Practice · When three lines fail to make a triangle · Sensim Math

Variant 1 easy

The two lines $y=x-2$ and $y=mx+4$ meet at the point $(3, 1)$ . A third line $y=ax-3$ is drawn. Find all values of the constant $a$ for which the three lines do not enclose a triangle. ( $m$ is a constant.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two lines already cross at $(3, 1)$ . A third line with $y$ -intercept $-3$ is added. For which slopes $a$ do the three lines fail to form a triangle?

Givens

$y=x-2$ and $y=mx+4$ both pass through $(3, 1)$ .
The third line is $y=ax-3$ .

Unknowns

Every value of $a$ that prevents a triangle.

Constraints

Three lines enclose a triangle only when each pair crosses and the three crossing points are different.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#1 Draw a Diagram

List the ways three lines can fail: the third line is parallel to the first, parallel to the second, or passes through the point where the first two already meet. (Two lines can't coincide, since the $y$ -intercepts $-2$ , $4$ , $-3$ are all different.) First find $m$ from the given crossing point, then turn each case into an equation for $a$ .

3 · Execute5 carry out the plan

1Find m

#13 Convert to Algebra 8.F.A.3

The line

y=mx+4

passes through

(3, 1)

.

1=3m+4 \Rightarrow m=-1

Now all three lines are known except

a

.

2Case 1: all three meet at one point

#13 Convert to Algebra 8.EE.C.8

If

y=ax-3

also passes through

(3, 1)

, the three crossing points collapse into one.

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

Three lines through one point make a star, not a triangle.

3Case 2: parallel to y = x - 2

#2 Make a Systematic List 8.EE.C.8

Equal slopes with different

y

-intercepts (

-3\ne -2

) means parallel, so those two never meet.

a=1

A missing corner means no triangle.

4Case 3: parallel to y = -x + 4

#2 Make a Systematic List 8.EE.C.8

Likewise, matching the slope

m=-1

makes the third line parallel to the second (

-3\ne 4

).

a=-1

Again one pair never crosses.

5Collect the cases

#2 Make a Systematic List 8.EE.C.8

For any other

a

, the third line crosses each of the others at a point different from

(3, 1)

, and those two points are different from each other, so a triangle forms.

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

Exactly three bad values.

Answer:

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

4 · Reviewdoes it hold up?

Check a good value, $a=0$ : $y=-3$ meets $y=x-2$ at $(-1, -3)$ and $y=-x+4$ at $(7, -3)$ ; with $(3, 1)$ that is three distinct corners, a real triangle. So only the special values break it.

Another way: Draw a Diagram: sketch the two lines through $(3, 1)$ and the fixed point $(0, -3)$ of the third line, then rotate the third line about $(0, -3)$ . It stops making a triangle exactly when it lines up with $(3, 1)$ or becomes parallel to one of the two lines.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Using slope and $y$-intercept to tell when lines are parallel.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Understanding that two lines meet at the solution of their system, and that parallel lines have none.

💡Takeaway. To make a triangle you need three different corners -- so hunt for the ways to lose a corner: parallel lines, or all three crossing at one spot.

Variant 2 easy

The two lines $y=x+4$ and $y=mx-4$ meet at the point $(-2, 2)$ . A third line $y=ax+3$ is drawn. Find all values of the constant $a$ for which the three lines do not enclose a triangle. ( $m$ is a constant.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two lines already cross at $(-2, 2)$ . A third line with $y$ -intercept $3$ is added. For which slopes $a$ do the three lines fail to form a triangle?

Givens

$y=x+4$ and $y=mx-4$ both pass through $(-2, 2)$ .
The third line is $y=ax+3$ .

Unknowns

Every value of $a$ that prevents a triangle.

Constraints

Three lines enclose a triangle only when each pair crosses and the three crossing points are different.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#1 Draw a Diagram

List the ways three lines can fail: the third line is parallel to the first, parallel to the second, or passes through the point where the first two already meet. (Two lines can't coincide, since the $y$ -intercepts $4$ , $-4$ , $3$ are all different.) First find $m$ from the given crossing point, then turn each case into an equation for $a$ .

3 · Execute5 carry out the plan

1Find m

#13 Convert to Algebra 8.F.A.3

The line

y=mx-4

passes through

(-2, 2)

.

2=-2m-4 \Rightarrow m=-3

Now all three lines are known except

a

.

2Case 1: all three meet at one point

#13 Convert to Algebra 8.EE.C.8

If

y=ax+3

also passes through

(-2, 2)

, the three crossing points collapse into one.

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

Three lines through one point make a star, not a triangle.

3Case 2: parallel to y = x + 4

#2 Make a Systematic List 8.EE.C.8

Equal slopes with different

y

-intercepts (

3\ne 4

) means parallel, so those two never meet.

a=1

A missing corner means no triangle.

4Case 3: parallel to y = -3x - 4

#2 Make a Systematic List 8.EE.C.8

Likewise, matching the slope

m=-3

makes the third line parallel to the second (

3\ne -4

).

a=-3

Again one pair never crosses.

5Collect the cases

#2 Make a Systematic List 8.EE.C.8

For any other

a

, the third line crosses each of the others at a point different from

(-2, 2)

, and those two points are different from each other, so a triangle forms.

a=-3,\ \frac{1}{2},\ 1

Exactly three bad values.

Answer:

a=-3,\ \frac{1}{2},\ 1

4 · Reviewdoes it hold up?

Check a good value, $a=0$ : $y=3$ meets $y=x+4$ at $(-1, 3)$ and $y=-3x-4$ at $(-\frac{7}{3}, 3)$ ; with $(-2, 2)$ that is three distinct corners, a real triangle. So only the special values break it.

Another way: Draw a Diagram: sketch the two lines through $(-2, 2)$ and the fixed point $(0, 3)$ of the third line, then rotate the third line about $(0, 3)$ . It stops making a triangle exactly when it lines up with $(-2, 2)$ or becomes parallel to one of the two lines.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Using slope and $y$-intercept to tell when lines are parallel.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Understanding that two lines meet at the solution of their system, and that parallel lines have none.

💡Takeaway. To make a triangle you need three different corners -- so hunt for the ways to lose a corner: parallel lines, or all three crossing at one spot.

Variant 3 easy

The two lines $y=-x+2$ and $y=mx+5$ meet at the point $(3, -1)$ . A third line $y=ax-4$ is drawn. Find all values of the constant $a$ for which the three lines do not enclose a triangle. ( $m$ is a constant.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two lines already cross at $(3, -1)$ . A third line with $y$ -intercept $-4$ is added. For which slopes $a$ do the three lines fail to form a triangle?

Givens

$y=-x+2$ and $y=mx+5$ both pass through $(3, -1)$ .
The third line is $y=ax-4$ .

Unknowns

Every value of $a$ that prevents a triangle.

Constraints

Three lines enclose a triangle only when each pair crosses and the three crossing points are different.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#1 Draw a Diagram

List the ways three lines can fail: the third line is parallel to the first, parallel to the second, or passes through the point where the first two already meet. (Two lines can't coincide, since the $y$ -intercepts $2$ , $5$ , $-4$ are all different.) First find $m$ from the given crossing point, then turn each case into an equation for $a$ .

3 · Execute5 carry out the plan

1Find m

#13 Convert to Algebra 8.F.A.3

The line

y=mx+5

passes through

(3, -1)

.

-1=3m+5 \Rightarrow m=-2

Now all three lines are known except

a

.

2Case 1: all three meet at one point

#13 Convert to Algebra 8.EE.C.8

If

y=ax-4

also passes through

(3, -1)

, the three crossing points collapse into one.

-1=3a-4 \Rightarrow a=1

Three lines through one point make a star, not a triangle.

3Case 2: parallel to y = -x + 2

#2 Make a Systematic List 8.EE.C.8

Equal slopes with different

y

-intercepts (

-4\ne 2

) means parallel, so those two never meet.

a=-1

A missing corner means no triangle.

4Case 3: parallel to y = -2x + 5

#2 Make a Systematic List 8.EE.C.8

Likewise, matching the slope

m=-2

makes the third line parallel to the second (

-4\ne 5

).

a=-2

Again one pair never crosses.

5Collect the cases

#2 Make a Systematic List 8.EE.C.8

For any other

a

, the third line crosses each of the others at a point different from

(3, -1)

, and those two points are different from each other, so a triangle forms.

a=-2,\ -1,\ 1

Exactly three bad values.

Answer:

a=-2,\ -1,\ 1

4 · Reviewdoes it hold up?

Check a good value, $a=0$ : $y=-4$ meets $y=-x+2$ at $(6, -4)$ and $y=-2x+5$ at $(\frac{9}{2}, -4)$ ; with $(3, -1)$ that is three distinct corners, a real triangle. So only the special values break it.

Another way: Draw a Diagram: sketch the two lines through $(3, -1)$ and the fixed point $(0, -4)$ of the third line, then rotate the third line about $(0, -4)$ . It stops making a triangle exactly when it lines up with $(3, -1)$ or becomes parallel to one of the two lines.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Using slope and $y$-intercept to tell when lines are parallel.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Understanding that two lines meet at the solution of their system, and that parallel lines have none.

💡Takeaway. To make a triangle you need three different corners -- so hunt for the ways to lose a corner: parallel lines, or all three crossing at one spot.

Variant 4 easy

The two lines $y=-x+6$ and $y=mx-2$ meet at the point $(2, 4)$ . A third line $y=ax+1$ is drawn. Find all values of the constant $a$ for which the three lines do not enclose a triangle. ( $m$ is a constant.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two lines already cross at $(2, 4)$ . A third line with $y$ -intercept $1$ is added. For which slopes $a$ do the three lines fail to form a triangle?

Givens

$y=-x+6$ and $y=mx-2$ both pass through $(2, 4)$ .
The third line is $y=ax+1$ .

Unknowns

Every value of $a$ that prevents a triangle.

Constraints

Three lines enclose a triangle only when each pair crosses and the three crossing points are different.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#1 Draw a Diagram

List the ways three lines can fail: the third line is parallel to the first, parallel to the second, or passes through the point where the first two already meet. (Two lines can't coincide, since the $y$ -intercepts $6$ , $-2$ , $1$ are all different.) First find $m$ from the given crossing point, then turn each case into an equation for $a$ .

3 · Execute5 carry out the plan

1Find m

#13 Convert to Algebra 8.F.A.3

The line

y=mx-2

passes through

(2, 4)

.

4=2m-2 \Rightarrow m=3

Now all three lines are known except

a

.

2Case 1: all three meet at one point

#13 Convert to Algebra 8.EE.C.8

If

y=ax+1

also passes through

(2, 4)

, the three crossing points collapse into one.

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

Three lines through one point make a star, not a triangle.

3Case 2: parallel to y = -x + 6

#2 Make a Systematic List 8.EE.C.8

Equal slopes with different

y

-intercepts (

1\ne 6

) means parallel, so those two never meet.

a=-1

A missing corner means no triangle.

4Case 3: parallel to y = 3x - 2

#2 Make a Systematic List 8.EE.C.8

Likewise, matching the slope

m=3

makes the third line parallel to the second (

1\ne -2

).

a=3

Again one pair never crosses.

5Collect the cases

#2 Make a Systematic List 8.EE.C.8

For any other

a

, the third line crosses each of the others at a point different from

(2, 4)

, and those two points are different from each other, so a triangle forms.

a=-1,\ \frac{3}{2},\ 3

Exactly three bad values.

Answer:

a=-1,\ \frac{3}{2},\ 3

4 · Reviewdoes it hold up?

Check a good value, $a=0$ : $y=1$ meets $y=-x+6$ at $(5, 1)$ and $y=3x-2$ at $(1, 1)$ ; with $(2, 4)$ that is three distinct corners, a real triangle. So only the special values break it.

Another way: Draw a Diagram: sketch the two lines through $(2, 4)$ and the fixed point $(0, 1)$ of the third line, then rotate the third line about $(0, 1)$ . It stops making a triangle exactly when it lines up with $(2, 4)$ or becomes parallel to one of the two lines.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Using slope and $y$-intercept to tell when lines are parallel.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Understanding that two lines meet at the solution of their system, and that parallel lines have none.

💡Takeaway. To make a triangle you need three different corners -- so hunt for the ways to lose a corner: parallel lines, or all three crossing at one spot.

Variant 5 medium

The two lines $y=2x-3$ and $y=mx+3$ meet at the point $(2, 1)$ . A third line $y=ax-2$ is drawn. Find all values of the constant $a$ for which the three lines do not enclose a triangle. ( $m$ is a constant.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two lines already cross at $(2, 1)$ . A third line with $y$ -intercept $-2$ is added. For which slopes $a$ do the three lines fail to form a triangle?

Givens

$y=2x-3$ and $y=mx+3$ both pass through $(2, 1)$ .
The third line is $y=ax-2$ .

Unknowns

Every value of $a$ that prevents a triangle.

Constraints

Three lines enclose a triangle only when each pair crosses and the three crossing points are different.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#1 Draw a Diagram

List the ways three lines can fail: the third line is parallel to the first, parallel to the second, or passes through the point where the first two already meet. (Two lines can't coincide, since the $y$ -intercepts $-3$ , $3$ , $-2$ are all different.) First find $m$ from the given crossing point, then turn each case into an equation for $a$ .

3 · Execute5 carry out the plan

1Find m

#13 Convert to Algebra 8.F.A.3

The line

y=mx+3

passes through

(2, 1)

.

1=2m+3 \Rightarrow m=-1

Now all three lines are known except

a

.

2Case 1: all three meet at one point

#13 Convert to Algebra 8.EE.C.8

If

y=ax-2

also passes through

(2, 1)

, the three crossing points collapse into one.

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

Three lines through one point make a star, not a triangle.

3Case 2: parallel to y = 2x - 3

#2 Make a Systematic List 8.EE.C.8

Equal slopes with different

y

-intercepts (

-2\ne -3

) means parallel, so those two never meet.

a=2

A missing corner means no triangle.

4Case 3: parallel to y = -x + 3

#2 Make a Systematic List 8.EE.C.8

Likewise, matching the slope

m=-1

makes the third line parallel to the second (

-2\ne 3

).

a=-1

Again one pair never crosses.

5Collect the cases

#2 Make a Systematic List 8.EE.C.8

For any other

a

, the third line crosses each of the others at a point different from

(2, 1)

, and those two points are different from each other, so a triangle forms.

a=-1,\ \frac{3}{2},\ 2

Exactly three bad values.

Answer:

a=-1,\ \frac{3}{2},\ 2

4 · Reviewdoes it hold up?

Check a good value, $a=0$ : $y=-2$ meets $y=2x-3$ at $(\frac{1}{2}, -2)$ and $y=-x+3$ at $(5, -2)$ ; with $(2, 1)$ that is three distinct corners, a real triangle. So only the special values break it.

Another way: Draw a Diagram: sketch the two lines through $(2, 1)$ and the fixed point $(0, -2)$ of the third line, then rotate the third line about $(0, -2)$ . It stops making a triangle exactly when it lines up with $(2, 1)$ or becomes parallel to one of the two lines.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Using slope and $y$-intercept to tell when lines are parallel.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Understanding that two lines meet at the solution of their system, and that parallel lines have none.

💡Takeaway. To make a triangle you need three different corners -- so hunt for the ways to lose a corner: parallel lines, or all three crossing at one spot.

Variant 6 medium

The two lines $y=3x+2$ and $y=mx+3$ meet at the point $(-1, -1)$ . A third line $y=ax-2$ is drawn. Find all values of the constant $a$ for which the three lines do not enclose a triangle. ( $m$ is a constant.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two lines already cross at $(-1, -1)$ . A third line with $y$ -intercept $-2$ is added. For which slopes $a$ do the three lines fail to form a triangle?

Givens

$y=3x+2$ and $y=mx+3$ both pass through $(-1, -1)$ .
The third line is $y=ax-2$ .

Unknowns

Every value of $a$ that prevents a triangle.

Constraints

Three lines enclose a triangle only when each pair crosses and the three crossing points are different.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#1 Draw a Diagram

List the ways three lines can fail: the third line is parallel to the first, parallel to the second, or passes through the point where the first two already meet. (Two lines can't coincide, since the $y$ -intercepts $2$ , $3$ , $-2$ are all different.) First find $m$ from the given crossing point, then turn each case into an equation for $a$ .

3 · Execute5 carry out the plan

1Find m

#13 Convert to Algebra 8.F.A.3

The line

y=mx+3

passes through

(-1, -1)

.

-1=-m+3 \Rightarrow m=4

Now all three lines are known except

a

.

2Case 1: all three meet at one point

#13 Convert to Algebra 8.EE.C.8

If

y=ax-2

also passes through

(-1, -1)

, the three crossing points collapse into one.

-1=-a-2 \Rightarrow a=-1

Three lines through one point make a star, not a triangle.

3Case 2: parallel to y = 3x + 2

#2 Make a Systematic List 8.EE.C.8

Equal slopes with different

y

-intercepts (

-2\ne 2

) means parallel, so those two never meet.

a=3

A missing corner means no triangle.

4Case 3: parallel to y = 4x + 3

#2 Make a Systematic List 8.EE.C.8

Likewise, matching the slope

m=4

makes the third line parallel to the second (

-2\ne 3

).

a=4

Again one pair never crosses.

5Collect the cases

#2 Make a Systematic List 8.EE.C.8

For any other

a

, the third line crosses each of the others at a point different from

(-1, -1)

, and those two points are different from each other, so a triangle forms.

a=-1,\ 3,\ 4

Exactly three bad values.

Answer:

a=-1,\ 3,\ 4

4 · Reviewdoes it hold up?

Check a good value, $a=0$ : $y=-2$ meets $y=3x+2$ at $(-\frac{4}{3}, -2)$ and $y=4x+3$ at $(-\frac{5}{4}, -2)$ ; with $(-1, -1)$ that is three distinct corners, a real triangle. So only the special values break it.

Another way: Draw a Diagram: sketch the two lines through $(-1, -1)$ and the fixed point $(0, -2)$ of the third line, then rotate the third line about $(0, -2)$ . It stops making a triangle exactly when it lines up with $(-1, -1)$ or becomes parallel to one of the two lines.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Using slope and $y$-intercept to tell when lines are parallel.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Understanding that two lines meet at the solution of their system, and that parallel lines have none.

💡Takeaway. To make a triangle you need three different corners -- so hunt for the ways to lose a corner: parallel lines, or all three crossing at one spot.

Variant 7 medium

The two lines $y=x+3$ and $y=mx-1$ meet at the point $(4, 7)$ . A third line $y=ax+5$ is drawn. Find all values of the constant $a$ for which the three lines do not enclose a triangle. ( $m$ is a constant.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two lines already cross at $(4, 7)$ . A third line with $y$ -intercept $5$ is added. For which slopes $a$ do the three lines fail to form a triangle?

Givens

$y=x+3$ and $y=mx-1$ both pass through $(4, 7)$ .
The third line is $y=ax+5$ .

Unknowns

Every value of $a$ that prevents a triangle.

Constraints

Three lines enclose a triangle only when each pair crosses and the three crossing points are different.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#1 Draw a Diagram

List the ways three lines can fail: the third line is parallel to the first, parallel to the second, or passes through the point where the first two already meet. (Two lines can't coincide, since the $y$ -intercepts $3$ , $-1$ , $5$ are all different.) First find $m$ from the given crossing point, then turn each case into an equation for $a$ .

3 · Execute5 carry out the plan

1Find m

#13 Convert to Algebra 8.F.A.3

The line

y=mx-1

passes through

(4, 7)

.

7=4m-1 \Rightarrow m=2

Now all three lines are known except

a

.

2Case 1: all three meet at one point

#13 Convert to Algebra 8.EE.C.8

If

y=ax+5

also passes through

(4, 7)

, the three crossing points collapse into one.

7=4a+5 \Rightarrow a=\frac{1}{2}

Three lines through one point make a star, not a triangle.

3Case 2: parallel to y = x + 3

#2 Make a Systematic List 8.EE.C.8

Equal slopes with different

y

-intercepts (

5\ne 3

) means parallel, so those two never meet.

a=1

A missing corner means no triangle.

4Case 3: parallel to y = 2x - 1

#2 Make a Systematic List 8.EE.C.8

Likewise, matching the slope

m=2

makes the third line parallel to the second (

5\ne -1

).

a=2

Again one pair never crosses.

5Collect the cases

#2 Make a Systematic List 8.EE.C.8

For any other

a

, the third line crosses each of the others at a point different from

(4, 7)

, and those two points are different from each other, so a triangle forms.

a=\frac{1}{2},\ 1,\ 2

Exactly three bad values.

Answer:

a=\frac{1}{2},\ 1,\ 2

4 · Reviewdoes it hold up?

Check a good value, $a=0$ : $y=5$ meets $y=x+3$ at $(2, 5)$ and $y=2x-1$ at $(3, 5)$ ; with $(4, 7)$ that is three distinct corners, a real triangle. So only the special values break it.

Another way: Draw a Diagram: sketch the two lines through $(4, 7)$ and the fixed point $(0, 5)$ of the third line, then rotate the third line about $(0, 5)$ . It stops making a triangle exactly when it lines up with $(4, 7)$ or becomes parallel to one of the two lines.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Using slope and $y$-intercept to tell when lines are parallel.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Understanding that two lines meet at the solution of their system, and that parallel lines have none.

💡Takeaway. To make a triangle you need three different corners -- so hunt for the ways to lose a corner: parallel lines, or all three crossing at one spot.

Variant 8 medium

The two lines $y=2x+2$ and $y=mx+1$ meet at the point $(1, 4)$ . A third line $y=ax-1$ is drawn. Find all values of the constant $a$ for which the three lines do not enclose a triangle. ( $m$ is a constant.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two lines already cross at $(1, 4)$ . A third line with $y$ -intercept $-1$ is added. For which slopes $a$ do the three lines fail to form a triangle?

Givens

$y=2x+2$ and $y=mx+1$ both pass through $(1, 4)$ .
The third line is $y=ax-1$ .

Unknowns

Every value of $a$ that prevents a triangle.

Constraints

Three lines enclose a triangle only when each pair crosses and the three crossing points are different.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#1 Draw a Diagram

List the ways three lines can fail: the third line is parallel to the first, parallel to the second, or passes through the point where the first two already meet. (Two lines can't coincide, since the $y$ -intercepts $2$ , $1$ , $-1$ are all different.) First find $m$ from the given crossing point, then turn each case into an equation for $a$ .

3 · Execute5 carry out the plan

1Find m

#13 Convert to Algebra 8.F.A.3

The line

y=mx+1

passes through

(1, 4)

.

4=m+1 \Rightarrow m=3

Now all three lines are known except

a

.

2Case 1: all three meet at one point

#13 Convert to Algebra 8.EE.C.8

If

y=ax-1

also passes through

(1, 4)

, the three crossing points collapse into one.

4=a-1 \Rightarrow a=5

Three lines through one point make a star, not a triangle.

3Case 2: parallel to y = 2x + 2

#2 Make a Systematic List 8.EE.C.8

Equal slopes with different

y

-intercepts (

-1\ne 2

) means parallel, so those two never meet.

a=2

A missing corner means no triangle.

4Case 3: parallel to y = 3x + 1

#2 Make a Systematic List 8.EE.C.8

Likewise, matching the slope

m=3

makes the third line parallel to the second (

-1\ne 1

).

a=3

Again one pair never crosses.

5Collect the cases

#2 Make a Systematic List 8.EE.C.8

For any other

a

, the third line crosses each of the others at a point different from

(1, 4)

, and those two points are different from each other, so a triangle forms.

a=2,\ 3,\ 5

Exactly three bad values.

Answer:

a=2,\ 3,\ 5

4 · Reviewdoes it hold up?

Check a good value, $a=0$ : $y=-1$ meets $y=2x+2$ at $(-\frac{3}{2}, -1)$ and $y=3x+1$ at $(-\frac{2}{3}, -1)$ ; with $(1, 4)$ that is three distinct corners, a real triangle. So only the special values break it.

Another way: Draw a Diagram: sketch the two lines through $(1, 4)$ and the fixed point $(0, -1)$ of the third line, then rotate the third line about $(0, -1)$ . It stops making a triangle exactly when it lines up with $(1, 4)$ or becomes parallel to one of the two lines.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Using slope and $y$-intercept to tell when lines are parallel.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Understanding that two lines meet at the solution of their system, and that parallel lines have none.

💡Takeaway. To make a triangle you need three different corners -- so hunt for the ways to lose a corner: parallel lines, or all three crossing at one spot.

Variant 9 hard

The two lines $y=3x-1$ and $y=mx+1$ meet at the point $(2, 5)$ . A third line $y=ax+2$ is drawn. Find all values of the constant $a$ for which the three lines do not enclose a triangle. ( $m$ is a constant.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two lines already cross at $(2, 5)$ . A third line with $y$ -intercept $2$ is added. For which slopes $a$ do the three lines fail to form a triangle?

Givens

$y=3x-1$ and $y=mx+1$ both pass through $(2, 5)$ .
The third line is $y=ax+2$ .

Unknowns

Every value of $a$ that prevents a triangle.

Constraints

Three lines enclose a triangle only when each pair crosses and the three crossing points are different.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#1 Draw a Diagram

List the ways three lines can fail: the third line is parallel to the first, parallel to the second, or passes through the point where the first two already meet. (Two lines can't coincide, since the $y$ -intercepts $-1$ , $1$ , $2$ are all different.) First find $m$ from the given crossing point, then turn each case into an equation for $a$ .

3 · Execute5 carry out the plan

1Find m

#13 Convert to Algebra 8.F.A.3

The line

y=mx+1

passes through

(2, 5)

.

5=2m+1 \Rightarrow m=2

Now all three lines are known except

a

.

2Case 1: all three meet at one point

#13 Convert to Algebra 8.EE.C.8

If

y=ax+2

also passes through

(2, 5)

, the three crossing points collapse into one.

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

Three lines through one point make a star, not a triangle.

3Case 2: parallel to y = 3x - 1

#2 Make a Systematic List 8.EE.C.8

Equal slopes with different

y

-intercepts (

2\ne -1

) means parallel, so those two never meet.

a=3

A missing corner means no triangle.

4Case 3: parallel to y = 2x + 1

#2 Make a Systematic List 8.EE.C.8

Likewise, matching the slope

m=2

makes the third line parallel to the second (

2\ne 1

).

a=2

Again one pair never crosses.

5Collect the cases

#2 Make a Systematic List 8.EE.C.8

For any other

a

, the third line crosses each of the others at a point different from

(2, 5)

, and those two points are different from each other, so a triangle forms.

a=\frac{3}{2},\ 2,\ 3

Exactly three bad values.

Answer:

a=\frac{3}{2},\ 2,\ 3

4 · Reviewdoes it hold up?

Check a good value, $a=0$ : $y=2$ meets $y=3x-1$ at $(1, 2)$ and $y=2x+1$ at $(\frac{1}{2}, 2)$ ; with $(2, 5)$ that is three distinct corners, a real triangle. So only the special values break it.

Another way: Draw a Diagram: sketch the two lines through $(2, 5)$ and the fixed point $(0, 2)$ of the third line, then rotate the third line about $(0, 2)$ . It stops making a triangle exactly when it lines up with $(2, 5)$ or becomes parallel to one of the two lines.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Using slope and $y$-intercept to tell when lines are parallel.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Understanding that two lines meet at the solution of their system, and that parallel lines have none.

💡Takeaway. To make a triangle you need three different corners -- so hunt for the ways to lose a corner: parallel lines, or all three crossing at one spot.

Variant 10 hard

The two lines $y=2x+1$ and $y=mx-2$ meet at the point $(3, 7)$ . A third line $y=ax+4$ is drawn. Find all values of the constant $a$ for which the three lines do not enclose a triangle. ( $m$ is a constant.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two lines already cross at $(3, 7)$ . A third line with $y$ -intercept $4$ is added. For which slopes $a$ do the three lines fail to form a triangle?

Givens

$y=2x+1$ and $y=mx-2$ both pass through $(3, 7)$ .
The third line is $y=ax+4$ .

Unknowns

Every value of $a$ that prevents a triangle.

Constraints

Three lines enclose a triangle only when each pair crosses and the three crossing points are different.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#1 Draw a Diagram

List the ways three lines can fail: the third line is parallel to the first, parallel to the second, or passes through the point where the first two already meet. (Two lines can't coincide, since the $y$ -intercepts $1$ , $-2$ , $4$ are all different.) First find $m$ from the given crossing point, then turn each case into an equation for $a$ .

3 · Execute5 carry out the plan

1Find m

#13 Convert to Algebra 8.F.A.3

The line

y=mx-2

passes through

(3, 7)

.

7=3m-2 \Rightarrow m=3

Now all three lines are known except

a

.

2Case 1: all three meet at one point

#13 Convert to Algebra 8.EE.C.8

If

y=ax+4

also passes through

(3, 7)

, the three crossing points collapse into one.

7=3a+4 \Rightarrow a=1

Three lines through one point make a star, not a triangle.

3Case 2: parallel to y = 2x + 1

#2 Make a Systematic List 8.EE.C.8

Equal slopes with different

y

-intercepts (

4\ne 1

) means parallel, so those two never meet.

a=2

A missing corner means no triangle.

4Case 3: parallel to y = 3x - 2

#2 Make a Systematic List 8.EE.C.8

Likewise, matching the slope

m=3

makes the third line parallel to the second (

4\ne -2

).

a=3

Again one pair never crosses.

5Collect the cases

#2 Make a Systematic List 8.EE.C.8

For any other

a

, the third line crosses each of the others at a point different from

(3, 7)

, and those two points are different from each other, so a triangle forms.

a=1,\ 2,\ 3

Exactly three bad values.

Answer:

a=1,\ 2,\ 3

4 · Reviewdoes it hold up?

Check a good value, $a=0$ : $y=4$ meets $y=2x+1$ at $(\frac{3}{2}, 4)$ and $y=3x-2$ at $(2, 4)$ ; with $(3, 7)$ that is three distinct corners, a real triangle. So only the special values break it.

Another way: Draw a Diagram: sketch the two lines through $(3, 7)$ and the fixed point $(0, 4)$ of the third line, then rotate the third line about $(0, 4)$ . It stops making a triangle exactly when it lines up with $(3, 7)$ or becomes parallel to one of the two lines.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Using slope and $y$-intercept to tell when lines are parallel.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Understanding that two lines meet at the solution of their system, and that parallel lines have none.

💡Takeaway. To make a triangle you need three different corners -- so hunt for the ways to lose a corner: parallel lines, or all three crossing at one spot.

Variant 11 hard

The two lines $y=-3x+1$ and $y=mx+3$ meet at the point $(-2, 7)$ . A third line $y=ax+4$ is drawn. Find all values of the constant $a$ for which the three lines do not enclose a triangle. ( $m$ is a constant.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two lines already cross at $(-2, 7)$ . A third line with $y$ -intercept $4$ is added. For which slopes $a$ do the three lines fail to form a triangle?

Givens

$y=-3x+1$ and $y=mx+3$ both pass through $(-2, 7)$ .
The third line is $y=ax+4$ .

Unknowns

Every value of $a$ that prevents a triangle.

Constraints

Three lines enclose a triangle only when each pair crosses and the three crossing points are different.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#1 Draw a Diagram

List the ways three lines can fail: the third line is parallel to the first, parallel to the second, or passes through the point where the first two already meet. (Two lines can't coincide, since the $y$ -intercepts $1$ , $3$ , $4$ are all different.) First find $m$ from the given crossing point, then turn each case into an equation for $a$ .

3 · Execute5 carry out the plan

1Find m

#13 Convert to Algebra 8.F.A.3

The line

y=mx+3

passes through

(-2, 7)

.

7=-2m+3 \Rightarrow m=-2

Now all three lines are known except

a

.

2Case 1: all three meet at one point

#13 Convert to Algebra 8.EE.C.8

If

y=ax+4

also passes through

(-2, 7)

, the three crossing points collapse into one.

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

Three lines through one point make a star, not a triangle.

3Case 2: parallel to y = -3x + 1

#2 Make a Systematic List 8.EE.C.8

Equal slopes with different

y

-intercepts (

4\ne 1

) means parallel, so those two never meet.

a=-3

A missing corner means no triangle.

4Case 3: parallel to y = -2x + 3

#2 Make a Systematic List 8.EE.C.8

Likewise, matching the slope

m=-2

makes the third line parallel to the second (

4\ne 3

).

a=-2

Again one pair never crosses.

5Collect the cases

#2 Make a Systematic List 8.EE.C.8

For any other

a

, the third line crosses each of the others at a point different from

(-2, 7)

, and those two points are different from each other, so a triangle forms.

a=-3,\ -2,\ -\frac{3}{2}

Exactly three bad values.

Answer:

a=-3,\ -2,\ -\frac{3}{2}

4 · Reviewdoes it hold up?

Check a good value, $a=0$ : $y=4$ meets $y=-3x+1$ at $(-1, 4)$ and $y=-2x+3$ at $(-\frac{1}{2}, 4)$ ; with $(-2, 7)$ that is three distinct corners, a real triangle. So only the special values break it.

Another way: Draw a Diagram: sketch the two lines through $(-2, 7)$ and the fixed point $(0, 4)$ of the third line, then rotate the third line about $(0, 4)$ . It stops making a triangle exactly when it lines up with $(-2, 7)$ or becomes parallel to one of the two lines.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Using slope and $y$-intercept to tell when lines are parallel.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Understanding that two lines meet at the solution of their system, and that parallel lines have none.

💡Takeaway. To make a triangle you need three different corners -- so hunt for the ways to lose a corner: parallel lines, or all three crossing at one spot.

Variant 12 hard

The two lines $y=-2x+5$ and $y=mx+4$ meet at the point $(-1, 7)$ . A third line $y=ax+1$ is drawn. Find all values of the constant $a$ for which the three lines do not enclose a triangle. ( $m$ is a constant.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two lines already cross at $(-1, 7)$ . A third line with $y$ -intercept $1$ is added. For which slopes $a$ do the three lines fail to form a triangle?

Givens

$y=-2x+5$ and $y=mx+4$ both pass through $(-1, 7)$ .
The third line is $y=ax+1$ .

Unknowns

Every value of $a$ that prevents a triangle.

Constraints

Three lines enclose a triangle only when each pair crosses and the three crossing points are different.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #13 Convert to Algebra#1 Draw a Diagram

List the ways three lines can fail: the third line is parallel to the first, parallel to the second, or passes through the point where the first two already meet. (Two lines can't coincide, since the $y$ -intercepts $5$ , $4$ , $1$ are all different.) First find $m$ from the given crossing point, then turn each case into an equation for $a$ .

3 · Execute5 carry out the plan

1Find m

#13 Convert to Algebra 8.F.A.3

The line

y=mx+4

passes through

(-1, 7)

.

7=-m+4 \Rightarrow m=-3

Now all three lines are known except

a

.

2Case 1: all three meet at one point

#13 Convert to Algebra 8.EE.C.8

If

y=ax+1

also passes through

(-1, 7)

, the three crossing points collapse into one.

7=-a+1 \Rightarrow a=-6

Three lines through one point make a star, not a triangle.

3Case 2: parallel to y = -2x + 5

#2 Make a Systematic List 8.EE.C.8

Equal slopes with different

y

-intercepts (

1\ne 5

) means parallel, so those two never meet.

a=-2

A missing corner means no triangle.

4Case 3: parallel to y = -3x + 4

#2 Make a Systematic List 8.EE.C.8

Likewise, matching the slope

m=-3

makes the third line parallel to the second (

1\ne 4

).

a=-3

Again one pair never crosses.

5Collect the cases

#2 Make a Systematic List 8.EE.C.8

For any other

a

, the third line crosses each of the others at a point different from

(-1, 7)

, and those two points are different from each other, so a triangle forms.

a=-6,\ -3,\ -2

Exactly three bad values.

Answer:

a=-6,\ -3,\ -2

4 · Reviewdoes it hold up?

Check a good value, $a=0$ : $y=1$ meets $y=-2x+5$ at $(2, 1)$ and $y=-3x+4$ at $(1, 1)$ ; with $(-1, 7)$ that is three distinct corners, a real triangle. So only the special values break it.

Another way: Draw a Diagram: sketch the two lines through $(-1, 7)$ and the fixed point $(0, 1)$ of the third line, then rotate the third line about $(0, 1)$ . It stops making a triangle exactly when it lines up with $(-1, 7)$ or becomes parallel to one of the two lines.

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Using slope and $y$-intercept to tell when lines are parallel.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Understanding that two lines meet at the solution of their system, and that parallel lines have none.

💡Takeaway. To make a triangle you need three different corners -- so hunt for the ways to lose a corner: parallel lines, or all three crossing at one spot.

When three lines fail to make a triangle · 12 practice problems

Generated variants — 12

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