Practice · Pivot a line past a triangle's extreme vertices · Sensim Math

Variant 1 easy

A line has slope $m$ and $y$ -intercept $-1$ . Find the range of values of $m$ for which this line does not meet the triangle with vertices $\text{A}(2, 3)$ , $\text{B}(4, 1)$ , and $\text{C}(3, -2)$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

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

Givens

The line has slope $m$ and $y$ -intercept $-1$ .
Triangle vertices: $\text{A}(2, 3)$ , $\text{B}(4, 1)$ , $\text{C}(3, -2)$ .

Unknowns

The range of $m$ for which the line and the triangle have no common point.

Constraints

The line always pivots about the fixed point $(0, -1)$ , which lies outside the triangle (all of the triangle has $x\ge 2$ ).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#13 Convert to Algebra

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.

3 · Execute4 carry out the plan

1Find the pivot

#1 Draw a Diagram 8.F.A.3

Whatever

m

is,

x=0

gives

y=-1

, so the line always passes through

(0, -1)

.

Changing

m

only turns the line about its

y

-intercept.

2Slopes to the vertices

#13 Convert to Algebra 8.EE.B.6

Compute the slope from

(0, -1)

to each vertex.

\text{A}: \frac{3-(-1)}{2-0}=2,\quad \text{B}: \frac{1-(-1)}{4-0}=\frac{1}{2},\quad \text{C}: \frac{-2-(-1)}{3-0}=-\frac{1}{3}

Only the largest and the smallest of these slopes will matter.

3Read off the hitting range

#16 Count the Complement 8.F.A.3

Every point of the triangle has

x>0

, so its slope from

(0, -1)

is between those of the vertices: from

-\frac{1}{3}

(at

\text{C}

) up to

2

(at

\text{A}

). The line meets the triangle exactly when

-\frac{1}{3}\le m\le 2

.

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

4Take the complement

#16 Count the Complement 8.F.A.3

The line misses the triangle for every other slope.

Steeper than the line to

\text{A}

passes above; tilted down more than the line to

\text{C}

passes below.

Answer:

m<-\frac{1}{3}

or

m>2

4 · Reviewdoes it hold up?

Test $m=3$ : $y=3x-1$ is $5$ at $x=2$ (above $\text{A}$ ), $11$ at $x=4$ (above $\text{B}$ ) and $8$ at $x=3$ (above $\text{C}$ ), so it passes above the triangle. Test $m=-1$ : $y=-x-1$ is $-3$ at $x=2$ (below $\text{A}$ ), $-5$ at $x=4$ (below $\text{B}$ ) and $-4$ at $x=3$ (below $\text{C}$ ), so it passes below. Test $m=0$ : $y=-1$ is $-1$ at $x=2$ (below $\text{A}$ ), $-1$ at $x=4$ (below $\text{B}$ ) and $-1$ at $x=3$ (above $\text{C}$ ), so it has vertices on both sides and crosses the triangle, as expected since $0$ is in the hitting range.

Another way: Vertex test: the line misses the triangle exactly when all three vertices lie strictly on the same side of it. All three above needs $3>2m-1$ , $1>4m-1$ , $-2>3m-1$ , that is $m<2$ , $m<\frac{1}{2}$ , $m<-\frac{1}{3}$ , so $m<-\frac{1}{3}$ . All three below reverses every inequality: $m>2$ , $m>\frac{1}{2}$ , $m>-\frac{1}{3}$ , so $m>2$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing $y=mx-1$ as a family of lines through $(0, -1)$ with slope $m$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing slopes between the pivot and each vertex.

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

Variant 2 easy

A line has slope $m$ and $y$ -intercept $2$ . Find the range of values of $m$ for which this line does not meet the triangle with vertices $\text{A}(1, 4)$ , $\text{B}(4, 5)$ , and $\text{C}(5, 1)$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=mx+2$ passes through $(0, 2)$ . For which slopes $m$ does such a line miss the triangle with corners $\text{A}(1, 4)$ , $\text{B}(4, 5)$ , $\text{C}(5, 1)$ entirely?

Givens

The line has slope $m$ and $y$ -intercept $2$ .
Triangle vertices: $\text{A}(1, 4)$ , $\text{B}(4, 5)$ , $\text{C}(5, 1)$ .

Unknowns

The range of $m$ for which the line and the triangle have no common point.

Constraints

The line always pivots about the fixed point $(0, 2)$ , which lies outside the triangle (all of the triangle has $x\ge 1$ ).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#13 Convert to Algebra

Draw the triangle and the pivot point $(0, 2)$ . As the line turns about $(0, 2)$ , it first touches the triangle at one extreme vertex and leaves it at another. So only the slopes from $(0, 2)$ to the three vertices matter: the line hits the triangle exactly when its slope lies between the smallest and largest of them.

3 · Execute4 carry out the plan

1Find the pivot

#1 Draw a Diagram 8.F.A.3

Whatever

m

is,

x=0

gives

y=2

, so the line always passes through

(0, 2)

.

Changing

m

only turns the line about its

y

-intercept.

2Slopes to the vertices

#13 Convert to Algebra 8.EE.B.6

Compute the slope from

(0, 2)

to each vertex.

\text{A}: \frac{4-2}{1-0}=2,\quad \text{B}: \frac{5-2}{4-0}=\frac{3}{4},\quad \text{C}: \frac{1-2}{5-0}=-\frac{1}{5}

Only the largest and the smallest of these slopes will matter.

3Read off the hitting range

#16 Count the Complement 8.F.A.3

Every point of the triangle has

x>0

, so its slope from

(0, 2)

is between those of the vertices: from

-\frac{1}{5}

(at

\text{C}

) up to

2

(at

\text{A}

). The line meets the triangle exactly when

-\frac{1}{5}\le m\le 2

.

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

4Take the complement

#16 Count the Complement 8.F.A.3

The line misses the triangle for every other slope.

Steeper than the line to

\text{A}

passes above; tilted down more than the line to

\text{C}

passes below.

Answer:

m<-\frac{1}{5}

or

m>2

4 · Reviewdoes it hold up?

Test $m=3$ : $y=3x+2$ is $5$ at $x=1$ (above $\text{A}$ ), $14$ at $x=4$ (above $\text{B}$ ) and $17$ at $x=5$ (above $\text{C}$ ), so it passes above the triangle. Test $m=-1$ : $y=-x+2$ is $1$ at $x=1$ (below $\text{A}$ ), $-2$ at $x=4$ (below $\text{B}$ ) and $-3$ at $x=5$ (below $\text{C}$ ), so it passes below. Test $m=0$ : $y=2$ is $2$ at $x=1$ (below $\text{A}$ ), $2$ at $x=4$ (below $\text{B}$ ) and $2$ at $x=5$ (above $\text{C}$ ), so it has vertices on both sides and crosses the triangle, as expected since $0$ is in the hitting range.

Another way: Vertex test: the line misses the triangle exactly when all three vertices lie strictly on the same side of it. All three above needs $4>m+2$ , $5>4m+2$ , $1>5m+2$ , that is $m<2$ , $m<\frac{3}{4}$ , $m<-\frac{1}{5}$ , so $m<-\frac{1}{5}$ . All three below reverses every inequality: $m>2$ , $m>\frac{3}{4}$ , $m>-\frac{1}{5}$ , so $m>2$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing $y=mx+2$ as a family of lines through $(0, 2)$ with slope $m$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing slopes between the pivot and each vertex.

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

Variant 3 easy

A line has slope $m$ and $y$ -intercept $1$ . Find the range of values of $m$ for which this line does not meet the triangle with vertices $\text{A}(2, 5)$ , $\text{B}(5, 3)$ , and $\text{C}(4, -1)$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=mx+1$ passes through $(0, 1)$ . For which slopes $m$ does such a line miss the triangle with corners $\text{A}(2, 5)$ , $\text{B}(5, 3)$ , $\text{C}(4, -1)$ entirely?

Givens

The line has slope $m$ and $y$ -intercept $1$ .
Triangle vertices: $\text{A}(2, 5)$ , $\text{B}(5, 3)$ , $\text{C}(4, -1)$ .

Unknowns

The range of $m$ for which the line and the triangle have no common point.

Constraints

The line always pivots about the fixed point $(0, 1)$ , which lies outside the triangle (all of the triangle has $x\ge 2$ ).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#13 Convert to Algebra

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.

3 · Execute4 carry out the plan

1Find the pivot

#1 Draw a Diagram 8.F.A.3

Whatever

m

is,

x=0

gives

y=1

, so the line always passes through

(0, 1)

.

Changing

m

only turns the line about its

y

-intercept.

2Slopes to the vertices

#13 Convert to Algebra 8.EE.B.6

Compute the slope from

(0, 1)

to each vertex.

\text{A}: \frac{5-1}{2-0}=2,\quad \text{B}: \frac{3-1}{5-0}=\frac{2}{5},\quad \text{C}: \frac{-1-1}{4-0}=-\frac{1}{2}

Only the largest and the smallest of these slopes will matter.

3Read off the hitting range

#16 Count the Complement 8.F.A.3

Every point of the triangle has

x>0

, so its slope from

(0, 1)

is between those of the vertices: from

-\frac{1}{2}

(at

\text{C}

) up to

2

(at

\text{A}

). The line meets the triangle exactly when

-\frac{1}{2}\le m\le 2

.

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

4Take the complement

#16 Count the Complement 8.F.A.3

The line misses the triangle for every other slope.

Steeper than the line to

\text{A}

passes above; tilted down more than the line to

\text{C}

passes below.

Answer:

m<-\frac{1}{2}

or

m>2

4 · Reviewdoes it hold up?

Test $m=3$ : $y=3x+1$ is $7$ at $x=2$ (above $\text{A}$ ), $16$ at $x=5$ (above $\text{B}$ ) and $13$ at $x=4$ (above $\text{C}$ ), so it passes above the triangle. Test $m=-1$ : $y=-x+1$ is $-1$ at $x=2$ (below $\text{A}$ ), $-4$ at $x=5$ (below $\text{B}$ ) and $-3$ at $x=4$ (below $\text{C}$ ), so it passes below. Test $m=0$ : $y=1$ is $1$ at $x=2$ (below $\text{A}$ ), $1$ at $x=5$ (below $\text{B}$ ) and $1$ at $x=4$ (above $\text{C}$ ), so it has vertices on both sides and crosses the triangle, as expected since $0$ is in the hitting range.

Another way: Vertex test: the line misses the triangle exactly when all three vertices lie strictly on the same side of it. All three above needs $5>2m+1$ , $3>5m+1$ , $-1>4m+1$ , that is $m<2$ , $m<\frac{2}{5}$ , $m<-\frac{1}{2}$ , so $m<-\frac{1}{2}$ . All three below reverses every inequality: $m>2$ , $m>\frac{2}{5}$ , $m>-\frac{1}{2}$ , so $m>2$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing $y=mx+1$ as a family of lines through $(0, 1)$ with slope $m$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing slopes between the pivot and each vertex.

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

Variant 4 easy

A line has slope $m$ and $y$ -intercept $-1$ . Find the range of values of $m$ for which this line does not meet the triangle with vertices $\text{A}(1, 2)$ , $\text{B}(4, -3)$ , and $\text{C}(2, -5)$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=mx-1$ passes through $(0, -1)$ . For which slopes $m$ does such a line miss the triangle with corners $\text{A}(1, 2)$ , $\text{B}(4, -3)$ , $\text{C}(2, -5)$ entirely?

Givens

The line has slope $m$ and $y$ -intercept $-1$ .
Triangle vertices: $\text{A}(1, 2)$ , $\text{B}(4, -3)$ , $\text{C}(2, -5)$ .

Unknowns

The range of $m$ for which the line and the triangle have no common point.

Constraints

The line always pivots about the fixed point $(0, -1)$ , which lies outside the triangle (all of the triangle has $x\ge 1$ ).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#13 Convert to Algebra

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.

3 · Execute4 carry out the plan

1Find the pivot

#1 Draw a Diagram 8.F.A.3

Whatever

m

is,

x=0

gives

y=-1

, so the line always passes through

(0, -1)

.

Changing

m

only turns the line about its

y

-intercept.

2Slopes to the vertices

#13 Convert to Algebra 8.EE.B.6

Compute the slope from

(0, -1)

to each vertex.

\text{A}: \frac{2-(-1)}{1-0}=3,\quad \text{B}: \frac{-3-(-1)}{4-0}=-\frac{1}{2},\quad \text{C}: \frac{-5-(-1)}{2-0}=-2

Only the largest and the smallest of these slopes will matter.

3Read off the hitting range

#16 Count the Complement 8.F.A.3

Every point of the triangle has

x>0

, so its slope from

(0, -1)

is between those of the vertices: from

-2

(at

\text{C}

) up to

3

(at

\text{A}

). The line meets the triangle exactly when

-2\le m\le 3

.

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

4Take the complement

#16 Count the Complement 8.F.A.3

The line misses the triangle for every other slope.

Steeper than the line to

\text{A}

passes above; tilted down more than the line to

\text{C}

passes below.

Answer:

m<-2

or

m>3

4 · Reviewdoes it hold up?

Test $m=4$ : $y=4x-1$ is $3$ at $x=1$ (above $\text{A}$ ), $15$ at $x=4$ (above $\text{B}$ ) and $7$ at $x=2$ (above $\text{C}$ ), so it passes above the triangle. Test $m=-3$ : $y=-3x-1$ is $-4$ at $x=1$ (below $\text{A}$ ), $-13$ at $x=4$ (below $\text{B}$ ) and $-7$ at $x=2$ (below $\text{C}$ ), so it passes below. Test $m=-1$ : $y=-x-1$ is $-2$ at $x=1$ (below $\text{A}$ ), $-5$ at $x=4$ (below $\text{B}$ ) and $-3$ at $x=2$ (above $\text{C}$ ), so it has vertices on both sides and crosses the triangle, as expected since $-1$ is in the hitting range.

Another way: Vertex test: the line misses the triangle exactly when all three vertices lie strictly on the same side of it. All three above needs $2>m-1$ , $-3>4m-1$ , $-5>2m-1$ , that is $m<3$ , $m<-\frac{1}{2}$ , $m<-2$ , so $m<-2$ . All three below reverses every inequality: $m>3$ , $m>-\frac{1}{2}$ , $m>-2$ , so $m>3$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing $y=mx-1$ as a family of lines through $(0, -1)$ with slope $m$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing slopes between the pivot and each vertex.

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

Variant 5 medium

A line has slope $m$ and $y$ -intercept $-3$ . Find the range of values of $m$ for which this line does not meet the triangle with vertices $\text{A}(2, 1)$ , $\text{B}(5, 2)$ , and $\text{C}(4, -5)$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=mx-3$ passes through $(0, -3)$ . For which slopes $m$ does such a line miss the triangle with corners $\text{A}(2, 1)$ , $\text{B}(5, 2)$ , $\text{C}(4, -5)$ entirely?

Givens

The line has slope $m$ and $y$ -intercept $-3$ .
Triangle vertices: $\text{A}(2, 1)$ , $\text{B}(5, 2)$ , $\text{C}(4, -5)$ .

Unknowns

The range of $m$ for which the line and the triangle have no common point.

Constraints

The line always pivots about the fixed point $(0, -3)$ , which lies outside the triangle (all of the triangle has $x\ge 2$ ).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#13 Convert to Algebra

Draw the triangle and the pivot point $(0, -3)$ . As the line turns about $(0, -3)$ , it first touches the triangle at one extreme vertex and leaves it at another. So only the slopes from $(0, -3)$ to the three vertices matter: the line hits the triangle exactly when its slope lies between the smallest and largest of them.

3 · Execute4 carry out the plan

1Find the pivot

#1 Draw a Diagram 8.F.A.3

Whatever

m

is,

x=0

gives

y=-3

, so the line always passes through

(0, -3)

.

Changing

m

only turns the line about its

y

-intercept.

2Slopes to the vertices

#13 Convert to Algebra 8.EE.B.6

Compute the slope from

(0, -3)

to each vertex.

\text{A}: \frac{1-(-3)}{2-0}=2,\quad \text{B}: \frac{2-(-3)}{5-0}=1,\quad \text{C}: \frac{-5-(-3)}{4-0}=-\frac{1}{2}

Only the largest and the smallest of these slopes will matter.

3Read off the hitting range

#16 Count the Complement 8.F.A.3

Every point of the triangle has

x>0

, so its slope from

(0, -3)

is between those of the vertices: from

-\frac{1}{2}

(at

\text{C}

) up to

2

(at

\text{A}

). The line meets the triangle exactly when

-\frac{1}{2}\le m\le 2

.

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

4Take the complement

#16 Count the Complement 8.F.A.3

The line misses the triangle for every other slope.

Steeper than the line to

\text{A}

passes above; tilted down more than the line to

\text{C}

passes below.

Answer:

m<-\frac{1}{2}

or

m>2

4 · Reviewdoes it hold up?

Test $m=3$ : $y=3x-3$ is $3$ at $x=2$ (above $\text{A}$ ), $12$ at $x=5$ (above $\text{B}$ ) and $9$ at $x=4$ (above $\text{C}$ ), so it passes above the triangle. Test $m=-1$ : $y=-x-3$ is $-5$ at $x=2$ (below $\text{A}$ ), $-8$ at $x=5$ (below $\text{B}$ ) and $-7$ at $x=4$ (below $\text{C}$ ), so it passes below. Test $m=0$ : $y=-3$ is $-3$ at $x=2$ (below $\text{A}$ ), $-3$ at $x=5$ (below $\text{B}$ ) and $-3$ at $x=4$ (above $\text{C}$ ), so it has vertices on both sides and crosses the triangle, as expected since $0$ is in the hitting range.

Another way: Vertex test: the line misses the triangle exactly when all three vertices lie strictly on the same side of it. All three above needs $1>2m-3$ , $2>5m-3$ , $-5>4m-3$ , that is $m<2$ , $m<1$ , $m<-\frac{1}{2}$ , so $m<-\frac{1}{2}$ . All three below reverses every inequality: $m>2$ , $m>1$ , $m>-\frac{1}{2}$ , so $m>2$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing $y=mx-3$ as a family of lines through $(0, -3)$ with slope $m$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing slopes between the pivot and each vertex.

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

Variant 6 medium

A line has slope $m$ and $y$ -intercept $3$ . Find the range of values of $m$ for which this line does not meet the triangle with vertices $\text{A}(2, 1)$ , $\text{B}(4, 5)$ , and $\text{C}(6, 6)$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=mx+3$ passes through $(0, 3)$ . For which slopes $m$ does such a line miss the triangle with corners $\text{A}(2, 1)$ , $\text{B}(4, 5)$ , $\text{C}(6, 6)$ entirely?

Givens

The line has slope $m$ and $y$ -intercept $3$ .
Triangle vertices: $\text{A}(2, 1)$ , $\text{B}(4, 5)$ , $\text{C}(6, 6)$ .

Unknowns

The range of $m$ for which the line and the triangle have no common point.

Constraints

The line always pivots about the fixed point $(0, 3)$ , which lies outside the triangle (all of the triangle has $x\ge 2$ ).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#13 Convert to Algebra

Draw the triangle and the pivot point $(0, 3)$ . As the line turns about $(0, 3)$ , it first touches the triangle at one extreme vertex and leaves it at another. So only the slopes from $(0, 3)$ to the three vertices matter: the line hits the triangle exactly when its slope lies between the smallest and largest of them.

3 · Execute4 carry out the plan

1Find the pivot

#1 Draw a Diagram 8.F.A.3

Whatever

m

is,

x=0

gives

y=3

, so the line always passes through

(0, 3)

.

Changing

m

only turns the line about its

y

-intercept.

2Slopes to the vertices

#13 Convert to Algebra 8.EE.B.6

Compute the slope from

(0, 3)

to each vertex.

\text{A}: \frac{1-3}{2-0}=-1,\quad \text{B}: \frac{5-3}{4-0}=\frac{1}{2},\quad \text{C}: \frac{6-3}{6-0}=\frac{1}{2}

\text{B}

and

\text{C}

give the same slope:

(0, 3)

,

\text{B}

,

\text{C}

lie on one line, so side

\text{BC}

points straight at the pivot.

3Read off the hitting range

#16 Count the Complement 8.F.A.3

Every point of the triangle has

x>0

, so its slope from

(0, 3)

is between those of the vertices: from

-1

(at

\text{A}

) up to

\frac{1}{2}

(along

\text{BC}

). The line meets the triangle exactly when

-1\le m\le \frac{1}{2}

.

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

4Take the complement

#16 Count the Complement 8.F.A.3

The line misses the triangle for every other slope.

Steeper than side

\text{BC}

passes above; tilted down more than the line to

\text{A}

passes below.

Answer:

m<-1

or

m>\frac{1}{2}

4 · Reviewdoes it hold up?

Test $m=1$ : $y=x+3$ is $5$ at $x=2$ (above $\text{A}$ ), $7$ at $x=4$ (above $\text{B}$ ) and $9$ at $x=6$ (above $\text{C}$ ), so it passes above the triangle. Test $m=-2$ : $y=-2x+3$ is $-1$ at $x=2$ (below $\text{A}$ ), $-5$ at $x=4$ (below $\text{B}$ ) and $-9$ at $x=6$ (below $\text{C}$ ), so it passes below. Test $m=0$ : $y=3$ is $3$ at $x=2$ (above $\text{A}$ ), $3$ at $x=4$ (below $\text{B}$ ) and $3$ at $x=6$ (below $\text{C}$ ), so it has vertices on both sides and crosses the triangle, as expected since $0$ is in the hitting range.

Another way: Vertex test: the line misses the triangle exactly when all three vertices lie strictly on the same side of it. All three above needs $1>2m+3$ , $5>4m+3$ , $6>6m+3$ , that is $m<-1$ , $m<\frac{1}{2}$ , $m<\frac{1}{2}$ , so $m<-1$ . All three below reverses every inequality: $m>-1$ , $m>\frac{1}{2}$ , $m>\frac{1}{2}$ , so $m>\frac{1}{2}$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing $y=mx+3$ as a family of lines through $(0, 3)$ with slope $m$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing slopes between the pivot and each vertex.

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

Variant 7 hard

A line has slope $m$ and $y$ -intercept $4$ . Find the range of values of $m$ for which this line does not meet the triangle with vertices $\text{A}(2, 6)$ , $\text{B}(6, 1)$ , and $\text{C}(4, -2)$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

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

Givens

The line has slope $m$ and $y$ -intercept $4$ .
Triangle vertices: $\text{A}(2, 6)$ , $\text{B}(6, 1)$ , $\text{C}(4, -2)$ .

Unknowns

The range of $m$ for which the line and the triangle have no common point.

Constraints

The line always pivots about the fixed point $(0, 4)$ , which lies outside the triangle (all of the triangle has $x\ge 2$ ).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#13 Convert to Algebra

Draw the triangle and the pivot point $(0, 4)$ . As the line turns about $(0, 4)$ , it first touches the triangle at one extreme vertex and leaves it at another. So only the slopes from $(0, 4)$ to the three vertices matter: the line hits the triangle exactly when its slope lies between the smallest and largest of them.

3 · Execute4 carry out the plan

1Find the pivot

#1 Draw a Diagram 8.F.A.3

Whatever

m

is,

x=0

gives

y=4

, so the line always passes through

(0, 4)

.

Changing

m

only turns the line about its

y

-intercept.

2Slopes to the vertices

#13 Convert to Algebra 8.EE.B.6

Compute the slope from

(0, 4)

to each vertex.

\text{A}: \frac{6-4}{2-0}=1,\quad \text{B}: \frac{1-4}{6-0}=-\frac{1}{2},\quad \text{C}: \frac{-2-4}{4-0}=-\frac{3}{2}

Only the largest and the smallest of these slopes will matter.

3Read off the hitting range

#16 Count the Complement 8.F.A.3

Every point of the triangle has

x>0

, so its slope from

(0, 4)

is between those of the vertices: from

-\frac{3}{2}

(at

\text{C}

) up to

1

(at

\text{A}

). The line meets the triangle exactly when

-\frac{3}{2}\le m\le 1

.

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

4Take the complement

#16 Count the Complement 8.F.A.3

The line misses the triangle for every other slope.

Steeper than the line to

\text{A}

passes above; tilted down more than the line to

\text{C}

passes below.

Answer:

m<-\frac{3}{2}

or

m>1

4 · Reviewdoes it hold up?

Test $m=2$ : $y=2x+4$ is $8$ at $x=2$ (above $\text{A}$ ), $16$ at $x=6$ (above $\text{B}$ ) and $12$ at $x=4$ (above $\text{C}$ ), so it passes above the triangle. Test $m=-2$ : $y=-2x+4$ is $0$ at $x=2$ (below $\text{A}$ ), $-8$ at $x=6$ (below $\text{B}$ ) and $-4$ at $x=4$ (below $\text{C}$ ), so it passes below. Test $m=-1$ : $y=-x+4$ is $2$ at $x=2$ (below $\text{A}$ ), $-2$ at $x=6$ (below $\text{B}$ ) and $0$ at $x=4$ (above $\text{C}$ ), so it has vertices on both sides and crosses the triangle, as expected since $-1$ is in the hitting range.

Another way: Vertex test: the line misses the triangle exactly when all three vertices lie strictly on the same side of it. All three above needs $6>2m+4$ , $1>6m+4$ , $-2>4m+4$ , that is $m<1$ , $m<-\frac{1}{2}$ , $m<-\frac{3}{2}$ , so $m<-\frac{3}{2}$ . All three below reverses every inequality: $m>1$ , $m>-\frac{1}{2}$ , $m>-\frac{3}{2}$ , so $m>1$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing $y=mx+4$ as a family of lines through $(0, 4)$ with slope $m$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing slopes between the pivot and each vertex.

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

Variant 8 medium

A line has slope $m$ and $y$ -intercept $1$ . Find the range of values of $m$ for which this line does not meet the triangle with vertices $\text{A}(1, 2)$ , $\text{B}(3, 4)$ , and $\text{C}(6, -1)$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

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

Givens

The line has slope $m$ and $y$ -intercept $1$ .
Triangle vertices: $\text{A}(1, 2)$ , $\text{B}(3, 4)$ , $\text{C}(6, -1)$ .

Unknowns

The range of $m$ for which the line and the triangle have no common point.

Constraints

The line always pivots about the fixed point $(0, 1)$ , which lies outside the triangle (all of the triangle has $x\ge 1$ ).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#13 Convert to Algebra

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.

3 · Execute4 carry out the plan

1Find the pivot

#1 Draw a Diagram 8.F.A.3

Whatever

m

is,

x=0

gives

y=1

, so the line always passes through

(0, 1)

.

Changing

m

only turns the line about its

y

-intercept.

2Slopes to the vertices

#13 Convert to Algebra 8.EE.B.6

Compute the slope from

(0, 1)

to each vertex.

\text{A}: \frac{2-1}{1-0}=1,\quad \text{B}: \frac{4-1}{3-0}=1,\quad \text{C}: \frac{-1-1}{6-0}=-\frac{1}{3}

\text{A}

and

\text{B}

give the same slope:

(0, 1)

,

\text{A}

,

\text{B}

lie on one line, so side

\text{AB}

points straight at the pivot.

3Read off the hitting range

#16 Count the Complement 8.F.A.3

Every point of the triangle has

x>0

, so its slope from

(0, 1)

is between those of the vertices: from

-\frac{1}{3}

(at

\text{C}

) up to

1

(along

\text{AB}

). The line meets the triangle exactly when

-\frac{1}{3}\le m\le 1

.

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

4Take the complement

#16 Count the Complement 8.F.A.3

The line misses the triangle for every other slope.

Steeper than side

\text{AB}

passes above; tilted down more than the line to

\text{C}

passes below.

Answer:

m<-\frac{1}{3}

or

m>1

4 · Reviewdoes it hold up?

Test $m=2$ : $y=2x+1$ is $3$ at $x=1$ (above $\text{A}$ ), $7$ at $x=3$ (above $\text{B}$ ) and $13$ at $x=6$ (above $\text{C}$ ), so it passes above the triangle. Test $m=-1$ : $y=-x+1$ is $0$ at $x=1$ (below $\text{A}$ ), $-2$ at $x=3$ (below $\text{B}$ ) and $-5$ at $x=6$ (below $\text{C}$ ), so it passes below. Test $m=0$ : $y=1$ is $1$ at $x=1$ (below $\text{A}$ ), $1$ at $x=3$ (below $\text{B}$ ) and $1$ at $x=6$ (above $\text{C}$ ), so it has vertices on both sides and crosses the triangle, as expected since $0$ is in the hitting range.

Another way: Vertex test: the line misses the triangle exactly when all three vertices lie strictly on the same side of it. All three above needs $2>m+1$ , $4>3m+1$ , $-1>6m+1$ , that is $m<1$ , $m<1$ , $m<-\frac{1}{3}$ , so $m<-\frac{1}{3}$ . All three below reverses every inequality: $m>1$ , $m>1$ , $m>-\frac{1}{3}$ , so $m>1$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing $y=mx+1$ as a family of lines through $(0, 1)$ with slope $m$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing slopes between the pivot and each vertex.

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

Variant 9 medium

A line has slope $m$ and $y$ -intercept $2$ . Find the range of values of $m$ for which this line does not meet the triangle with vertices $\text{A}(3, 1)$ , $\text{B}(3, 5)$ , and $\text{C}(6, 4)$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=mx+2$ passes through $(0, 2)$ . For which slopes $m$ does such a line miss the triangle with corners $\text{A}(3, 1)$ , $\text{B}(3, 5)$ , $\text{C}(6, 4)$ entirely?

Givens

The line has slope $m$ and $y$ -intercept $2$ .
Triangle vertices: $\text{A}(3, 1)$ , $\text{B}(3, 5)$ , $\text{C}(6, 4)$ .

Unknowns

The range of $m$ for which the line and the triangle have no common point.

Constraints

The line always pivots about the fixed point $(0, 2)$ , which lies outside the triangle (all of the triangle has $x\ge 3$ ).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#13 Convert to Algebra

Draw the triangle and the pivot point $(0, 2)$ . As the line turns about $(0, 2)$ , it first touches the triangle at one extreme vertex and leaves it at another. So only the slopes from $(0, 2)$ to the three vertices matter: the line hits the triangle exactly when its slope lies between the smallest and largest of them.

3 · Execute4 carry out the plan

1Find the pivot

#1 Draw a Diagram 8.F.A.3

Whatever

m

is,

x=0

gives

y=2

, so the line always passes through

(0, 2)

.

Changing

m

only turns the line about its

y

-intercept.

2Slopes to the vertices

#13 Convert to Algebra 8.EE.B.6

Compute the slope from

(0, 2)

to each vertex.

\text{A}: \frac{1-2}{3-0}=-\frac{1}{3},\quad \text{B}: \frac{5-2}{3-0}=1,\quad \text{C}: \frac{4-2}{6-0}=\frac{1}{3}

Only the largest and the smallest of these slopes will matter.

3Read off the hitting range

#16 Count the Complement 8.F.A.3

Every point of the triangle has

x>0

, so its slope from

(0, 2)

is between those of the vertices: from

-\frac{1}{3}

(at

\text{A}

) up to

1

(at

\text{B}

). The line meets the triangle exactly when

-\frac{1}{3}\le m\le 1

.

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

4Take the complement

#16 Count the Complement 8.F.A.3

The line misses the triangle for every other slope.

Steeper than the line to

\text{B}

passes above; tilted down more than the line to

\text{A}

passes below.

Answer:

m<-\frac{1}{3}

or

m>1

4 · Reviewdoes it hold up?

Test $m=2$ : $y=2x+2$ is $8$ at $x=3$ (above $\text{A}$ ), $8$ at $x=3$ (above $\text{B}$ ) and $14$ at $x=6$ (above $\text{C}$ ), so it passes above the triangle. Test $m=-1$ : $y=-x+2$ is $-1$ at $x=3$ (below $\text{A}$ ), $-1$ at $x=3$ (below $\text{B}$ ) and $-4$ at $x=6$ (below $\text{C}$ ), so it passes below. Test $m=0$ : $y=2$ is $2$ at $x=3$ (above $\text{A}$ ), $2$ at $x=3$ (below $\text{B}$ ) and $2$ at $x=6$ (below $\text{C}$ ), so it has vertices on both sides and crosses the triangle, as expected since $0$ is in the hitting range.

Another way: Vertex test: the line misses the triangle exactly when all three vertices lie strictly on the same side of it. All three above needs $1>3m+2$ , $5>3m+2$ , $4>6m+2$ , that is $m<-\frac{1}{3}$ , $m<1$ , $m<\frac{1}{3}$ , so $m<-\frac{1}{3}$ . All three below reverses every inequality: $m>-\frac{1}{3}$ , $m>1$ , $m>\frac{1}{3}$ , so $m>1$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing $y=mx+2$ as a family of lines through $(0, 2)$ with slope $m$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing slopes between the pivot and each vertex.

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

Variant 10 hard

A line has slope $m$ and $y$ -intercept $1$ . Find the range of values of $m$ for which this line does not meet the triangle with vertices $\text{A}(3, 7)$ , $\text{B}(6, 3)$ , and $\text{C}(2, -1)$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=mx+1$ passes through $(0, 1)$ . For which slopes $m$ does such a line miss the triangle with corners $\text{A}(3, 7)$ , $\text{B}(6, 3)$ , $\text{C}(2, -1)$ entirely?

Givens

The line has slope $m$ and $y$ -intercept $1$ .
Triangle vertices: $\text{A}(3, 7)$ , $\text{B}(6, 3)$ , $\text{C}(2, -1)$ .

Unknowns

The range of $m$ for which the line and the triangle have no common point.

Constraints

The line always pivots about the fixed point $(0, 1)$ , which lies outside the triangle (all of the triangle has $x\ge 2$ ).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#13 Convert to Algebra

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.

3 · Execute4 carry out the plan

1Find the pivot

#1 Draw a Diagram 8.F.A.3

Whatever

m

is,

x=0

gives

y=1

, so the line always passes through

(0, 1)

.

Changing

m

only turns the line about its

y

-intercept.

2Slopes to the vertices

#13 Convert to Algebra 8.EE.B.6

Compute the slope from

(0, 1)

to each vertex.

\text{A}: \frac{7-1}{3-0}=2,\quad \text{B}: \frac{3-1}{6-0}=\frac{1}{3},\quad \text{C}: \frac{-1-1}{2-0}=-1

Only the largest and the smallest of these slopes will matter.

3Read off the hitting range

#16 Count the Complement 8.F.A.3

Every point of the triangle has

x>0

, so its slope from

(0, 1)

is between those of the vertices: from

-1

(at

\text{C}

) up to

2

(at

\text{A}

). The line meets the triangle exactly when

-1\le m\le 2

.

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

4Take the complement

#16 Count the Complement 8.F.A.3

The line misses the triangle for every other slope.

Steeper than the line to

\text{A}

passes above; tilted down more than the line to

\text{C}

passes below.

Answer:

m<-1

or

m>2

4 · Reviewdoes it hold up?

Test $m=3$ : $y=3x+1$ is $10$ at $x=3$ (above $\text{A}$ ), $19$ at $x=6$ (above $\text{B}$ ) and $7$ at $x=2$ (above $\text{C}$ ), so it passes above the triangle. Test $m=-2$ : $y=-2x+1$ is $-5$ at $x=3$ (below $\text{A}$ ), $-11$ at $x=6$ (below $\text{B}$ ) and $-3$ at $x=2$ (below $\text{C}$ ), so it passes below. Test $m=0$ : $y=1$ is $1$ at $x=3$ (below $\text{A}$ ), $1$ at $x=6$ (below $\text{B}$ ) and $1$ at $x=2$ (above $\text{C}$ ), so it has vertices on both sides and crosses the triangle, as expected since $0$ is in the hitting range.

Another way: Vertex test: the line misses the triangle exactly when all three vertices lie strictly on the same side of it. All three above needs $7>3m+1$ , $3>6m+1$ , $-1>2m+1$ , that is $m<2$ , $m<\frac{1}{3}$ , $m<-1$ , so $m<-1$ . All three below reverses every inequality: $m>2$ , $m>\frac{1}{3}$ , $m>-1$ , so $m>2$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing $y=mx+1$ as a family of lines through $(0, 1)$ with slope $m$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing slopes between the pivot and each vertex.

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

Variant 11 hard

A line has slope $m$ and $y$ -intercept $-2$ . Find the range of values of $m$ for which this line does not meet the triangle with vertices $\text{A}(1, 1)$ , $\text{B}(3, 7)$ , and $\text{C}(6, 2)$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=mx-2$ passes through $(0, -2)$ . For which slopes $m$ does such a line miss the triangle with corners $\text{A}(1, 1)$ , $\text{B}(3, 7)$ , $\text{C}(6, 2)$ entirely?

Givens

The line has slope $m$ and $y$ -intercept $-2$ .
Triangle vertices: $\text{A}(1, 1)$ , $\text{B}(3, 7)$ , $\text{C}(6, 2)$ .

Unknowns

The range of $m$ for which the line and the triangle have no common point.

Constraints

The line always pivots about the fixed point $(0, -2)$ , which lies outside the triangle (all of the triangle has $x\ge 1$ ).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#13 Convert to Algebra

Draw the triangle and the pivot point $(0, -2)$ . As the line turns about $(0, -2)$ , it first touches the triangle at one extreme vertex and leaves it at another. So only the slopes from $(0, -2)$ to the three vertices matter: the line hits the triangle exactly when its slope lies between the smallest and largest of them.

3 · Execute4 carry out the plan

1Find the pivot

#1 Draw a Diagram 8.F.A.3

Whatever

m

is,

x=0

gives

y=-2

, so the line always passes through

(0, -2)

.

Changing

m

only turns the line about its

y

-intercept.

2Slopes to the vertices

#13 Convert to Algebra 8.EE.B.6

Compute the slope from

(0, -2)

to each vertex.

\text{A}: \frac{1-(-2)}{1-0}=3,\quad \text{B}: \frac{7-(-2)}{3-0}=3,\quad \text{C}: \frac{2-(-2)}{6-0}=\frac{2}{3}

\text{A}

and

\text{B}

give the same slope:

(0, -2)

,

\text{A}

,

\text{B}

lie on one line, so side

\text{AB}

points straight at the pivot.

3Read off the hitting range

#16 Count the Complement 8.F.A.3

Every point of the triangle has

x>0

, so its slope from

(0, -2)

is between those of the vertices: from

\frac{2}{3}

(at

\text{C}

) up to

3

(along

\text{AB}

). The line meets the triangle exactly when

\frac{2}{3}\le m\le 3

.

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

4Take the complement

#16 Count the Complement 8.F.A.3

The line misses the triangle for every other slope.

Steeper than side

\text{AB}

passes above; tilted down more than the line to

\text{C}

passes below.

Answer:

m<\frac{2}{3}

or

m>3

4 · Reviewdoes it hold up?

Test $m=4$ : $y=4x-2$ is $2$ at $x=1$ (above $\text{A}$ ), $10$ at $x=3$ (above $\text{B}$ ) and $22$ at $x=6$ (above $\text{C}$ ), so it passes above the triangle. Test $m=0$ : $y=-2$ is $-2$ at $x=1$ (below $\text{A}$ ), $-2$ at $x=3$ (below $\text{B}$ ) and $-2$ at $x=6$ (below $\text{C}$ ), so it passes below. Test $m=1$ : $y=x-2$ is $-1$ at $x=1$ (below $\text{A}$ ), $1$ at $x=3$ (below $\text{B}$ ) and $4$ at $x=6$ (above $\text{C}$ ), so it has vertices on both sides and crosses the triangle, as expected since $1$ is in the hitting range.

Another way: Vertex test: the line misses the triangle exactly when all three vertices lie strictly on the same side of it. All three above needs $1>m-2$ , $7>3m-2$ , $2>6m-2$ , that is $m<3$ , $m<3$ , $m<\frac{2}{3}$ , so $m<\frac{2}{3}$ . All three below reverses every inequality: $m>3$ , $m>3$ , $m>\frac{2}{3}$ , so $m>3$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing $y=mx-2$ as a family of lines through $(0, -2)$ with slope $m$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing slopes between the pivot and each vertex.

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

Variant 12 hard

A line has slope $m$ and $y$ -intercept $2$ . Find the range of values of $m$ for which this line does not meet the triangle with vertices $\text{A}(1, 5)$ , $\text{B}(2, 8)$ , and $\text{C}(6, 5)$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=mx+2$ passes through $(0, 2)$ . For which slopes $m$ does such a line miss the triangle with corners $\text{A}(1, 5)$ , $\text{B}(2, 8)$ , $\text{C}(6, 5)$ entirely?

Givens

The line has slope $m$ and $y$ -intercept $2$ .
Triangle vertices: $\text{A}(1, 5)$ , $\text{B}(2, 8)$ , $\text{C}(6, 5)$ .

Unknowns

The range of $m$ for which the line and the triangle have no common point.

Constraints

The line always pivots about the fixed point $(0, 2)$ , which lies outside the triangle (all of the triangle has $x\ge 1$ ).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #16 Count the Complement#13 Convert to Algebra

Draw the triangle and the pivot point $(0, 2)$ . As the line turns about $(0, 2)$ , it first touches the triangle at one extreme vertex and leaves it at another. So only the slopes from $(0, 2)$ to the three vertices matter: the line hits the triangle exactly when its slope lies between the smallest and largest of them.

3 · Execute4 carry out the plan

1Find the pivot

#1 Draw a Diagram 8.F.A.3

Whatever

m

is,

x=0

gives

y=2

, so the line always passes through

(0, 2)

.

Changing

m

only turns the line about its

y

-intercept.

2Slopes to the vertices

#13 Convert to Algebra 8.EE.B.6

Compute the slope from

(0, 2)

to each vertex.

\text{A}: \frac{5-2}{1-0}=3,\quad \text{B}: \frac{8-2}{2-0}=3,\quad \text{C}: \frac{5-2}{6-0}=\frac{1}{2}

\text{A}

and

\text{B}

give the same slope:

(0, 2)

,

\text{A}

,

\text{B}

lie on one line, so side

\text{AB}

points straight at the pivot.

3Read off the hitting range

#16 Count the Complement 8.F.A.3

Every point of the triangle has

x>0

, so its slope from

(0, 2)

is between those of the vertices: from

\frac{1}{2}

(at

\text{C}

) up to

3

(along

\text{AB}

). The line meets the triangle exactly when

\frac{1}{2}\le m\le 3

.

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

4Take the complement

#16 Count the Complement 8.F.A.3

The line misses the triangle for every other slope.

Steeper than side

\text{AB}

passes above; tilted down more than the line to

\text{C}

passes below.

Answer:

m<\frac{1}{2}

or

m>3

4 · Reviewdoes it hold up?

Test $m=4$ : $y=4x+2$ is $6$ at $x=1$ (above $\text{A}$ ), $10$ at $x=2$ (above $\text{B}$ ) and $26$ at $x=6$ (above $\text{C}$ ), so it passes above the triangle. Test $m=0$ : $y=2$ is $2$ at $x=1$ (below $\text{A}$ ), $2$ at $x=2$ (below $\text{B}$ ) and $2$ at $x=6$ (below $\text{C}$ ), so it passes below. Test $m=1$ : $y=x+2$ is $3$ at $x=1$ (below $\text{A}$ ), $4$ at $x=2$ (below $\text{B}$ ) and $8$ at $x=6$ (above $\text{C}$ ), so it has vertices on both sides and crosses the triangle, as expected since $1$ is in the hitting range.

Another way: Vertex test: the line misses the triangle exactly when all three vertices lie strictly on the same side of it. All three above needs $5>m+2$ , $8>2m+2$ , $5>6m+2$ , that is $m<3$ , $m<3$ , $m<\frac{1}{2}$ , so $m<\frac{1}{2}$ . All three below reverses every inequality: $m>3$ , $m>3$ , $m>\frac{1}{2}$ , so $m>3$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing $y=mx+2$ as a family of lines through $(0, 2)$ with slope $m$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing slopes between the pivot and each vertex.

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

Pivot a line past a triangle's extreme vertices · 12 practice problems

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