# The curve y=x^2-7x+6 crosses the x-axis at P and Q. The tangents to the curve P and Q meet at R. The normals to the curve at P and Q meet at S. Find the distance RS

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### 1 Answer

The function `y=x^2-7x+6` factors `y=(x-6)(x-1)` into which means that the points P and Q are P(1,0) and Q(6,0). The derivative of the function is `y'=2x-7` so the slope of the tangent at P is 2(1)-7=-5 and the slope of the tangent at Q is 2(6)-7=5.

This means the tangent line at P is `y=-5x+b` and we can find b by solving:

`0=-5+b` to get: `y=-5x+5` .

Similarly, the tangent line at Q is `0=5(6)+b` which gives `y=5x-30` .

These two lines meet at

`5x-30=-5x+5`

`10x=35`

`x=7/2`

which means `y=-5(7/2)+5=-35/2+10/2=-25/2` .

The normal to P has slope `1/5` , which gives:

`0=1/5(1)+b` so `y=1/5x-1/5`

The normal to Q has slope `-1/5` , which gives:

`0=-1/5(6)+b` so `y=-1/5x+6/5`

The intersection at S is found by meeting these lines to get:

`1/5x-1/5=-1/5x+6/5` multiply by 5

`x-1=-x+6`

`2x=7`

`x=7/2`

which means the y-value is `y=1/5(7/2)-1/5=1/5(7/2-2/2)=1/5(5/2)=1/2` .

Since the x-values are the same between R and S, then the distance between them is

`1/2-(-25/2)=26/2=13`

**The distance RS is 13 units.**