Practice · Rotate a line about its fixed point · Sensim Math

Variant 1 easy

The graph of the linear function $y=ax-4$ intersects the line segment joining the two points $A(-2, 3)$ and $B(-4, -1)$ . Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=ax-4$ passes through $(0, -4)$ , whatever $a$ is. Find all slopes $a$ for which this line touches the segment from $A(-2, 3)$ to $B(-4, -1)$ .

Givens

The line is $y=ax-4$ .
The segment joins $A(-2, 3)$ and $B(-4, -1)$ .

Unknowns

The range of values of $a$ .

Constraints

The line must meet the segment, endpoints included.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

Setting $x=0$ shows every such line goes through $P(0,-4)$ , so changing $a$ just rotates the line about $P$ . Draw it: the line sweeps across the segment between the position through $B$ and the position through $A$ . Find those two boundary slopes by substituting each endpoint, then take everything in between.

3 · Execute4 carry out the plan

1Find the fixed point

#9 Solve an Easier Related Problem 8.F.A.3

Put

x=0

: the

a

term vanishes, so every line in the family passes through the same point.

x=0 \Rightarrow y=a\cdot 0-4=-4

The lines are spokes turning around

P(0, -4)

.

2Slope when the line hits A

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

Substitute

A(-2, 3)

into

y=ax-4

.

3=-2a-4 \Rightarrow a=-\frac{7}{2}

That is the slope from

P

to

A

:

\frac{3-(-4)}{-2-0}=-\frac{7}{2}

.

3Slope when the line hits B

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

Substitute

B(-4, -1)

into

y=ax-4

.

-1=-4a-4 \Rightarrow a=-\frac{3}{4}

That is the slope from

P

to

B

:

\frac{-1-(-4)}{-4-0}=-\frac{3}{4}

.

4Sweep between the two positions

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

The whole segment lies to the left of the

y

-axis, so as the line turns from passing through

A

to passing through

B

, it crosses every point of the segment, and its slope runs through every value from

-\frac{7}{2}

to

-\frac{3}{4}

. Any steeper or flatter line misses.

-\frac{7}{2}\le a\le -\frac{3}{4}

The two endpoints mark the extreme positions of the turning line.

Answer:

-\frac{7}{2}\le a\le -\frac{3}{4}

4 · Reviewdoes it hold up?

Try $a=-2$ (inside the range): $y=-2x-4$ meets segment $AB$ , which lies on $y=2x+7$ , where $-2x-4=2x+7$ , so $x=-\frac{11}{4}$ , between $-4$ and $-2$ . Try $a=-4$ (outside): $-4x-4=2x+7$ gives $x=-\frac{11}{6}$ , which is not between $-4$ and $-2$ , so it misses.

Another way: Algebra all the way: segment $AB$ lies on $y=2x+7$ for $-4\le x\le -2$ . Setting $ax-4=2x+7$ gives $x=\frac{11}{a-2}$ , and requiring $-4\le \frac{11}{a-2}\le -2$ leads to the same range.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing that every line $y=ax-4$ passes through $(0, -4)$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing the slope between the fixed point and each endpoint.

💡Takeaway. When a line is pinned at one point, changing its slope just swings it like a door -- the ends of the target tell you how far it can swing.

Variant 2 easy

The graph of the linear function $y=ax+1$ intersects the line segment joining the two points $A(-3, -3)$ and $B(-4, 0)$ . Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=ax+1$ passes through $(0, 1)$ , whatever $a$ is. Find all slopes $a$ for which this line touches the segment from $A(-3, -3)$ to $B(-4, 0)$ .

Givens

The line is $y=ax+1$ .
The segment joins $A(-3, -3)$ and $B(-4, 0)$ .

Unknowns

The range of values of $a$ .

Constraints

The line must meet the segment, endpoints included.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

Setting $x=0$ shows every such line goes through $P(0,1)$ , so changing $a$ just rotates the line about $P$ . Draw it: the line sweeps across the segment between the position through $B$ and the position through $A$ . Find those two boundary slopes by substituting each endpoint, then take everything in between.

3 · Execute4 carry out the plan

1Find the fixed point

#9 Solve an Easier Related Problem 8.F.A.3

Put

x=0

: the

a

term vanishes, so every line in the family passes through the same point.

x=0 \Rightarrow y=a\cdot 0+1=1

The lines are spokes turning around

P(0, 1)

.

2Slope when the line hits A

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

Substitute

A(-3, -3)

into

y=ax+1

.

-3=-3a+1 \Rightarrow a=\frac{4}{3}

That is the slope from

P

to

A

:

\frac{-3-1}{-3-0}=\frac{4}{3}

.

3Slope when the line hits B

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

Substitute

B(-4, 0)

into

y=ax+1

.

0=-4a+1 \Rightarrow a=\frac{1}{4}

That is the slope from

P

to

B

:

\frac{0-1}{-4-0}=\frac{1}{4}

.

4Sweep between the two positions

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

The whole segment lies to the left of the

y

-axis, so as the line turns from passing through

A

to passing through

B

, it crosses every point of the segment, and its slope runs through every value from

\frac{1}{4}

to

\frac{4}{3}

. Any steeper or flatter line misses.

\frac{1}{4}\le a\le \frac{4}{3}

The two endpoints mark the extreme positions of the turning line.

Answer:

\frac{1}{4}\le a\le \frac{4}{3}

4 · Reviewdoes it hold up?

Try $a=1$ (inside the range): $y=x+1$ meets segment $AB$ , which lies on $y=-3x-12$ , where $x+1=-3x-12$ , so $x=-\frac{13}{4}$ , between $-4$ and $-3$ . Try $a=0$ (outside): $1=-3x-12$ gives $x=-\frac{13}{3}$ , which is not between $-4$ and $-3$ , so it misses.

Another way: Algebra all the way: segment $AB$ lies on $y=-3x-12$ for $-4\le x\le -3$ . Setting $ax+1=-3x-12$ gives $x=\frac{-13}{a+3}$ , and requiring $-4\le \frac{-13}{a+3}\le -3$ leads to the same range.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing that every line $y=ax+1$ passes through $(0, 1)$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing the slope between the fixed point and each endpoint.

💡Takeaway. When a line is pinned at one point, changing its slope just swings it like a door -- the ends of the target tell you how far it can swing.

Variant 3 easy

The graph of the linear function $y=ax-2$ intersects the line segment joining the two points $A(-2, 3)$ and $B(-4, -1)$ . Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=ax-2$ passes through $(0, -2)$ , whatever $a$ is. Find all slopes $a$ for which this line touches the segment from $A(-2, 3)$ to $B(-4, -1)$ .

Givens

The line is $y=ax-2$ .
The segment joins $A(-2, 3)$ and $B(-4, -1)$ .

Unknowns

The range of values of $a$ .

Constraints

The line must meet the segment, endpoints included.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

Setting $x=0$ shows every such line goes through $P(0,-2)$ , so changing $a$ just rotates the line about $P$ . Draw it: the line sweeps across the segment between the position through $B$ and the position through $A$ . Find those two boundary slopes by substituting each endpoint, then take everything in between.

3 · Execute4 carry out the plan

1Find the fixed point

#9 Solve an Easier Related Problem 8.F.A.3

Put

x=0

: the

a

term vanishes, so every line in the family passes through the same point.

x=0 \Rightarrow y=a\cdot 0-2=-2

The lines are spokes turning around

P(0, -2)

.

2Slope when the line hits A

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

Substitute

A(-2, 3)

into

y=ax-2

.

3=-2a-2 \Rightarrow a=-\frac{5}{2}

That is the slope from

P

to

A

:

\frac{3-(-2)}{-2-0}=-\frac{5}{2}

.

3Slope when the line hits B

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

Substitute

B(-4, -1)

into

y=ax-2

.

-1=-4a-2 \Rightarrow a=-\frac{1}{4}

That is the slope from

P

to

B

:

\frac{-1-(-2)}{-4-0}=-\frac{1}{4}

.

4Sweep between the two positions

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

The whole segment lies to the left of the

y

-axis, so as the line turns from passing through

A

to passing through

B

, it crosses every point of the segment, and its slope runs through every value from

-\frac{5}{2}

to

-\frac{1}{4}

. Any steeper or flatter line misses.

-\frac{5}{2}\le a\le -\frac{1}{4}

The two endpoints mark the extreme positions of the turning line.

Answer:

-\frac{5}{2}\le a\le -\frac{1}{4}

4 · Reviewdoes it hold up?

Try $a=-1$ (inside the range): $y=-x-2$ meets segment $AB$ , which lies on $y=2x+7$ , where $-x-2=2x+7$ , so $x=-3$ , between $-4$ and $-2$ . Try $a=-3$ (outside): $-3x-2=2x+7$ gives $x=-\frac{9}{5}$ , which is not between $-4$ and $-2$ , so it misses.

Another way: Algebra all the way: segment $AB$ lies on $y=2x+7$ for $-4\le x\le -2$ . Setting $ax-2=2x+7$ gives $x=\frac{9}{a-2}$ , and requiring $-4\le \frac{9}{a-2}\le -2$ leads to the same range.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing that every line $y=ax-2$ passes through $(0, -2)$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing the slope between the fixed point and each endpoint.

💡Takeaway. When a line is pinned at one point, changing its slope just swings it like a door -- the ends of the target tell you how far it can swing.

Variant 4 medium

The graph of the linear function $y=ax+3$ intersects the line segment joining the two points $A(-2, 2)$ and $B(-3, -5)$ . Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=ax+3$ passes through $(0, 3)$ , whatever $a$ is. Find all slopes $a$ for which this line touches the segment from $A(-2, 2)$ to $B(-3, -5)$ .

Givens

The line is $y=ax+3$ .
The segment joins $A(-2, 2)$ and $B(-3, -5)$ .

Unknowns

The range of values of $a$ .

Constraints

The line must meet the segment, endpoints included.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

Setting $x=0$ shows every such line goes through $P(0,3)$ , so changing $a$ just rotates the line about $P$ . Draw it: the line sweeps across the segment between the position through $B$ and the position through $A$ . Find those two boundary slopes by substituting each endpoint, then take everything in between.

3 · Execute4 carry out the plan

1Find the fixed point

#9 Solve an Easier Related Problem 8.F.A.3

Put

x=0

: the

a

term vanishes, so every line in the family passes through the same point.

x=0 \Rightarrow y=a\cdot 0+3=3

The lines are spokes turning around

P(0, 3)

.

2Slope when the line hits A

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

Substitute

A(-2, 2)

into

y=ax+3

.

2=-2a+3 \Rightarrow a=\frac{1}{2}

That is the slope from

P

to

A

:

\frac{2-3}{-2-0}=\frac{1}{2}

.

3Slope when the line hits B

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

Substitute

B(-3, -5)

into

y=ax+3

.

-5=-3a+3 \Rightarrow a=\frac{8}{3}

That is the slope from

P

to

B

:

\frac{-5-3}{-3-0}=\frac{8}{3}

.

4Sweep between the two positions

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

The whole segment lies to the left of the

y

-axis, so as the line turns from passing through

A

to passing through

B

, it crosses every point of the segment, and its slope runs through every value from

\frac{1}{2}

to

\frac{8}{3}

. Any steeper or flatter line misses.

\frac{1}{2}\le a\le \frac{8}{3}

The two endpoints mark the extreme positions of the turning line.

Answer:

\frac{1}{2}\le a\le \frac{8}{3}

4 · Reviewdoes it hold up?

Try $a=2$ (inside the range): $y=2x+3$ meets segment $AB$ , which lies on $y=7x+16$ , where $2x+3=7x+16$ , so $x=-\frac{13}{5}$ , between $-3$ and $-2$ . Try $a=0$ (outside): $3=7x+16$ gives $x=-\frac{13}{7}$ , which is not between $-3$ and $-2$ , so it misses.

Another way: Algebra all the way: segment $AB$ lies on $y=7x+16$ for $-3\le x\le -2$ . Setting $ax+3=7x+16$ gives $x=\frac{13}{a-7}$ , and requiring $-3\le \frac{13}{a-7}\le -2$ leads to the same range.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing that every line $y=ax+3$ passes through $(0, 3)$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing the slope between the fixed point and each endpoint.

💡Takeaway. When a line is pinned at one point, changing its slope just swings it like a door -- the ends of the target tell you how far it can swing.

Variant 5 medium

The graph of the linear function $y=ax-4$ intersects the line segment joining the two points $A(-1, -2)$ and $B(-5, -2)$ . Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=ax-4$ passes through $(0, -4)$ , whatever $a$ is. Find all slopes $a$ for which this line touches the segment from $A(-1, -2)$ to $B(-5, -2)$ .

Givens

The line is $y=ax-4$ .
The segment joins $A(-1, -2)$ and $B(-5, -2)$ .

Unknowns

The range of values of $a$ .

Constraints

The line must meet the segment, endpoints included.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

Setting $x=0$ shows every such line goes through $P(0,-4)$ , so changing $a$ just rotates the line about $P$ . Draw it: the line sweeps across the segment between the position through $B$ and the position through $A$ . Find those two boundary slopes by substituting each endpoint, then take everything in between.

3 · Execute4 carry out the plan

1Find the fixed point

#9 Solve an Easier Related Problem 8.F.A.3

Put

x=0

: the

a

term vanishes, so every line in the family passes through the same point.

x=0 \Rightarrow y=a\cdot 0-4=-4

The lines are spokes turning around

P(0, -4)

.

2Slope when the line hits A

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

Substitute

A(-1, -2)

into

y=ax-4

.

-2=-a-4 \Rightarrow a=-2

That is the slope from

P

to

A

:

\frac{-2-(-4)}{-1-0}=-2

.

3Slope when the line hits B

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

Substitute

B(-5, -2)

into

y=ax-4

.

-2=-5a-4 \Rightarrow a=-\frac{2}{5}

That is the slope from

P

to

B

:

\frac{-2-(-4)}{-5-0}=-\frac{2}{5}

.

4Sweep between the two positions

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

The whole segment lies to the left of the

y

-axis, so as the line turns from passing through

A

to passing through

B

, it crosses every point of the segment, and its slope runs through every value from

-2

to

-\frac{2}{5}

. Any steeper or flatter line misses.

-2\le a\le -\frac{2}{5}

The two endpoints mark the extreme positions of the turning line.

Answer:

-2\le a\le -\frac{2}{5}

4 · Reviewdoes it hold up?

Try $a=-1$ (inside the range): $y=-x-4$ meets segment $AB$ , which lies on $y=-2$ , where $-x-4=-2$ , so $x=-2$ , between $-5$ and $-1$ . Try $a=-3$ (outside): $-3x-4=-2$ gives $x=-\frac{2}{3}$ , which is not between $-5$ and $-1$ , so it misses.

Another way: Algebra all the way: segment $AB$ lies on $y=-2$ for $-5\le x\le -1$ . Setting $ax-4=-2$ gives $x=\frac{2}{a}$ , and requiring $-5\le \frac{2}{a}\le -1$ leads to the same range.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing that every line $y=ax-4$ passes through $(0, -4)$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing the slope between the fixed point and each endpoint.

💡Takeaway. When a line is pinned at one point, changing its slope just swings it like a door -- the ends of the target tell you how far it can swing.

Variant 6 medium

The graph of the linear function $y=ax-3$ intersects the line segment joining the two points $A(-4, 5)$ and $B(-5, 1)$ . Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=ax-3$ passes through $(0, -3)$ , whatever $a$ is. Find all slopes $a$ for which this line touches the segment from $A(-4, 5)$ to $B(-5, 1)$ .

Givens

The line is $y=ax-3$ .
The segment joins $A(-4, 5)$ and $B(-5, 1)$ .

Unknowns

The range of values of $a$ .

Constraints

The line must meet the segment, endpoints included.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

Setting $x=0$ shows every such line goes through $P(0,-3)$ , so changing $a$ just rotates the line about $P$ . Draw it: the line sweeps across the segment between the position through $B$ and the position through $A$ . Find those two boundary slopes by substituting each endpoint, then take everything in between.

3 · Execute4 carry out the plan

1Find the fixed point

#9 Solve an Easier Related Problem 8.F.A.3

Put

x=0

: the

a

term vanishes, so every line in the family passes through the same point.

x=0 \Rightarrow y=a\cdot 0-3=-3

The lines are spokes turning around

P(0, -3)

.

2Slope when the line hits A

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

Substitute

A(-4, 5)

into

y=ax-3

.

5=-4a-3 \Rightarrow a=-2

That is the slope from

P

to

A

:

\frac{5-(-3)}{-4-0}=-2

.

3Slope when the line hits B

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

Substitute

B(-5, 1)

into

y=ax-3

.

1=-5a-3 \Rightarrow a=-\frac{4}{5}

That is the slope from

P

to

B

:

\frac{1-(-3)}{-5-0}=-\frac{4}{5}

.

4Sweep between the two positions

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

The whole segment lies to the left of the

y

-axis, so as the line turns from passing through

A

to passing through

B

, it crosses every point of the segment, and its slope runs through every value from

-2

to

-\frac{4}{5}

. Any steeper or flatter line misses.

-2\le a\le -\frac{4}{5}

The two endpoints mark the extreme positions of the turning line.

Answer:

-2\le a\le -\frac{4}{5}

4 · Reviewdoes it hold up?

Try $a=-1$ (inside the range): $y=-x-3$ meets segment $AB$ , which lies on $y=4x+21$ , where $-x-3=4x+21$ , so $x=-\frac{24}{5}$ , between $-5$ and $-4$ . Try $a=-3$ (outside): $-3x-3=4x+21$ gives $x=-\frac{24}{7}$ , which is not between $-5$ and $-4$ , so it misses.

Another way: Algebra all the way: segment $AB$ lies on $y=4x+21$ for $-5\le x\le -4$ . Setting $ax-3=4x+21$ gives $x=\frac{24}{a-4}$ , and requiring $-5\le \frac{24}{a-4}\le -4$ leads to the same range.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing that every line $y=ax-3$ passes through $(0, -3)$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing the slope between the fixed point and each endpoint.

💡Takeaway. When a line is pinned at one point, changing its slope just swings it like a door -- the ends of the target tell you how far it can swing.

Variant 7 easy

The graph of the linear function $y=ax-1$ intersects the line segment joining the two points $A(-2, 4)$ and $B(-5, 1)$ . Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=ax-1$ passes through $(0, -1)$ , whatever $a$ is. Find all slopes $a$ for which this line touches the segment from $A(-2, 4)$ to $B(-5, 1)$ .

Givens

The line is $y=ax-1$ .
The segment joins $A(-2, 4)$ and $B(-5, 1)$ .

Unknowns

The range of values of $a$ .

Constraints

The line must meet the segment, endpoints included.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

Setting $x=0$ shows every such line goes through $P(0,-1)$ , so changing $a$ just rotates the line about $P$ . Draw it: the line sweeps across the segment between the position through $B$ and the position through $A$ . Find those two boundary slopes by substituting each endpoint, then take everything in between.

3 · Execute4 carry out the plan

1Find the fixed point

#9 Solve an Easier Related Problem 8.F.A.3

Put

x=0

: the

a

term vanishes, so every line in the family passes through the same point.

x=0 \Rightarrow y=a\cdot 0-1=-1

The lines are spokes turning around

P(0, -1)

.

2Slope when the line hits A

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

Substitute

A(-2, 4)

into

y=ax-1

.

4=-2a-1 \Rightarrow a=-\frac{5}{2}

That is the slope from

P

to

A

:

\frac{4-(-1)}{-2-0}=-\frac{5}{2}

.

3Slope when the line hits B

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

Substitute

B(-5, 1)

into

y=ax-1

.

1=-5a-1 \Rightarrow a=-\frac{2}{5}

That is the slope from

P

to

B

:

\frac{1-(-1)}{-5-0}=-\frac{2}{5}

.

4Sweep between the two positions

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

The whole segment lies to the left of the

y

-axis, so as the line turns from passing through

A

to passing through

B

, it crosses every point of the segment, and its slope runs through every value from

-\frac{5}{2}

to

-\frac{2}{5}

. Any steeper or flatter line misses.

-\frac{5}{2}\le a\le -\frac{2}{5}

The two endpoints mark the extreme positions of the turning line.

Answer:

-\frac{5}{2}\le a\le -\frac{2}{5}

4 · Reviewdoes it hold up?

Try $a=-1$ (inside the range): $y=-x-1$ meets segment $AB$ , which lies on $y=x+6$ , where $-x-1=x+6$ , so $x=-\frac{7}{2}$ , between $-5$ and $-2$ . Try $a=-3$ (outside): $-3x-1=x+6$ gives $x=-\frac{7}{4}$ , which is not between $-5$ and $-2$ , so it misses.

Another way: Algebra all the way: segment $AB$ lies on $y=x+6$ for $-5\le x\le -2$ . Setting $ax-1=x+6$ gives $x=\frac{7}{a-1}$ , and requiring $-5\le \frac{7}{a-1}\le -2$ leads to the same range.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing that every line $y=ax-1$ passes through $(0, -1)$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing the slope between the fixed point and each endpoint.

💡Takeaway. When a line is pinned at one point, changing its slope just swings it like a door -- the ends of the target tell you how far it can swing.

Variant 8 hard

The graph of the linear function $y=ax+1$ intersects the line segment joining the two points $A(-5, 3)$ and $B(-6, -3)$ . Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=ax+1$ passes through $(0, 1)$ , whatever $a$ is. Find all slopes $a$ for which this line touches the segment from $A(-5, 3)$ to $B(-6, -3)$ .

Givens

The line is $y=ax+1$ .
The segment joins $A(-5, 3)$ and $B(-6, -3)$ .

Unknowns

The range of values of $a$ .

Constraints

The line must meet the segment, endpoints included.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

Setting $x=0$ shows every such line goes through $P(0,1)$ , so changing $a$ just rotates the line about $P$ . Draw it: the line sweeps across the segment between the position through $B$ and the position through $A$ . Find those two boundary slopes by substituting each endpoint, then take everything in between.

3 · Execute4 carry out the plan

1Find the fixed point

#9 Solve an Easier Related Problem 8.F.A.3

Put

x=0

: the

a

term vanishes, so every line in the family passes through the same point.

x=0 \Rightarrow y=a\cdot 0+1=1

The lines are spokes turning around

P(0, 1)

.

2Slope when the line hits A

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

Substitute

A(-5, 3)

into

y=ax+1

.

3=-5a+1 \Rightarrow a=-\frac{2}{5}

That is the slope from

P

to

A

:

\frac{3-1}{-5-0}=-\frac{2}{5}

.

3Slope when the line hits B

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

Substitute

B(-6, -3)

into

y=ax+1

.

-3=-6a+1 \Rightarrow a=\frac{2}{3}

That is the slope from

P

to

B

:

\frac{-3-1}{-6-0}=\frac{2}{3}

.

4Sweep between the two positions

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

The whole segment lies to the left of the

y

-axis, so as the line turns from passing through

A

to passing through

B

, it crosses every point of the segment, and its slope runs through every value from

-\frac{2}{5}

to

\frac{2}{3}

. Any steeper or flatter line misses.

-\frac{2}{5}\le a\le \frac{2}{3}

The two endpoints mark the extreme positions of the turning line.

Answer:

-\frac{2}{5}\le a\le \frac{2}{3}

4 · Reviewdoes it hold up?

Try $a=0$ (inside the range): $y=1$ meets segment $AB$ , which lies on $y=6x+33$ , where $1=6x+33$ , so $x=-\frac{16}{3}$ , between $-6$ and $-5$ . Try $a=-1$ (outside): $-x+1=6x+33$ gives $x=-\frac{32}{7}$ , which is not between $-6$ and $-5$ , so it misses.

Another way: Algebra all the way: segment $AB$ lies on $y=6x+33$ for $-6\le x\le -5$ . Setting $ax+1=6x+33$ gives $x=\frac{32}{a-6}$ , and requiring $-6\le \frac{32}{a-6}\le -5$ leads to the same range.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing that every line $y=ax+1$ passes through $(0, 1)$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing the slope between the fixed point and each endpoint.

💡Takeaway. When a line is pinned at one point, changing its slope just swings it like a door -- the ends of the target tell you how far it can swing.

Variant 9 hard

The graph of the linear function $y=ax-4$ intersects the line segment joining the two points $A(-3, 0)$ and $B(-6, -3)$ . Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=ax-4$ passes through $(0, -4)$ , whatever $a$ is. Find all slopes $a$ for which this line touches the segment from $A(-3, 0)$ to $B(-6, -3)$ .

Givens

The line is $y=ax-4$ .
The segment joins $A(-3, 0)$ and $B(-6, -3)$ .

Unknowns

The range of values of $a$ .

Constraints

The line must meet the segment, endpoints included.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

Setting $x=0$ shows every such line goes through $P(0,-4)$ , so changing $a$ just rotates the line about $P$ . Draw it: the line sweeps across the segment between the position through $B$ and the position through $A$ . Find those two boundary slopes by substituting each endpoint, then take everything in between.

3 · Execute4 carry out the plan

1Find the fixed point

#9 Solve an Easier Related Problem 8.F.A.3

Put

x=0

: the

a

term vanishes, so every line in the family passes through the same point.

x=0 \Rightarrow y=a\cdot 0-4=-4

The lines are spokes turning around

P(0, -4)

.

2Slope when the line hits A

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

Substitute

A(-3, 0)

into

y=ax-4

.

0=-3a-4 \Rightarrow a=-\frac{4}{3}

That is the slope from

P

to

A

:

\frac{0-(-4)}{-3-0}=-\frac{4}{3}

.

3Slope when the line hits B

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

Substitute

B(-6, -3)

into

y=ax-4

.

-3=-6a-4 \Rightarrow a=-\frac{1}{6}

That is the slope from

P

to

B

:

\frac{-3-(-4)}{-6-0}=-\frac{1}{6}

.

4Sweep between the two positions

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

The whole segment lies to the left of the

y

-axis, so as the line turns from passing through

A

to passing through

B

, it crosses every point of the segment, and its slope runs through every value from

-\frac{4}{3}

to

-\frac{1}{6}

. Any steeper or flatter line misses.

-\frac{4}{3}\le a\le -\frac{1}{6}

The two endpoints mark the extreme positions of the turning line.

Answer:

-\frac{4}{3}\le a\le -\frac{1}{6}

4 · Reviewdoes it hold up?

Try $a=-1$ (inside the range): $y=-x-4$ meets segment $AB$ , which lies on $y=x+3$ , where $-x-4=x+3$ , so $x=-\frac{7}{2}$ , between $-6$ and $-3$ . Try $a=-2$ (outside): $-2x-4=x+3$ gives $x=-\frac{7}{3}$ , which is not between $-6$ and $-3$ , so it misses.

Another way: Algebra all the way: segment $AB$ lies on $y=x+3$ for $-6\le x\le -3$ . Setting $ax-4=x+3$ gives $x=\frac{7}{a-1}$ , and requiring $-6\le \frac{7}{a-1}\le -3$ leads to the same range.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing that every line $y=ax-4$ passes through $(0, -4)$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing the slope between the fixed point and each endpoint.

💡Takeaway. When a line is pinned at one point, changing its slope just swings it like a door -- the ends of the target tell you how far it can swing.

Variant 10 medium

The graph of the linear function $y=ax+4$ intersects the line segment joining the two points $A(-2, 3)$ and $B(-6, -5)$ . Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=ax+4$ passes through $(0, 4)$ , whatever $a$ is. Find all slopes $a$ for which this line touches the segment from $A(-2, 3)$ to $B(-6, -5)$ .

Givens

The line is $y=ax+4$ .
The segment joins $A(-2, 3)$ and $B(-6, -5)$ .

Unknowns

The range of values of $a$ .

Constraints

The line must meet the segment, endpoints included.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

Setting $x=0$ shows every such line goes through $P(0,4)$ , so changing $a$ just rotates the line about $P$ . Draw it: the line sweeps across the segment between the position through $B$ and the position through $A$ . Find those two boundary slopes by substituting each endpoint, then take everything in between.

3 · Execute4 carry out the plan

1Find the fixed point

#9 Solve an Easier Related Problem 8.F.A.3

Put

x=0

: the

a

term vanishes, so every line in the family passes through the same point.

x=0 \Rightarrow y=a\cdot 0+4=4

The lines are spokes turning around

P(0, 4)

.

2Slope when the line hits A

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

Substitute

A(-2, 3)

into

y=ax+4

.

3=-2a+4 \Rightarrow a=\frac{1}{2}

That is the slope from

P

to

A

:

\frac{3-4}{-2-0}=\frac{1}{2}

.

3Slope when the line hits B

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

Substitute

B(-6, -5)

into

y=ax+4

.

-5=-6a+4 \Rightarrow a=\frac{3}{2}

That is the slope from

P

to

B

:

\frac{-5-4}{-6-0}=\frac{3}{2}

.

4Sweep between the two positions

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

The whole segment lies to the left of the

y

-axis, so as the line turns from passing through

A

to passing through

B

, it crosses every point of the segment, and its slope runs through every value from

\frac{1}{2}

to

\frac{3}{2}

. Any steeper or flatter line misses.

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

The two endpoints mark the extreme positions of the turning line.

Answer:

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

4 · Reviewdoes it hold up?

Try $a=1$ (inside the range): $y=x+4$ meets segment $AB$ , which lies on $y=2x+7$ , where $x+4=2x+7$ , so $x=-3$ , between $-6$ and $-2$ . Try $a=0$ (outside): $4=2x+7$ gives $x=-\frac{3}{2}$ , which is not between $-6$ and $-2$ , so it misses.

Another way: Algebra all the way: segment $AB$ lies on $y=2x+7$ for $-6\le x\le -2$ . Setting $ax+4=2x+7$ gives $x=\frac{3}{a-2}$ , and requiring $-6\le \frac{3}{a-2}\le -2$ leads to the same range.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing that every line $y=ax+4$ passes through $(0, 4)$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing the slope between the fixed point and each endpoint.

💡Takeaway. When a line is pinned at one point, changing its slope just swings it like a door -- the ends of the target tell you how far it can swing.

Variant 11 hard

The graph of the linear function $y=ax-1$ intersects the line segment joining the two points $A(-4, 0)$ and $B(-6, 6)$ . Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=ax-1$ passes through $(0, -1)$ , whatever $a$ is. Find all slopes $a$ for which this line touches the segment from $A(-4, 0)$ to $B(-6, 6)$ .

Givens

The line is $y=ax-1$ .
The segment joins $A(-4, 0)$ and $B(-6, 6)$ .

Unknowns

The range of values of $a$ .

Constraints

The line must meet the segment, endpoints included.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

Setting $x=0$ shows every such line goes through $P(0,-1)$ , so changing $a$ just rotates the line about $P$ . Draw it: the line sweeps across the segment between the position through $B$ and the position through $A$ . Find those two boundary slopes by substituting each endpoint, then take everything in between.

3 · Execute4 carry out the plan

1Find the fixed point

#9 Solve an Easier Related Problem 8.F.A.3

Put

x=0

: the

a

term vanishes, so every line in the family passes through the same point.

x=0 \Rightarrow y=a\cdot 0-1=-1

The lines are spokes turning around

P(0, -1)

.

2Slope when the line hits A

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

Substitute

A(-4, 0)

into

y=ax-1

.

0=-4a-1 \Rightarrow a=-\frac{1}{4}

That is the slope from

P

to

A

:

\frac{0-(-1)}{-4-0}=-\frac{1}{4}

.

3Slope when the line hits B

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

Substitute

B(-6, 6)

into

y=ax-1

.

6=-6a-1 \Rightarrow a=-\frac{7}{6}

That is the slope from

P

to

B

:

\frac{6-(-1)}{-6-0}=-\frac{7}{6}

.

4Sweep between the two positions

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

The whole segment lies to the left of the

y

-axis, so as the line turns from passing through

A

to passing through

B

, it crosses every point of the segment, and its slope runs through every value from

-\frac{7}{6}

to

-\frac{1}{4}

. Any steeper or flatter line misses.

-\frac{7}{6}\le a\le -\frac{1}{4}

The two endpoints mark the extreme positions of the turning line.

Answer:

-\frac{7}{6}\le a\le -\frac{1}{4}

4 · Reviewdoes it hold up?

Try $a=-1$ (inside the range): $y=-x-1$ meets segment $AB$ , which lies on $y=-3x-12$ , where $-x-1=-3x-12$ , so $x=-\frac{11}{2}$ , between $-6$ and $-4$ . Try $a=-2$ (outside): $-2x-1=-3x-12$ gives $x=-11$ , which is not between $-6$ and $-4$ , so it misses.

Another way: Algebra all the way: segment $AB$ lies on $y=-3x-12$ for $-6\le x\le -4$ . Setting $ax-1=-3x-12$ gives $x=\frac{-11}{a+3}$ , and requiring $-6\le \frac{-11}{a+3}\le -4$ leads to the same range.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing that every line $y=ax-1$ passes through $(0, -1)$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing the slope between the fixed point and each endpoint.

💡Takeaway. When a line is pinned at one point, changing its slope just swings it like a door -- the ends of the target tell you how far it can swing.

Variant 12 hard

The graph of the linear function $y=ax+2$ intersects the line segment joining the two points $A(-4, 7)$ and $B(-6, -1)$ . Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Every line $y=ax+2$ passes through $(0, 2)$ , whatever $a$ is. Find all slopes $a$ for which this line touches the segment from $A(-4, 7)$ to $B(-6, -1)$ .

Givens

The line is $y=ax+2$ .
The segment joins $A(-4, 7)$ and $B(-6, -1)$ .

Unknowns

The range of values of $a$ .

Constraints

The line must meet the segment, endpoints included.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

Setting $x=0$ shows every such line goes through $P(0,2)$ , so changing $a$ just rotates the line about $P$ . Draw it: the line sweeps across the segment between the position through $B$ and the position through $A$ . Find those two boundary slopes by substituting each endpoint, then take everything in between.

3 · Execute4 carry out the plan

1Find the fixed point

#9 Solve an Easier Related Problem 8.F.A.3

Put

x=0

: the

a

term vanishes, so every line in the family passes through the same point.

x=0 \Rightarrow y=a\cdot 0+2=2

The lines are spokes turning around

P(0, 2)

.

2Slope when the line hits A

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

Substitute

A(-4, 7)

into

y=ax+2

.

7=-4a+2 \Rightarrow a=-\frac{5}{4}

That is the slope from

P

to

A

:

\frac{7-2}{-4-0}=-\frac{5}{4}

.

3Slope when the line hits B

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

Substitute

B(-6, -1)

into

y=ax+2

.

-1=-6a+2 \Rightarrow a=\frac{1}{2}

That is the slope from

P

to

B

:

\frac{-1-2}{-6-0}=\frac{1}{2}

.

4Sweep between the two positions

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

The whole segment lies to the left of the

y

-axis, so as the line turns from passing through

A

to passing through

B

, it crosses every point of the segment, and its slope runs through every value from

-\frac{5}{4}

to

\frac{1}{2}

. Any steeper or flatter line misses.

-\frac{5}{4}\le a\le \frac{1}{2}

The two endpoints mark the extreme positions of the turning line.

Answer:

-\frac{5}{4}\le a\le \frac{1}{2}

4 · Reviewdoes it hold up?

Try $a=0$ (inside the range): $y=2$ meets segment $AB$ , which lies on $y=4x+23$ , where $2=4x+23$ , so $x=-\frac{21}{4}$ , between $-6$ and $-4$ . Try $a=-2$ (outside): $-2x+2=4x+23$ gives $x=-\frac{7}{2}$ , which is not between $-6$ and $-4$ , so it misses.

Another way: Algebra all the way: segment $AB$ lies on $y=4x+23$ for $-6\le x\le -4$ . Setting $ax+2=4x+23$ gives $x=\frac{21}{a-4}$ , and requiring $-6\le \frac{21}{a-4}\le -4$ leads to the same range.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Seeing that every line $y=ax+2$ passes through $(0, 2)$.
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points — Computing the slope between the fixed point and each endpoint.

💡Takeaway. When a line is pinned at one point, changing its slope just swings it like a door -- the ends of the target tell you how far it can swing.

Rotate a line about its fixed point · 12 practice problems

Generated variants — 12

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