{1}^{-1}∨ -1/m_{1}**

m

_{2}= -1/(-2/3)

∴ m

_{2}= 3/2

### Solving the Equation of the Line

Now we are ready to solve the equation of the line for |B*c*|, remembering that B is (5,3).

**y=mx+c**

*3 = ( ^{3}⁄_{2})(5) + c*

*3 = 7.5 + c*

*3 – 7.5 = c*

*∴ c = -4.5*

In this case the equation of the line (** y=mx+c**) may be expressed

**. And with this our problem has been solved. As always, if you have any comments or problems please post them in the**

*3=(3/2)(5)-4.5**Thoughts and Questions*section below. Thanks for reading.