Depth · one concept from first principles · step 2 of 4

Pivot a line past a triangle's extreme vertices

Every line y=mx+1 passes through (0, 1). For which slopes m does such a line miss the triangle with corners A(1, 2), B(3, 4), C(6, -1) entirely?

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How to solve

Strategy Draw a Diagram — Draw the triangle and the pivot point (0, 1). As the line turns about (0, 1), it first touches the triangle at one extreme vertex and leaves it at another. So only the slopes from (0, 1) to the three vertices matter: the line hits the triangle exactly when its slope lies between the smallest and largest of them.

1STEP 1

Find the pivot

Whatever m is, x=0 gives y=1, so the line always passes through (0, 1).

2STEP 2

Slopes to the vertices

Compute the slope from (0, 1) to each vertex.

A: (2-1)/(1-0)=1, B: (4-1)/(3-0)=1, C: (-1-1)/(6-0)=-1/3

3STEP 3

Read off the hitting range

Every point of the triangle has x ＞ 0, so slopes from (0, 1) run from -1/3 (C) to 1 (AB): it hits when -1/3 ≤ m ≤ 1.

Find when the line DOES hit, then take the rest.

8.F.A.3Change Focus Count The Complement

The line meets the triangle exactly when $-\frac{1}{3}\le m\le1$ .

Why?

At $x=0$ the term $m\times0$ is $0$ and adding $0$ to $1$ leaves $1$ , so every such line passes through $(0,1)$ and changing $m$ only turns it about that point.

🧱Additive identity (zero)Adding zero adds nothing, so the number stays exactly the same.

Why?

The line passes through a point exactly when that point's coordinates make $y=mx+1$ true, so its slope must equal the slope from $(0,1)$ to that point.

🧱A solution makes the equation trueA solution is a number that, put in for the unknown, makes both sides come out equal; that is all being a solution means.

Why?

The triangle lies to the right of the $y$ -axis, so the line turning about $(0,1)$ never passes vertical while it sweeps across it, and the lines that hit it are exactly those with slopes between the slope to $C$ and the slope along $AB$ , from $-\frac{1}{3}$ to $1$ .

Why?

For a point to the right of $(0,1)$ the run is positive, so multiplying by it keeps the order, and the point lies above the line of slope $m$ exactly when its own slope from $(0,1)$ is bigger than $m$ .

🧱Scaling an inequalityMultiplying both sides by the same positive number stretches or shrinks the gaps on the number line but keeps the order; multiplying by the same negative number also flips the line around zero, so the inequality turns over.

Why?

The corners have slopes $1$ , $1$ and $-\frac{1}{3}$ from $(0,1)$ , so for $-\frac{1}{3}\le m\le1$ the corner $A$ is on or above the line and $C$ is on or below it, while for $m$ outside that range all three corners sit strictly on the same side.

🧱Transitivity of orderIf the first is smaller than the second and the second smaller than the third, the first is smaller than the third, so pairwise comparisons chain into one full order.

Why?

Along each side the gap between its height and the line's height changes by the same amount at every step, so side $AC$ reaches the line in the first case, and in the second no side does and the line cannot get inside the triangle without crossing a side.

🧱A constant step reaches any termWhen a list climbs by the same amount every time, any term is the starting value plus that step repeated, so you never have to write out the ones between.

4STEP 4

Take the complement

The line misses the triangle for every other slope.

Answer

m ＜ -1/3 or m ＞ 1

Test m=2: y=2x+1 is 3 at x=1 (above A) and 7 at x=3 (above B), so it passes above the triangle. Test m=-1: y=-x+1 is -5 at x=6, below C, and 0 at x=1, below A, so it passes below. Test m=0: y=1 meets side BC, as expected since 0 is in the hitting range.

Takeaway

A line spinning about a fixed point only needs to be checked at the corners of the shape.

Find the pivot
Slopes to the vertices
Read off the hitting range
Take the complement

Where next?

Another one like thissuggested

12 practice problems

Step back and solidifysuggested

Rotate a line about its fixed point Find which side the bisector hits first

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▶ Practice — 12 problems