Practice · No triangle: a parallel pair or one common point · Sensim Math

Variant 1 easy

The three distinct lines $x-y=0$ , $x+y=4$ , and $mx-3y=6$ do not form a triangle. Find all possible values of the constant $m$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three different lines are $x-y=0$ , $x+y=4$ , and $mx-3y=6$ . For which values of $m$ do they fail to enclose a triangle?

Givens

Line 1: $x-y=0$ .
Line 2: $x+y=4$ .
Line 3: $mx-3y=6$ , with $m$ a constant.
The three lines are distinct.

Unknowns

Every value of $m$ for which no triangle is formed.

Constraints

Three lines enclose a triangle exactly when every pair meets and the three meeting points are different.

2 · Planchoose the strategy

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

List every way a triangle can fail: some pair of lines is parallel (that pair never meets), or all three pass through one point (the three corners collapse). Line 3 is the only one that moves with $m$ , so check it against each fixed line and against their crossing point.

3 · Execute4 carry out the plan

1Write each line in slope form

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

Solve each equation for

y

to read off the slopes.

y=x,\quad y=-x+4,\quad y=\frac{m}{3}x-2

Slope form makes parallel lines easy to spot.

2Case 1: line 3 parallel to line 1

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

Parallel lines have equal slopes and different intercepts (

-2\ne 0

).

\frac{m}{3}=1,\ m=3

Two parallel lines never meet, so one corner is missing.

3Case 2: line 3 parallel to line 2

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

Again equal slopes, and the intercepts

-2

and

4

differ.

\frac{m}{3}=-1,\ m=-3

Lines 1 and 2 have different slopes, so this list of parallel pairs is complete.

4Case 3: all three through one point

#1 Draw a Diagram 8.EE.C.8

Lines 1 and 2 cross where

x=-x+4

, so

x=2

and

y=2

. Line 3 must pass through

(2, 2)

.

m\times 2-3\times 2=6,\ m=6

If the third line runs through the crossing of the other two, all three corners are the same point.

Answer:

m=-3, 3, 6

4 · Reviewdoes it hold up?

Check each value: $m=3$ gives $y=x-2$ , parallel to $y=x$ ; $m=-3$ gives $y=-x-2$ , parallel to $y=-x+4$ ; $m=6$ gives $y=2x-2$ , and $2\times 2-2=2$ , so it passes through $(2, 2)$ . In every case the three lines are still distinct.

Another way: Count corners instead: line 3 meets line 1 at $x=\frac{6}{m-3}$ and line 2 at $x=\frac{18}{m+3}$ . A triangle fails when either crossing does not exist ( $m=3$ or $m=-3$ ) or when the two crossings coincide at $x=2$ , which gives $m=6$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Reading each line's slope and intercept from $y=mx+b$ form.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using parallel lines (no solution) and a common solution of two equations.

💡Takeaway. Three lines miss a triangle only two ways: two run parallel, or all three meet at one point.

Variant 2 easy

The three distinct lines $2x+y=4$ , $x-y=-1$ , and $mx-3y=3$ do not form a triangle. Find all possible values of the constant $m$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three different lines are $2x+y=4$ , $x-y=-1$ , and $mx-3y=3$ . For which values of $m$ do they fail to enclose a triangle?

Givens

Line 1: $2x+y=4$ .
Line 2: $x-y=-1$ .
Line 3: $mx-3y=3$ , with $m$ a constant.
The three lines are distinct.

Unknowns

Every value of $m$ for which no triangle is formed.

Constraints

Three lines enclose a triangle exactly when every pair meets and the three meeting points are different.

2 · Planchoose the strategy

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

List every way a triangle can fail: some pair of lines is parallel (that pair never meets), or all three pass through one point (the three corners collapse). Line 3 is the only one that moves with $m$ , so check it against each fixed line and against their crossing point.

3 · Execute4 carry out the plan

1Write each line in slope form

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

Solve each equation for

y

to read off the slopes.

y=-2x+4,\quad y=x+1,\quad y=\frac{m}{3}x-1

Slope form makes parallel lines easy to spot.

2Case 1: line 3 parallel to line 1

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

Parallel lines have equal slopes and different intercepts (

-1\ne 4

).

\frac{m}{3}=-2,\ m=-6

Two parallel lines never meet, so one corner is missing.

3Case 2: line 3 parallel to line 2

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

Again equal slopes, and the intercepts

-1

and

1

differ.

\frac{m}{3}=1,\ m=3

Lines 1 and 2 have different slopes, so this list of parallel pairs is complete.

4Case 3: all three through one point

#1 Draw a Diagram 8.EE.C.8

Lines 1 and 2 cross where

-2x+4=x+1

, so

x=1

and

y=2

. Line 3 must pass through

(1, 2)

.

m\times 1-3\times 2=3,\ m=9

If the third line runs through the crossing of the other two, all three corners are the same point.

Answer:

m=-6, 3, 9

4 · Reviewdoes it hold up?

Check each value: $m=-6$ gives $y=-2x-1$ , parallel to $y=-2x+4$ ; $m=3$ gives $y=x-1$ , parallel to $y=x+1$ ; $m=9$ gives $y=3x-1$ , and $3\times 1-1=2$ , so it passes through $(1, 2)$ . In every case the three lines are still distinct.

Another way: Count corners instead: line 3 meets line 1 at $x=\frac{15}{m+6}$ and line 2 at $x=\frac{6}{m-3}$ . A triangle fails when either crossing does not exist ( $m=-6$ or $m=3$ ) or when the two crossings coincide at $x=1$ , which gives $m=9$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Reading each line's slope and intercept from $y=mx+b$ form.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using parallel lines (no solution) and a common solution of two equations.

💡Takeaway. Three lines miss a triangle only two ways: two run parallel, or all three meet at one point.

Variant 3 easy

The three distinct lines $2x-y=0$ , $x+y=3$ , and $mx-2y=4$ do not form a triangle. Find all possible values of the constant $m$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three different lines are $2x-y=0$ , $x+y=3$ , and $mx-2y=4$ . For which values of $m$ do they fail to enclose a triangle?

Givens

Line 1: $2x-y=0$ .
Line 2: $x+y=3$ .
Line 3: $mx-2y=4$ , with $m$ a constant.
The three lines are distinct.

Unknowns

Every value of $m$ for which no triangle is formed.

Constraints

Three lines enclose a triangle exactly when every pair meets and the three meeting points are different.

2 · Planchoose the strategy

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

List every way a triangle can fail: some pair of lines is parallel (that pair never meets), or all three pass through one point (the three corners collapse). Line 3 is the only one that moves with $m$ , so check it against each fixed line and against their crossing point.

3 · Execute4 carry out the plan

1Write each line in slope form

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

Solve each equation for

y

to read off the slopes.

y=2x,\quad y=-x+3,\quad y=\frac{m}{2}x-2

Slope form makes parallel lines easy to spot.

2Case 1: line 3 parallel to line 1

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

Parallel lines have equal slopes and different intercepts (

-2\ne 0

).

\frac{m}{2}=2,\ m=4

Two parallel lines never meet, so one corner is missing.

3Case 2: line 3 parallel to line 2

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

Again equal slopes, and the intercepts

-2

and

3

differ.

\frac{m}{2}=-1,\ m=-2

Lines 1 and 2 have different slopes, so this list of parallel pairs is complete.

4Case 3: all three through one point

#1 Draw a Diagram 8.EE.C.8

Lines 1 and 2 cross where

2x=-x+3

, so

x=1

and

y=2

. Line 3 must pass through

(1, 2)

.

m\times 1-2\times 2=4,\ m=8

If the third line runs through the crossing of the other two, all three corners are the same point.

Answer:

m=-2, 4, 8

4 · Reviewdoes it hold up?

Check each value: $m=4$ gives $y=2x-2$ , parallel to $y=2x$ ; $m=-2$ gives $y=-x-2$ , parallel to $y=-x+3$ ; $m=8$ gives $y=4x-2$ , and $4\times 1-2=2$ , so it passes through $(1, 2)$ . In every case the three lines are still distinct.

Another way: Count corners instead: line 3 meets line 1 at $x=\frac{4}{m-4}$ and line 2 at $x=\frac{10}{m+2}$ . A triangle fails when either crossing does not exist ( $m=4$ or $m=-2$ ) or when the two crossings coincide at $x=1$ , which gives $m=8$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Reading each line's slope and intercept from $y=mx+b$ form.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using parallel lines (no solution) and a common solution of two equations.

💡Takeaway. Three lines miss a triangle only two ways: two run parallel, or all three meet at one point.

Variant 4 easy

The three distinct lines $4x-y=0$ , $x+y=5$ , and $mx-2y=2$ do not form a triangle. Find all possible values of the constant $m$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three different lines are $4x-y=0$ , $x+y=5$ , and $mx-2y=2$ . For which values of $m$ do they fail to enclose a triangle?

Givens

Line 1: $4x-y=0$ .
Line 2: $x+y=5$ .
Line 3: $mx-2y=2$ , with $m$ a constant.
The three lines are distinct.

Unknowns

Every value of $m$ for which no triangle is formed.

Constraints

Three lines enclose a triangle exactly when every pair meets and the three meeting points are different.

2 · Planchoose the strategy

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

List every way a triangle can fail: some pair of lines is parallel (that pair never meets), or all three pass through one point (the three corners collapse). Line 3 is the only one that moves with $m$ , so check it against each fixed line and against their crossing point.

3 · Execute4 carry out the plan

1Write each line in slope form

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

Solve each equation for

y

to read off the slopes.

y=4x,\quad y=-x+5,\quad y=\frac{m}{2}x-1

Slope form makes parallel lines easy to spot.

2Case 1: line 3 parallel to line 1

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

Parallel lines have equal slopes and different intercepts (

-1\ne 0

).

\frac{m}{2}=4,\ m=8

Two parallel lines never meet, so one corner is missing.

3Case 2: line 3 parallel to line 2

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

Again equal slopes, and the intercepts

-1

and

5

differ.

\frac{m}{2}=-1,\ m=-2

Lines 1 and 2 have different slopes, so this list of parallel pairs is complete.

4Case 3: all three through one point

#1 Draw a Diagram 8.EE.C.8

Lines 1 and 2 cross where

4x=-x+5

, so

x=1

and

y=4

. Line 3 must pass through

(1, 4)

.

m\times 1-2\times 4=2,\ m=10

If the third line runs through the crossing of the other two, all three corners are the same point.

Answer:

m=-2, 8, 10

4 · Reviewdoes it hold up?

Check each value: $m=8$ gives $y=4x-1$ , parallel to $y=4x$ ; $m=-2$ gives $y=-x-1$ , parallel to $y=-x+5$ ; $m=10$ gives $y=5x-1$ , and $5\times 1-1=4$ , so it passes through $(1, 4)$ . In every case the three lines are still distinct.

Another way: Count corners instead: line 3 meets line 1 at $x=\frac{2}{m-8}$ and line 2 at $x=\frac{12}{m+2}$ . A triangle fails when either crossing does not exist ( $m=8$ or $m=-2$ ) or when the two crossings coincide at $x=1$ , which gives $m=10$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Reading each line's slope and intercept from $y=mx+b$ form.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using parallel lines (no solution) and a common solution of two equations.

💡Takeaway. Three lines miss a triangle only two ways: two run parallel, or all three meet at one point.

Variant 5 medium

The three distinct lines $x+y=2$ , $3x-y=2$ , and $mx-2y=6$ do not form a triangle. Find all possible values of the constant $m$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three different lines are $x+y=2$ , $3x-y=2$ , and $mx-2y=6$ . For which values of $m$ do they fail to enclose a triangle?

Givens

Line 1: $x+y=2$ .
Line 2: $3x-y=2$ .
Line 3: $mx-2y=6$ , with $m$ a constant.
The three lines are distinct.

Unknowns

Every value of $m$ for which no triangle is formed.

Constraints

Three lines enclose a triangle exactly when every pair meets and the three meeting points are different.

2 · Planchoose the strategy

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

List every way a triangle can fail: some pair of lines is parallel (that pair never meets), or all three pass through one point (the three corners collapse). Line 3 is the only one that moves with $m$ , so check it against each fixed line and against their crossing point.

3 · Execute4 carry out the plan

1Write each line in slope form

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

Solve each equation for

y

to read off the slopes.

y=-x+2,\quad y=3x-2,\quad y=\frac{m}{2}x-3

Slope form makes parallel lines easy to spot.

2Case 1: line 3 parallel to line 1

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

Parallel lines have equal slopes and different intercepts (

-3\ne 2

).

\frac{m}{2}=-1,\ m=-2

Two parallel lines never meet, so one corner is missing.

3Case 2: line 3 parallel to line 2

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

Again equal slopes, and the intercepts

-3

and

-2

differ.

\frac{m}{2}=3,\ m=6

Lines 1 and 2 have different slopes, so this list of parallel pairs is complete.

4Case 3: all three through one point

#1 Draw a Diagram 8.EE.C.8

Lines 1 and 2 cross where

-x+2=3x-2

, so

x=1

and

y=1

. Line 3 must pass through

(1, 1)

.

m\times 1-2\times 1=6,\ m=8

If the third line runs through the crossing of the other two, all three corners are the same point.

Answer:

m=-2, 6, 8

4 · Reviewdoes it hold up?

Check each value: $m=-2$ gives $y=-x-3$ , parallel to $y=-x+2$ ; $m=6$ gives $y=3x-3$ , parallel to $y=3x-2$ ; $m=8$ gives $y=4x-3$ , and $4\times 1-3=1$ , so it passes through $(1, 1)$ . In every case the three lines are still distinct.

Another way: Count corners instead: line 3 meets line 1 at $x=\frac{10}{m+2}$ and line 2 at $x=\frac{2}{m-6}$ . A triangle fails when either crossing does not exist ( $m=-2$ or $m=6$ ) or when the two crossings coincide at $x=1$ , which gives $m=8$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Reading each line's slope and intercept from $y=mx+b$ form.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using parallel lines (no solution) and a common solution of two equations.

💡Takeaway. Three lines miss a triangle only two ways: two run parallel, or all three meet at one point.

Variant 6 medium

The three distinct lines $x-y=-2$ , $3x+y=6$ , and $mx-3y=6$ do not form a triangle. Find all possible values of the constant $m$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three different lines are $x-y=-2$ , $3x+y=6$ , and $mx-3y=6$ . For which values of $m$ do they fail to enclose a triangle?

Givens

Line 1: $x-y=-2$ .
Line 2: $3x+y=6$ .
Line 3: $mx-3y=6$ , with $m$ a constant.
The three lines are distinct.

Unknowns

Every value of $m$ for which no triangle is formed.

Constraints

Three lines enclose a triangle exactly when every pair meets and the three meeting points are different.

2 · Planchoose the strategy

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

List every way a triangle can fail: some pair of lines is parallel (that pair never meets), or all three pass through one point (the three corners collapse). Line 3 is the only one that moves with $m$ , so check it against each fixed line and against their crossing point.

3 · Execute4 carry out the plan

1Write each line in slope form

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

Solve each equation for

y

to read off the slopes.

y=x+2,\quad y=-3x+6,\quad y=\frac{m}{3}x-2

Slope form makes parallel lines easy to spot.

2Case 1: line 3 parallel to line 1

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

Parallel lines have equal slopes and different intercepts (

-2\ne 2

).

\frac{m}{3}=1,\ m=3

Two parallel lines never meet, so one corner is missing.

3Case 2: line 3 parallel to line 2

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

Again equal slopes, and the intercepts

-2

and

6

differ.

\frac{m}{3}=-3,\ m=-9

Lines 1 and 2 have different slopes, so this list of parallel pairs is complete.

4Case 3: all three through one point

#1 Draw a Diagram 8.EE.C.8

Lines 1 and 2 cross where

x+2=-3x+6

, so

x=1

and

y=3

. Line 3 must pass through

(1, 3)

.

m\times 1-3\times 3=6,\ m=15

If the third line runs through the crossing of the other two, all three corners are the same point.

Answer:

m=-9, 3, 15

4 · Reviewdoes it hold up?

Check each value: $m=3$ gives $y=x-2$ , parallel to $y=x+2$ ; $m=-9$ gives $y=-3x-2$ , parallel to $y=-3x+6$ ; $m=15$ gives $y=5x-2$ , and $5\times 1-2=3$ , so it passes through $(1, 3)$ . In every case the three lines are still distinct.

Another way: Count corners instead: line 3 meets line 1 at $x=\frac{12}{m-3}$ and line 2 at $x=\frac{24}{m+9}$ . A triangle fails when either crossing does not exist ( $m=3$ or $m=-9$ ) or when the two crossings coincide at $x=1$ , which gives $m=15$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Reading each line's slope and intercept from $y=mx+b$ form.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using parallel lines (no solution) and a common solution of two equations.

💡Takeaway. Three lines miss a triangle only two ways: two run parallel, or all three meet at one point.

Variant 7 medium

The three distinct lines $x+y=0$ , $2x-y=6$ , and $mx-4y=8$ do not form a triangle. Find all possible values of the constant $m$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three different lines are $x+y=0$ , $2x-y=6$ , and $mx-4y=8$ . For which values of $m$ do they fail to enclose a triangle?

Givens

Line 1: $x+y=0$ .
Line 2: $2x-y=6$ .
Line 3: $mx-4y=8$ , with $m$ a constant.
The three lines are distinct.

Unknowns

Every value of $m$ for which no triangle is formed.

Constraints

Three lines enclose a triangle exactly when every pair meets and the three meeting points are different.

2 · Planchoose the strategy

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

List every way a triangle can fail: some pair of lines is parallel (that pair never meets), or all three pass through one point (the three corners collapse). Line 3 is the only one that moves with $m$ , so check it against each fixed line and against their crossing point.

3 · Execute4 carry out the plan

1Write each line in slope form

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

Solve each equation for

y

to read off the slopes.

y=-x,\quad y=2x-6,\quad y=\frac{m}{4}x-2

Slope form makes parallel lines easy to spot.

2Case 1: line 3 parallel to line 1

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

Parallel lines have equal slopes and different intercepts (

-2\ne 0

).

\frac{m}{4}=-1,\ m=-4

Two parallel lines never meet, so one corner is missing.

3Case 2: line 3 parallel to line 2

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

Again equal slopes, and the intercepts

-2

and

-6

differ.

\frac{m}{4}=2,\ m=8

Lines 1 and 2 have different slopes, so this list of parallel pairs is complete.

4Case 3: all three through one point

#1 Draw a Diagram 8.EE.C.8

Lines 1 and 2 cross where

-x=2x-6

, so

x=2

and

y=-2

. Line 3 must pass through

(2, -2)

.

m\times 2-4\times (-2)=8,\ m=0

If the third line runs through the crossing of the other two, all three corners are the same point.

Answer:

m=-4, 0, 8

4 · Reviewdoes it hold up?

Check each value: $m=-4$ gives $y=-x-2$ , parallel to $y=-x$ ; $m=8$ gives $y=2x-2$ , parallel to $y=2x-6$ ; $m=0$ gives $y=-2$ , and $0\times 2-2=-2$ , so it passes through $(2, -2)$ . In every case the three lines are still distinct.

Another way: Count corners instead: line 3 meets line 1 at $x=\frac{8}{m+4}$ and line 2 at $x=-\frac{16}{m-8}$ . A triangle fails when either crossing does not exist ( $m=-4$ or $m=8$ ) or when the two crossings coincide at $x=2$ , which gives $m=0$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Reading each line's slope and intercept from $y=mx+b$ form.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using parallel lines (no solution) and a common solution of two equations.

💡Takeaway. Three lines miss a triangle only two ways: two run parallel, or all three meet at one point.

Variant 8 medium

The three distinct lines $3x-y=0$ , $x+y=8$ , and $mx-2y=-4$ do not form a triangle. Find all possible values of the constant $m$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three different lines are $3x-y=0$ , $x+y=8$ , and $mx-2y=-4$ . For which values of $m$ do they fail to enclose a triangle?

Givens

Line 1: $3x-y=0$ .
Line 2: $x+y=8$ .
Line 3: $mx-2y=-4$ , with $m$ a constant.
The three lines are distinct.

Unknowns

Every value of $m$ for which no triangle is formed.

Constraints

Three lines enclose a triangle exactly when every pair meets and the three meeting points are different.

2 · Planchoose the strategy

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

List every way a triangle can fail: some pair of lines is parallel (that pair never meets), or all three pass through one point (the three corners collapse). Line 3 is the only one that moves with $m$ , so check it against each fixed line and against their crossing point.

3 · Execute4 carry out the plan

1Write each line in slope form

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

Solve each equation for

y

to read off the slopes.

y=3x,\quad y=-x+8,\quad y=\frac{m}{2}x+2

Slope form makes parallel lines easy to spot.

2Case 1: line 3 parallel to line 1

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

Parallel lines have equal slopes and different intercepts (

2\ne 0

).

\frac{m}{2}=3,\ m=6

Two parallel lines never meet, so one corner is missing.

3Case 2: line 3 parallel to line 2

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

Again equal slopes, and the intercepts

2

and

8

differ.

\frac{m}{2}=-1,\ m=-2

Lines 1 and 2 have different slopes, so this list of parallel pairs is complete.

4Case 3: all three through one point

#1 Draw a Diagram 8.EE.C.8

Lines 1 and 2 cross where

3x=-x+8

, so

x=2

and

y=6

. Line 3 must pass through

(2, 6)

.

m\times 2-2\times 6=-4,\ m=4

If the third line runs through the crossing of the other two, all three corners are the same point.

Answer:

m=-2, 4, 6

4 · Reviewdoes it hold up?

Check each value: $m=6$ gives $y=3x+2$ , parallel to $y=3x$ ; $m=-2$ gives $y=-x+2$ , parallel to $y=-x+8$ ; $m=4$ gives $y=2x+2$ , and $2\times 2+2=6$ , so it passes through $(2, 6)$ . In every case the three lines are still distinct.

Another way: Count corners instead: line 3 meets line 1 at $x=-\frac{4}{m-6}$ and line 2 at $x=\frac{12}{m+2}$ . A triangle fails when either crossing does not exist ( $m=6$ or $m=-2$ ) or when the two crossings coincide at $x=2$ , which gives $m=4$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Reading each line's slope and intercept from $y=mx+b$ form.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using parallel lines (no solution) and a common solution of two equations.

💡Takeaway. Three lines miss a triangle only two ways: two run parallel, or all three meet at one point.

Variant 9 hard

The three distinct lines $x+y=-2$ , $2x-y=-4$ , and $mx-4y=-8$ do not form a triangle. Find all possible values of the constant $m$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three different lines are $x+y=-2$ , $2x-y=-4$ , and $mx-4y=-8$ . For which values of $m$ do they fail to enclose a triangle?

Givens

Line 1: $x+y=-2$ .
Line 2: $2x-y=-4$ .
Line 3: $mx-4y=-8$ , with $m$ a constant.
The three lines are distinct.

Unknowns

Every value of $m$ for which no triangle is formed.

Constraints

Three lines enclose a triangle exactly when every pair meets and the three meeting points are different.

2 · Planchoose the strategy

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

List every way a triangle can fail: some pair of lines is parallel (that pair never meets), or all three pass through one point (the three corners collapse). Line 3 is the only one that moves with $m$ , so check it against each fixed line and against their crossing point.

3 · Execute4 carry out the plan

1Write each line in slope form

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

Solve each equation for

y

to read off the slopes.

y=-x-2,\quad y=2x+4,\quad y=\frac{m}{4}x+2

Slope form makes parallel lines easy to spot.

2Case 1: line 3 parallel to line 1

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

Parallel lines have equal slopes and different intercepts (

2\ne -2

).

\frac{m}{4}=-1,\ m=-4

Two parallel lines never meet, so one corner is missing.

3Case 2: line 3 parallel to line 2

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

Again equal slopes, and the intercepts

2

and

4

differ.

\frac{m}{4}=2,\ m=8

Lines 1 and 2 have different slopes, so this list of parallel pairs is complete.

4Case 3: all three through one point

#1 Draw a Diagram 8.EE.C.8

Lines 1 and 2 cross where

-x-2=2x+4

, so

x=-2

and

y=0

. Line 3 must pass through

(-2, 0)

.

m\times (-2)-4\times 0=-8,\ m=4

If the third line runs through the crossing of the other two, all three corners are the same point.

Answer:

m=-4, 4, 8

4 · Reviewdoes it hold up?

Check each value: $m=-4$ gives $y=-x+2$ , parallel to $y=-x-2$ ; $m=8$ gives $y=2x+2$ , parallel to $y=2x+4$ ; $m=4$ gives $y=x+2$ , and $1\times (-2)+2=0$ , so it passes through $(-2, 0)$ . In every case the three lines are still distinct.

Another way: Count corners instead: line 3 meets line 1 at $x=-\frac{16}{m+4}$ and line 2 at $x=\frac{8}{m-8}$ . A triangle fails when either crossing does not exist ( $m=-4$ or $m=8$ ) or when the two crossings coincide at $x=-2$ , which gives $m=4$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Reading each line's slope and intercept from $y=mx+b$ form.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using parallel lines (no solution) and a common solution of two equations.

💡Takeaway. Three lines miss a triangle only two ways: two run parallel, or all three meet at one point.

Variant 10 hard

The three distinct lines $x-y=1$ , $2x+y=5$ , and $mx-3y=9$ do not form a triangle. Find all possible values of the constant $m$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three different lines are $x-y=1$ , $2x+y=5$ , and $mx-3y=9$ . For which values of $m$ do they fail to enclose a triangle?

Givens

Line 1: $x-y=1$ .
Line 2: $2x+y=5$ .
Line 3: $mx-3y=9$ , with $m$ a constant.
The three lines are distinct.

Unknowns

Every value of $m$ for which no triangle is formed.

Constraints

Three lines enclose a triangle exactly when every pair meets and the three meeting points are different.

2 · Planchoose the strategy

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

List every way a triangle can fail: some pair of lines is parallel (that pair never meets), or all three pass through one point (the three corners collapse). Line 3 is the only one that moves with $m$ , so check it against each fixed line and against their crossing point.

3 · Execute4 carry out the plan

1Write each line in slope form

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

Solve each equation for

y

to read off the slopes.

y=x-1,\quad y=-2x+5,\quad y=\frac{m}{3}x-3

Slope form makes parallel lines easy to spot.

2Case 1: line 3 parallel to line 1

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

Parallel lines have equal slopes and different intercepts (

-3\ne -1

).

\frac{m}{3}=1,\ m=3

Two parallel lines never meet, so one corner is missing.

3Case 2: line 3 parallel to line 2

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

Again equal slopes, and the intercepts

-3

and

5

differ.

\frac{m}{3}=-2,\ m=-6

Lines 1 and 2 have different slopes, so this list of parallel pairs is complete.

4Case 3: all three through one point

#1 Draw a Diagram 8.EE.C.8

Lines 1 and 2 cross where

x-1=-2x+5

, so

x=2

and

y=1

. Line 3 must pass through

(2, 1)

.

m\times 2-3\times 1=9,\ m=6

If the third line runs through the crossing of the other two, all three corners are the same point.

Answer:

m=-6, 3, 6

4 · Reviewdoes it hold up?

Check each value: $m=3$ gives $y=x-3$ , parallel to $y=x-1$ ; $m=-6$ gives $y=-2x-3$ , parallel to $y=-2x+5$ ; $m=6$ gives $y=2x-3$ , and $2\times 2-3=1$ , so it passes through $(2, 1)$ . In every case the three lines are still distinct.

Another way: Count corners instead: line 3 meets line 1 at $x=\frac{6}{m-3}$ and line 2 at $x=\frac{24}{m+6}$ . A triangle fails when either crossing does not exist ( $m=3$ or $m=-6$ ) or when the two crossings coincide at $x=2$ , which gives $m=6$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Reading each line's slope and intercept from $y=mx+b$ form.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using parallel lines (no solution) and a common solution of two equations.

💡Takeaway. Three lines miss a triangle only two ways: two run parallel, or all three meet at one point.

Variant 11 hard

The three distinct lines $x-y=3$ , $x+y=-1$ , and $mx-5y=20$ do not form a triangle. Find all possible values of the constant $m$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three different lines are $x-y=3$ , $x+y=-1$ , and $mx-5y=20$ . For which values of $m$ do they fail to enclose a triangle?

Givens

Line 1: $x-y=3$ .
Line 2: $x+y=-1$ .
Line 3: $mx-5y=20$ , with $m$ a constant.
The three lines are distinct.

Unknowns

Every value of $m$ for which no triangle is formed.

Constraints

Three lines enclose a triangle exactly when every pair meets and the three meeting points are different.

2 · Planchoose the strategy

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

List every way a triangle can fail: some pair of lines is parallel (that pair never meets), or all three pass through one point (the three corners collapse). Line 3 is the only one that moves with $m$ , so check it against each fixed line and against their crossing point.

3 · Execute4 carry out the plan

1Write each line in slope form

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

Solve each equation for

y

to read off the slopes.

y=x-3,\quad y=-x-1,\quad y=\frac{m}{5}x-4

Slope form makes parallel lines easy to spot.

2Case 1: line 3 parallel to line 1

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

Parallel lines have equal slopes and different intercepts (

-4\ne -3

).

\frac{m}{5}=1,\ m=5

Two parallel lines never meet, so one corner is missing.

3Case 2: line 3 parallel to line 2

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

Again equal slopes, and the intercepts

-4

and

-1

differ.

\frac{m}{5}=-1,\ m=-5

Lines 1 and 2 have different slopes, so this list of parallel pairs is complete.

4Case 3: all three through one point

#1 Draw a Diagram 8.EE.C.8

Lines 1 and 2 cross where

x-3=-x-1

, so

x=1

and

y=-2

. Line 3 must pass through

(1, -2)

.

m\times 1-5\times (-2)=20,\ m=10

If the third line runs through the crossing of the other two, all three corners are the same point.

Answer:

m=-5, 5, 10

4 · Reviewdoes it hold up?

Check each value: $m=5$ gives $y=x-4$ , parallel to $y=x-3$ ; $m=-5$ gives $y=-x-4$ , parallel to $y=-x-1$ ; $m=10$ gives $y=2x-4$ , and $2\times 1-4=-2$ , so it passes through $(1, -2)$ . In every case the three lines are still distinct.

Another way: Count corners instead: line 3 meets line 1 at $x=\frac{5}{m-5}$ and line 2 at $x=\frac{15}{m+5}$ . A triangle fails when either crossing does not exist ( $m=5$ or $m=-5$ ) or when the two crossings coincide at $x=1$ , which gives $m=10$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Reading each line's slope and intercept from $y=mx+b$ form.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using parallel lines (no solution) and a common solution of two equations.

💡Takeaway. Three lines miss a triangle only two ways: two run parallel, or all three meet at one point.

Variant 12 hard

The three distinct lines $x-y=0$ , $2x+y=3$ , and $mx-4y=16$ do not form a triangle. Find all possible values of the constant $m$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three different lines are $x-y=0$ , $2x+y=3$ , and $mx-4y=16$ . For which values of $m$ do they fail to enclose a triangle?

Givens

Line 1: $x-y=0$ .
Line 2: $2x+y=3$ .
Line 3: $mx-4y=16$ , with $m$ a constant.
The three lines are distinct.

Unknowns

Every value of $m$ for which no triangle is formed.

Constraints

Three lines enclose a triangle exactly when every pair meets and the three meeting points are different.

2 · Planchoose the strategy

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

List every way a triangle can fail: some pair of lines is parallel (that pair never meets), or all three pass through one point (the three corners collapse). Line 3 is the only one that moves with $m$ , so check it against each fixed line and against their crossing point.

3 · Execute4 carry out the plan

1Write each line in slope form

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

Solve each equation for

y

to read off the slopes.

y=x,\quad y=-2x+3,\quad y=\frac{m}{4}x-4

Slope form makes parallel lines easy to spot.

2Case 1: line 3 parallel to line 1

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

Parallel lines have equal slopes and different intercepts (

-4\ne 0

).

\frac{m}{4}=1,\ m=4

Two parallel lines never meet, so one corner is missing.

3Case 2: line 3 parallel to line 2

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

Again equal slopes, and the intercepts

-4

and

3

differ.

\frac{m}{4}=-2,\ m=-8

Lines 1 and 2 have different slopes, so this list of parallel pairs is complete.

4Case 3: all three through one point

#1 Draw a Diagram 8.EE.C.8

Lines 1 and 2 cross where

x=-2x+3

, so

x=1

and

y=1

. Line 3 must pass through

(1, 1)

.

m\times 1-4\times 1=16,\ m=20

If the third line runs through the crossing of the other two, all three corners are the same point.

Answer:

m=-8, 4, 20

4 · Reviewdoes it hold up?

Check each value: $m=4$ gives $y=x-4$ , parallel to $y=x$ ; $m=-8$ gives $y=-2x-4$ , parallel to $y=-2x+3$ ; $m=20$ gives $y=5x-4$ , and $5\times 1-4=1$ , so it passes through $(1, 1)$ . In every case the three lines are still distinct.

Another way: Count corners instead: line 3 meets line 1 at $x=\frac{16}{m-4}$ and line 2 at $x=\frac{28}{m+8}$ . A triangle fails when either crossing does not exist ( $m=4$ or $m=-8$ ) or when the two crossings coincide at $x=1$ , which gives $m=20$ .

Standardsmin grade 8

8.F.A.3 Interpret the equation y = mx + b as defining a linear function — Reading each line's slope and intercept from $y=mx+b$ form.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using parallel lines (no solution) and a common solution of two equations.

💡Takeaway. Three lines miss a triangle only two ways: two run parallel, or all three meet at one point.

No triangle: a parallel pair or one common point · 12 practice problems

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