Practice · Area bisectors of a square pass through its center · Sensim Math

Variant 1 easy

As shown in the figure, square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(6, 0)$ , $\text{B}(6, 6)$ , and $\text{C}(0, 6)$ . A line $l$ passes through the point $\text{P}(1, 2)$ inside the square and divides the area of square $\text{OABC}$ into two equal parts. Find the equation of line $l$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through the point $\text{P}(1, 2)$ inside the square with corners $(0, 0)$ , $(6, 0)$ , $(6, 6)$ , $(0, 6)$ cuts the square into two pieces of equal area. We need the equation of that line.

Givens

Square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(6, 0)$ , $\text{B}(6, 6)$ , $\text{C}(0, 6)$ .
Line $l$ passes through $\text{P}(1, 2)$ .
Line $l$ splits the square's area in half.

Unknowns

The equation of line $l$ .

Constraints

One point on the line is known; a second point must come from the equal-area condition.

2 · Planchoose the strategy

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

Draw the square and notice it is symmetric about its center: turning it halfway around the center $(3, 3)$ lands it on itself. Any line through the center splits it into two matching halves, and a line that misses the center cannot split it evenly. So the center is the missing second point, and two points fix the line.

3 · Execute4 carry out the plan

1Find the center

#1 Draw a Diagram 6.G.A.1

The diagonals of the square meet at the midpoint of

\text{O}(0, 0)

and

\text{B}(6, 6)

.

\left(\frac{0+6}{2}, \frac{0+6}{2}\right)=(3, 3)

The center is the point every half-turn of the square keeps fixed.

2Why the line must pass through the center

#16 Count the Complement 6.G.A.1

A line through the center cuts the square into two pieces that are half-turns of each other, so they have equal area. If a line missed the center, the parallel line through the center would already split the area evenly, and our line would add or remove the strip between them from one side. So the bisecting line through

\text{P}

must also pass through

(3, 3)

.

Instead of measuring areas, ask which point every bisector must share.

3Slope through two points

#13 Convert to Algebra 8.F.B.4

Line

l

passes through

\text{P}(1, 2)

and

(3, 3)

.

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

Rise over run between the two known points.

4Find the y-intercept

#13 Convert to Algebra 8.F.B.4

Write

y=\frac{1}{2}x+b

and substitute

x=3

,

y=3

.

3=\frac{1}{2}\times 3+b,\ b=\frac{3}{2}

One known point pins down the intercept once the slope is known.

Answer:

y=\frac{1}{2}x+\frac{3}{2}

4 · Reviewdoes it hold up?

At $x=0$ the line is at $y=\frac{3}{2}$ (on side $\text{OC}$ ) and at $x=6$ it is at $y=\frac{9}{2}$ (on side $\text{AB}$ ). The two pieces are trapezoids with parallel sides $\frac{3}{2}$ and $\frac{9}{2}$ (bottom piece) and $\frac{9}{2}$ and $\frac{3}{2}$ (top piece), each of width $6$ , so each has area $\frac{1}{2}\times\left(\frac{3}{2}+\frac{9}{2}\right)\times 6=18$ , half of $36$ .

Another way: Algebra only: let the line meet $x=0$ at height $p$ and $x=6$ at height $q$ . The lower piece is a trapezoid of area $\frac{1}{2}(p+q)\times 6=18$ , so $p+q=6$ , which says the line's height at $x=3$ is $\frac{p+q}{2}=3$ -- the center again.

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Building the line's equation from two points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Comparing the areas of the two pieces of the square.

💡Takeaway. A shape with a center of symmetry is cut in half by exactly the lines through that center.

Variant 2 easy

As shown in the figure, square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(8, 0)$ , $\text{B}(8, 8)$ , and $\text{C}(0, 8)$ . A line $l$ passes through the point $\text{P}(7, 5)$ inside the square and divides the area of square $\text{OABC}$ into two equal parts. Find the equation of line $l$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through the point $\text{P}(7, 5)$ inside the square with corners $(0, 0)$ , $(8, 0)$ , $(8, 8)$ , $(0, 8)$ cuts the square into two pieces of equal area. We need the equation of that line.

Givens

Square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(8, 0)$ , $\text{B}(8, 8)$ , $\text{C}(0, 8)$ .
Line $l$ passes through $\text{P}(7, 5)$ .
Line $l$ splits the square's area in half.

Unknowns

The equation of line $l$ .

Constraints

One point on the line is known; a second point must come from the equal-area condition.

2 · Planchoose the strategy

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

Draw the square and notice it is symmetric about its center: turning it halfway around the center $(4, 4)$ lands it on itself. Any line through the center splits it into two matching halves, and a line that misses the center cannot split it evenly. So the center is the missing second point, and two points fix the line.

3 · Execute4 carry out the plan

1Find the center

#1 Draw a Diagram 6.G.A.1

The diagonals of the square meet at the midpoint of

\text{O}(0, 0)

and

\text{B}(8, 8)

.

\left(\frac{0+8}{2}, \frac{0+8}{2}\right)=(4, 4)

The center is the point every half-turn of the square keeps fixed.

2Why the line must pass through the center

#16 Count the Complement 6.G.A.1

A line through the center cuts the square into two pieces that are half-turns of each other, so they have equal area. If a line missed the center, the parallel line through the center would already split the area evenly, and our line would add or remove the strip between them from one side. So the bisecting line through

\text{P}

must also pass through

(4, 4)

.

Instead of measuring areas, ask which point every bisector must share.

3Slope through two points

#13 Convert to Algebra 8.F.B.4

Line

l

passes through

\text{P}(7, 5)

and

(4, 4)

.

\frac{4-5}{4-7}=\frac{1}{3}

Rise over run between the two known points.

4Find the y-intercept

#13 Convert to Algebra 8.F.B.4

Write

y=\frac{1}{3}x+b

and substitute

x=4

,

y=4

.

4=\frac{1}{3}\times 4+b,\ b=\frac{8}{3}

One known point pins down the intercept once the slope is known.

Answer:

y=\frac{1}{3}x+\frac{8}{3}

4 · Reviewdoes it hold up?

At $x=0$ the line is at $y=\frac{8}{3}$ (on side $\text{OC}$ ) and at $x=8$ it is at $y=\frac{16}{3}$ (on side $\text{AB}$ ). The two pieces are trapezoids with parallel sides $\frac{8}{3}$ and $\frac{16}{3}$ (bottom piece) and $\frac{16}{3}$ and $\frac{8}{3}$ (top piece), each of width $8$ , so each has area $\frac{1}{2}\times\left(\frac{8}{3}+\frac{16}{3}\right)\times 8=32$ , half of $64$ .

Another way: Algebra only: let the line meet $x=0$ at height $p$ and $x=8$ at height $q$ . The lower piece is a trapezoid of area $\frac{1}{2}(p+q)\times 8=32$ , so $p+q=8$ , which says the line's height at $x=4$ is $\frac{p+q}{2}=4$ -- the center again.

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Building the line's equation from two points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Comparing the areas of the two pieces of the square.

💡Takeaway. A shape with a center of symmetry is cut in half by exactly the lines through that center.

Variant 3 easy

As shown in the figure, square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(8, 0)$ , $\text{B}(8, 8)$ , and $\text{C}(0, 8)$ . A line $l$ passes through the point $\text{P}(2, 3)$ inside the square and divides the area of square $\text{OABC}$ into two equal parts. Find the equation of line $l$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through the point $\text{P}(2, 3)$ inside the square with corners $(0, 0)$ , $(8, 0)$ , $(8, 8)$ , $(0, 8)$ cuts the square into two pieces of equal area. We need the equation of that line.

Givens

Square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(8, 0)$ , $\text{B}(8, 8)$ , $\text{C}(0, 8)$ .
Line $l$ passes through $\text{P}(2, 3)$ .
Line $l$ splits the square's area in half.

Unknowns

The equation of line $l$ .

Constraints

One point on the line is known; a second point must come from the equal-area condition.

2 · Planchoose the strategy

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

Draw the square and notice it is symmetric about its center: turning it halfway around the center $(4, 4)$ lands it on itself. Any line through the center splits it into two matching halves, and a line that misses the center cannot split it evenly. So the center is the missing second point, and two points fix the line.

3 · Execute4 carry out the plan

1Find the center

#1 Draw a Diagram 6.G.A.1

The diagonals of the square meet at the midpoint of

\text{O}(0, 0)

and

\text{B}(8, 8)

.

\left(\frac{0+8}{2}, \frac{0+8}{2}\right)=(4, 4)

The center is the point every half-turn of the square keeps fixed.

2Why the line must pass through the center

#16 Count the Complement 6.G.A.1

A line through the center cuts the square into two pieces that are half-turns of each other, so they have equal area. If a line missed the center, the parallel line through the center would already split the area evenly, and our line would add or remove the strip between them from one side. So the bisecting line through

\text{P}

must also pass through

(4, 4)

.

Instead of measuring areas, ask which point every bisector must share.

3Slope through two points

#13 Convert to Algebra 8.F.B.4

Line

l

passes through

\text{P}(2, 3)

and

(4, 4)

.

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

Rise over run between the two known points.

4Find the y-intercept

#13 Convert to Algebra 8.F.B.4

Write

y=\frac{1}{2}x+b

and substitute

x=4

,

y=4

.

4=\frac{1}{2}\times 4+b,\ b=2

One known point pins down the intercept once the slope is known.

Answer:

y=\frac{1}{2}x+2

4 · Reviewdoes it hold up?

At $x=0$ the line is at $y=2$ (on side $\text{OC}$ ) and at $x=8$ it is at $y=6$ (on side $\text{AB}$ ). The two pieces are trapezoids with parallel sides $2$ and $6$ (bottom piece) and $6$ and $2$ (top piece), each of width $8$ , so each has area $\frac{1}{2}\times\left(2+6\right)\times 8=32$ , half of $64$ .

Another way: Algebra only: let the line meet $x=0$ at height $p$ and $x=8$ at height $q$ . The lower piece is a trapezoid of area $\frac{1}{2}(p+q)\times 8=32$ , so $p+q=8$ , which says the line's height at $x=4$ is $\frac{p+q}{2}=4$ -- the center again.

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Building the line's equation from two points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Comparing the areas of the two pieces of the square.

💡Takeaway. A shape with a center of symmetry is cut in half by exactly the lines through that center.

Variant 4 easy

As shown in the figure, square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(10, 0)$ , $\text{B}(10, 10)$ , and $\text{C}(0, 10)$ . A line $l$ passes through the point $\text{P}(3, 4)$ inside the square and divides the area of square $\text{OABC}$ into two equal parts. Find the equation of line $l$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through the point $\text{P}(3, 4)$ inside the square with corners $(0, 0)$ , $(10, 0)$ , $(10, 10)$ , $(0, 10)$ cuts the square into two pieces of equal area. We need the equation of that line.

Givens

Square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(10, 0)$ , $\text{B}(10, 10)$ , $\text{C}(0, 10)$ .
Line $l$ passes through $\text{P}(3, 4)$ .
Line $l$ splits the square's area in half.

Unknowns

The equation of line $l$ .

Constraints

One point on the line is known; a second point must come from the equal-area condition.

2 · Planchoose the strategy

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

Draw the square and notice it is symmetric about its center: turning it halfway around the center $(5, 5)$ lands it on itself. Any line through the center splits it into two matching halves, and a line that misses the center cannot split it evenly. So the center is the missing second point, and two points fix the line.

3 · Execute4 carry out the plan

1Find the center

#1 Draw a Diagram 6.G.A.1

The diagonals of the square meet at the midpoint of

\text{O}(0, 0)

and

\text{B}(10, 10)

.

\left(\frac{0+10}{2}, \frac{0+10}{2}\right)=(5, 5)

The center is the point every half-turn of the square keeps fixed.

2Why the line must pass through the center

#16 Count the Complement 6.G.A.1

A line through the center cuts the square into two pieces that are half-turns of each other, so they have equal area. If a line missed the center, the parallel line through the center would already split the area evenly, and our line would add or remove the strip between them from one side. So the bisecting line through

\text{P}

must also pass through

(5, 5)

.

Instead of measuring areas, ask which point every bisector must share.

3Slope through two points

#13 Convert to Algebra 8.F.B.4

Line

l

passes through

\text{P}(3, 4)

and

(5, 5)

.

\frac{5-4}{5-3}=\frac{1}{2}

Rise over run between the two known points.

4Find the y-intercept

#13 Convert to Algebra 8.F.B.4

Write

y=\frac{1}{2}x+b

and substitute

x=5

,

y=5

.

5=\frac{1}{2}\times 5+b,\ b=\frac{5}{2}

One known point pins down the intercept once the slope is known.

Answer:

y=\frac{1}{2}x+\frac{5}{2}

4 · Reviewdoes it hold up?

At $x=0$ the line is at $y=\frac{5}{2}$ (on side $\text{OC}$ ) and at $x=10$ it is at $y=\frac{15}{2}$ (on side $\text{AB}$ ). The two pieces are trapezoids with parallel sides $\frac{5}{2}$ and $\frac{15}{2}$ (bottom piece) and $\frac{15}{2}$ and $\frac{5}{2}$ (top piece), each of width $10$ , so each has area $\frac{1}{2}\times\left(\frac{5}{2}+\frac{15}{2}\right)\times 10=50$ , half of $100$ .

Another way: Algebra only: let the line meet $x=0$ at height $p$ and $x=10$ at height $q$ . The lower piece is a trapezoid of area $\frac{1}{2}(p+q)\times 10=50$ , so $p+q=10$ , which says the line's height at $x=5$ is $\frac{p+q}{2}=5$ -- the center again.

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Building the line's equation from two points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Comparing the areas of the two pieces of the square.

💡Takeaway. A shape with a center of symmetry is cut in half by exactly the lines through that center.

Variant 5 medium

As shown in the figure, square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(4, 0)$ , $\text{B}(4, 4)$ , and $\text{C}(0, 4)$ . A line $l$ passes through the point $\text{P}\left(1, \frac{3}{2}\right)$ inside the square and divides the area of square $\text{OABC}$ into two equal parts. Find the equation of line $l$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through the point $\text{P}\left(1, \frac{3}{2}\right)$ inside the square with corners $(0, 0)$ , $(4, 0)$ , $(4, 4)$ , $(0, 4)$ cuts the square into two pieces of equal area. We need the equation of that line.

Givens

Square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(4, 0)$ , $\text{B}(4, 4)$ , $\text{C}(0, 4)$ .
Line $l$ passes through $\text{P}\left(1, \frac{3}{2}\right)$ .
Line $l$ splits the square's area in half.

Unknowns

The equation of line $l$ .

Constraints

One point on the line is known; a second point must come from the equal-area condition.

2 · Planchoose the strategy

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

Draw the square and notice it is symmetric about its center: turning it halfway around the center $(2, 2)$ lands it on itself. Any line through the center splits it into two matching halves, and a line that misses the center cannot split it evenly. So the center is the missing second point, and two points fix the line.

3 · Execute4 carry out the plan

1Find the center

#1 Draw a Diagram 6.G.A.1

The diagonals of the square meet at the midpoint of

\text{O}(0, 0)

and

\text{B}(4, 4)

.

\left(\frac{0+4}{2}, \frac{0+4}{2}\right)=(2, 2)

The center is the point every half-turn of the square keeps fixed.

2Why the line must pass through the center

#16 Count the Complement 6.G.A.1

A line through the center cuts the square into two pieces that are half-turns of each other, so they have equal area. If a line missed the center, the parallel line through the center would already split the area evenly, and our line would add or remove the strip between them from one side. So the bisecting line through

\text{P}

must also pass through

(2, 2)

.

Instead of measuring areas, ask which point every bisector must share.

3Slope through two points

#13 Convert to Algebra 8.F.B.4

Line

l

passes through

\text{P}\left(1, \frac{3}{2}\right)

and

(2, 2)

.

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

Rise over run between the two known points.

4Find the y-intercept

#13 Convert to Algebra 8.F.B.4

Write

y=\frac{1}{2}x+b

and substitute

x=2

,

y=2

.

2=\frac{1}{2}\times 2+b,\ b=1

One known point pins down the intercept once the slope is known.

Answer:

y=\frac{1}{2}x+1

4 · Reviewdoes it hold up?

At $x=0$ the line is at $y=1$ (on side $\text{OC}$ ) and at $x=4$ it is at $y=3$ (on side $\text{AB}$ ). The two pieces are trapezoids with parallel sides $1$ and $3$ (bottom piece) and $3$ and $1$ (top piece), each of width $4$ , so each has area $\frac{1}{2}\times\left(1+3\right)\times 4=8$ , half of $16$ .

Another way: Algebra only: let the line meet $x=0$ at height $p$ and $x=4$ at height $q$ . The lower piece is a trapezoid of area $\frac{1}{2}(p+q)\times 4=8$ , so $p+q=4$ , which says the line's height at $x=2$ is $\frac{p+q}{2}=2$ -- the center again.

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Building the line's equation from two points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Comparing the areas of the two pieces of the square.

💡Takeaway. A shape with a center of symmetry is cut in half by exactly the lines through that center.

Variant 6 medium

As shown in the figure, square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(4, 0)$ , $\text{B}(4, 4)$ , and $\text{C}(0, 4)$ . A line $l$ passes through the point $\text{P}\left(3, \frac{5}{2}\right)$ inside the square and divides the area of square $\text{OABC}$ into two equal parts. Find the equation of line $l$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through the point $\text{P}\left(3, \frac{5}{2}\right)$ inside the square with corners $(0, 0)$ , $(4, 0)$ , $(4, 4)$ , $(0, 4)$ cuts the square into two pieces of equal area. We need the equation of that line.

Givens

Square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(4, 0)$ , $\text{B}(4, 4)$ , $\text{C}(0, 4)$ .
Line $l$ passes through $\text{P}\left(3, \frac{5}{2}\right)$ .
Line $l$ splits the square's area in half.

Unknowns

The equation of line $l$ .

Constraints

One point on the line is known; a second point must come from the equal-area condition.

2 · Planchoose the strategy

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

Draw the square and notice it is symmetric about its center: turning it halfway around the center $(2, 2)$ lands it on itself. Any line through the center splits it into two matching halves, and a line that misses the center cannot split it evenly. So the center is the missing second point, and two points fix the line.

3 · Execute4 carry out the plan

1Find the center

#1 Draw a Diagram 6.G.A.1

The diagonals of the square meet at the midpoint of

\text{O}(0, 0)

and

\text{B}(4, 4)

.

\left(\frac{0+4}{2}, \frac{0+4}{2}\right)=(2, 2)

The center is the point every half-turn of the square keeps fixed.

2Why the line must pass through the center

#16 Count the Complement 6.G.A.1

A line through the center cuts the square into two pieces that are half-turns of each other, so they have equal area. If a line missed the center, the parallel line through the center would already split the area evenly, and our line would add or remove the strip between them from one side. So the bisecting line through

\text{P}

must also pass through

(2, 2)

.

Instead of measuring areas, ask which point every bisector must share.

3Slope through two points

#13 Convert to Algebra 8.F.B.4

Line

l

passes through

\text{P}\left(3, \frac{5}{2}\right)

and

(2, 2)

.

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

Rise over run between the two known points.

4Find the y-intercept

#13 Convert to Algebra 8.F.B.4

Write

y=\frac{1}{2}x+b

and substitute

x=2

,

y=2

.

2=\frac{1}{2}\times 2+b,\ b=1

One known point pins down the intercept once the slope is known.

Answer:

y=\frac{1}{2}x+1

4 · Reviewdoes it hold up?

At $x=0$ the line is at $y=1$ (on side $\text{OC}$ ) and at $x=4$ it is at $y=3$ (on side $\text{AB}$ ). The two pieces are trapezoids with parallel sides $1$ and $3$ (bottom piece) and $3$ and $1$ (top piece), each of width $4$ , so each has area $\frac{1}{2}\times\left(1+3\right)\times 4=8$ , half of $16$ .

Another way: Algebra only: let the line meet $x=0$ at height $p$ and $x=4$ at height $q$ . The lower piece is a trapezoid of area $\frac{1}{2}(p+q)\times 4=8$ , so $p+q=4$ , which says the line's height at $x=2$ is $\frac{p+q}{2}=2$ -- the center again.

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Building the line's equation from two points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Comparing the areas of the two pieces of the square.

💡Takeaway. A shape with a center of symmetry is cut in half by exactly the lines through that center.

Variant 7 medium

As shown in the figure, square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(6, 0)$ , $\text{B}(6, 6)$ , and $\text{C}(0, 6)$ . A line $l$ passes through the point $\text{P}\left(5, \frac{7}{2}\right)$ inside the square and divides the area of square $\text{OABC}$ into two equal parts. Find the equation of line $l$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through the point $\text{P}\left(5, \frac{7}{2}\right)$ inside the square with corners $(0, 0)$ , $(6, 0)$ , $(6, 6)$ , $(0, 6)$ cuts the square into two pieces of equal area. We need the equation of that line.

Givens

Square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(6, 0)$ , $\text{B}(6, 6)$ , $\text{C}(0, 6)$ .
Line $l$ passes through $\text{P}\left(5, \frac{7}{2}\right)$ .
Line $l$ splits the square's area in half.

Unknowns

The equation of line $l$ .

Constraints

One point on the line is known; a second point must come from the equal-area condition.

2 · Planchoose the strategy

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

Draw the square and notice it is symmetric about its center: turning it halfway around the center $(3, 3)$ lands it on itself. Any line through the center splits it into two matching halves, and a line that misses the center cannot split it evenly. So the center is the missing second point, and two points fix the line.

3 · Execute4 carry out the plan

1Find the center

#1 Draw a Diagram 6.G.A.1

The diagonals of the square meet at the midpoint of

\text{O}(0, 0)

and

\text{B}(6, 6)

.

\left(\frac{0+6}{2}, \frac{0+6}{2}\right)=(3, 3)

The center is the point every half-turn of the square keeps fixed.

2Why the line must pass through the center

#16 Count the Complement 6.G.A.1

A line through the center cuts the square into two pieces that are half-turns of each other, so they have equal area. If a line missed the center, the parallel line through the center would already split the area evenly, and our line would add or remove the strip between them from one side. So the bisecting line through

\text{P}

must also pass through

(3, 3)

.

Instead of measuring areas, ask which point every bisector must share.

3Slope through two points

#13 Convert to Algebra 8.F.B.4

Line

l

passes through

\text{P}\left(5, \frac{7}{2}\right)

and

(3, 3)

.

\frac{3-\frac{7}{2}}{3-5}=\frac{1}{4}

Rise over run between the two known points.

4Find the y-intercept

#13 Convert to Algebra 8.F.B.4

Write

y=\frac{1}{4}x+b

and substitute

x=3

,

y=3

.

3=\frac{1}{4}\times 3+b,\ b=\frac{9}{4}

One known point pins down the intercept once the slope is known.

Answer:

y=\frac{1}{4}x+\frac{9}{4}

4 · Reviewdoes it hold up?

At $x=0$ the line is at $y=\frac{9}{4}$ (on side $\text{OC}$ ) and at $x=6$ it is at $y=\frac{15}{4}$ (on side $\text{AB}$ ). The two pieces are trapezoids with parallel sides $\frac{9}{4}$ and $\frac{15}{4}$ (bottom piece) and $\frac{15}{4}$ and $\frac{9}{4}$ (top piece), each of width $6$ , so each has area $\frac{1}{2}\times\left(\frac{9}{4}+\frac{15}{4}\right)\times 6=18$ , half of $36$ .

Another way: Algebra only: let the line meet $x=0$ at height $p$ and $x=6$ at height $q$ . The lower piece is a trapezoid of area $\frac{1}{2}(p+q)\times 6=18$ , so $p+q=6$ , which says the line's height at $x=3$ is $\frac{p+q}{2}=3$ -- the center again.

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Building the line's equation from two points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Comparing the areas of the two pieces of the square.

💡Takeaway. A shape with a center of symmetry is cut in half by exactly the lines through that center.

Variant 8 medium

As shown in the figure, square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(8, 0)$ , $\text{B}(8, 8)$ , and $\text{C}(0, 8)$ . A line $l$ passes through the point $\text{P}\left(1, \frac{7}{2}\right)$ inside the square and divides the area of square $\text{OABC}$ into two equal parts. Find the equation of line $l$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through the point $\text{P}\left(1, \frac{7}{2}\right)$ inside the square with corners $(0, 0)$ , $(8, 0)$ , $(8, 8)$ , $(0, 8)$ cuts the square into two pieces of equal area. We need the equation of that line.

Givens

Square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(8, 0)$ , $\text{B}(8, 8)$ , $\text{C}(0, 8)$ .
Line $l$ passes through $\text{P}\left(1, \frac{7}{2}\right)$ .
Line $l$ splits the square's area in half.

Unknowns

The equation of line $l$ .

Constraints

One point on the line is known; a second point must come from the equal-area condition.

2 · Planchoose the strategy

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

Draw the square and notice it is symmetric about its center: turning it halfway around the center $(4, 4)$ lands it on itself. Any line through the center splits it into two matching halves, and a line that misses the center cannot split it evenly. So the center is the missing second point, and two points fix the line.

3 · Execute4 carry out the plan

1Find the center

#1 Draw a Diagram 6.G.A.1

The diagonals of the square meet at the midpoint of

\text{O}(0, 0)

and

\text{B}(8, 8)

.

\left(\frac{0+8}{2}, \frac{0+8}{2}\right)=(4, 4)

The center is the point every half-turn of the square keeps fixed.

2Why the line must pass through the center

#16 Count the Complement 6.G.A.1

A line through the center cuts the square into two pieces that are half-turns of each other, so they have equal area. If a line missed the center, the parallel line through the center would already split the area evenly, and our line would add or remove the strip between them from one side. So the bisecting line through

\text{P}

must also pass through

(4, 4)

.

Instead of measuring areas, ask which point every bisector must share.

3Slope through two points

#13 Convert to Algebra 8.F.B.4

Line

l

passes through

\text{P}\left(1, \frac{7}{2}\right)

and

(4, 4)

.

\frac{4-\frac{7}{2}}{4-1}=\frac{1}{6}

Rise over run between the two known points.

4Find the y-intercept

#13 Convert to Algebra 8.F.B.4

Write

y=\frac{1}{6}x+b

and substitute

x=4

,

y=4

.

4=\frac{1}{6}\times 4+b,\ b=\frac{10}{3}

One known point pins down the intercept once the slope is known.

Answer:

y=\frac{1}{6}x+\frac{10}{3}

4 · Reviewdoes it hold up?

At $x=0$ the line is at $y=\frac{10}{3}$ (on side $\text{OC}$ ) and at $x=8$ it is at $y=\frac{14}{3}$ (on side $\text{AB}$ ). The two pieces are trapezoids with parallel sides $\frac{10}{3}$ and $\frac{14}{3}$ (bottom piece) and $\frac{14}{3}$ and $\frac{10}{3}$ (top piece), each of width $8$ , so each has area $\frac{1}{2}\times\left(\frac{10}{3}+\frac{14}{3}\right)\times 8=32$ , half of $64$ .

Another way: Algebra only: let the line meet $x=0$ at height $p$ and $x=8$ at height $q$ . The lower piece is a trapezoid of area $\frac{1}{2}(p+q)\times 8=32$ , so $p+q=8$ , which says the line's height at $x=4$ is $\frac{p+q}{2}=4$ -- the center again.

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Building the line's equation from two points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Comparing the areas of the two pieces of the square.

💡Takeaway. A shape with a center of symmetry is cut in half by exactly the lines through that center.

Variant 9 hard

As shown in the figure, square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(10, 0)$ , $\text{B}(10, 10)$ , and $\text{C}(0, 10)$ . A line $l$ passes through the point $\text{P}\left(8, \frac{9}{2}\right)$ inside the square and divides the area of square $\text{OABC}$ into two equal parts. Find the equation of line $l$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through the point $\text{P}\left(8, \frac{9}{2}\right)$ inside the square with corners $(0, 0)$ , $(10, 0)$ , $(10, 10)$ , $(0, 10)$ cuts the square into two pieces of equal area. We need the equation of that line.

Givens

Square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(10, 0)$ , $\text{B}(10, 10)$ , $\text{C}(0, 10)$ .
Line $l$ passes through $\text{P}\left(8, \frac{9}{2}\right)$ .
Line $l$ splits the square's area in half.

Unknowns

The equation of line $l$ .

Constraints

One point on the line is known; a second point must come from the equal-area condition.

2 · Planchoose the strategy

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

Draw the square and notice it is symmetric about its center: turning it halfway around the center $(5, 5)$ lands it on itself. Any line through the center splits it into two matching halves, and a line that misses the center cannot split it evenly. So the center is the missing second point, and two points fix the line.

3 · Execute4 carry out the plan

1Find the center

#1 Draw a Diagram 6.G.A.1

The diagonals of the square meet at the midpoint of

\text{O}(0, 0)

and

\text{B}(10, 10)

.

\left(\frac{0+10}{2}, \frac{0+10}{2}\right)=(5, 5)

The center is the point every half-turn of the square keeps fixed.

2Why the line must pass through the center

#16 Count the Complement 6.G.A.1

A line through the center cuts the square into two pieces that are half-turns of each other, so they have equal area. If a line missed the center, the parallel line through the center would already split the area evenly, and our line would add or remove the strip between them from one side. So the bisecting line through

\text{P}

must also pass through

(5, 5)

.

Instead of measuring areas, ask which point every bisector must share.

3Slope through two points

#13 Convert to Algebra 8.F.B.4

Line

l

passes through

\text{P}\left(8, \frac{9}{2}\right)

and

(5, 5)

.

\frac{5-\frac{9}{2}}{5-8}=-\frac{1}{6}

Rise over run between the two known points.

4Find the y-intercept

#13 Convert to Algebra 8.F.B.4

Write

y=-\frac{1}{6}x+b

and substitute

x=5

,

y=5

.

5=-\frac{1}{6}\times 5+b,\ b=\frac{35}{6}

One known point pins down the intercept once the slope is known.

Answer:

y=-\frac{1}{6}x+\frac{35}{6}

4 · Reviewdoes it hold up?

At $x=0$ the line is at $y=\frac{35}{6}$ (on side $\text{OC}$ ) and at $x=10$ it is at $y=\frac{25}{6}$ (on side $\text{AB}$ ). The two pieces are trapezoids with parallel sides $\frac{35}{6}$ and $\frac{25}{6}$ (bottom piece) and $\frac{25}{6}$ and $\frac{35}{6}$ (top piece), each of width $10$ , so each has area $\frac{1}{2}\times\left(\frac{35}{6}+\frac{25}{6}\right)\times 10=50$ , half of $100$ .

Another way: Algebra only: let the line meet $x=0$ at height $p$ and $x=10$ at height $q$ . The lower piece is a trapezoid of area $\frac{1}{2}(p+q)\times 10=50$ , so $p+q=10$ , which says the line's height at $x=5$ is $\frac{p+q}{2}=5$ -- the center again.

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Building the line's equation from two points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Comparing the areas of the two pieces of the square.

💡Takeaway. A shape with a center of symmetry is cut in half by exactly the lines through that center.

Variant 10 hard

As shown in the figure, square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(12, 0)$ , $\text{B}(12, 12)$ , and $\text{C}(0, 12)$ . A line $l$ passes through the point $\text{P}\left(4, \frac{9}{2}\right)$ inside the square and divides the area of square $\text{OABC}$ into two equal parts. Find the equation of line $l$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through the point $\text{P}\left(4, \frac{9}{2}\right)$ inside the square with corners $(0, 0)$ , $(12, 0)$ , $(12, 12)$ , $(0, 12)$ cuts the square into two pieces of equal area. We need the equation of that line.

Givens

Square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(12, 0)$ , $\text{B}(12, 12)$ , $\text{C}(0, 12)$ .
Line $l$ passes through $\text{P}\left(4, \frac{9}{2}\right)$ .
Line $l$ splits the square's area in half.

Unknowns

The equation of line $l$ .

Constraints

One point on the line is known; a second point must come from the equal-area condition.

2 · Planchoose the strategy

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

Draw the square and notice it is symmetric about its center: turning it halfway around the center $(6, 6)$ lands it on itself. Any line through the center splits it into two matching halves, and a line that misses the center cannot split it evenly. So the center is the missing second point, and two points fix the line.

3 · Execute4 carry out the plan

1Find the center

#1 Draw a Diagram 6.G.A.1

The diagonals of the square meet at the midpoint of

\text{O}(0, 0)

and

\text{B}(12, 12)

.

\left(\frac{0+12}{2}, \frac{0+12}{2}\right)=(6, 6)

The center is the point every half-turn of the square keeps fixed.

2Why the line must pass through the center

#16 Count the Complement 6.G.A.1

A line through the center cuts the square into two pieces that are half-turns of each other, so they have equal area. If a line missed the center, the parallel line through the center would already split the area evenly, and our line would add or remove the strip between them from one side. So the bisecting line through

\text{P}

must also pass through

(6, 6)

.

Instead of measuring areas, ask which point every bisector must share.

3Slope through two points

#13 Convert to Algebra 8.F.B.4

Line

l

passes through

\text{P}\left(4, \frac{9}{2}\right)

and

(6, 6)

.

\frac{6-\frac{9}{2}}{6-4}=\frac{3}{4}

Rise over run between the two known points.

4Find the y-intercept

#13 Convert to Algebra 8.F.B.4

Write

y=\frac{3}{4}x+b

and substitute

x=6

,

y=6

.

6=\frac{3}{4}\times 6+b,\ b=\frac{3}{2}

One known point pins down the intercept once the slope is known.

Answer:

y=\frac{3}{4}x+\frac{3}{2}

4 · Reviewdoes it hold up?

At $x=0$ the line is at $y=\frac{3}{2}$ (on side $\text{OC}$ ) and at $x=12$ it is at $y=\frac{21}{2}$ (on side $\text{AB}$ ). The two pieces are trapezoids with parallel sides $\frac{3}{2}$ and $\frac{21}{2}$ (bottom piece) and $\frac{21}{2}$ and $\frac{3}{2}$ (top piece), each of width $12$ , so each has area $\frac{1}{2}\times\left(\frac{3}{2}+\frac{21}{2}\right)\times 12=72$ , half of $144$ .

Another way: Algebra only: let the line meet $x=0$ at height $p$ and $x=12$ at height $q$ . The lower piece is a trapezoid of area $\frac{1}{2}(p+q)\times 12=72$ , so $p+q=12$ , which says the line's height at $x=6$ is $\frac{p+q}{2}=6$ -- the center again.

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Building the line's equation from two points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Comparing the areas of the two pieces of the square.

💡Takeaway. A shape with a center of symmetry is cut in half by exactly the lines through that center.

Variant 11 hard

As shown in the figure, square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(4, 0)$ , $\text{B}(4, 4)$ , and $\text{C}(0, 4)$ . A line $l$ passes through the point $\text{P}\left(\frac{1}{2}, \frac{5}{2}\right)$ inside the square and divides the area of square $\text{OABC}$ into two equal parts. Find the equation of line $l$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through the point $\text{P}\left(\frac{1}{2}, \frac{5}{2}\right)$ inside the square with corners $(0, 0)$ , $(4, 0)$ , $(4, 4)$ , $(0, 4)$ cuts the square into two pieces of equal area. We need the equation of that line.

Givens

Square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(4, 0)$ , $\text{B}(4, 4)$ , $\text{C}(0, 4)$ .
Line $l$ passes through $\text{P}\left(\frac{1}{2}, \frac{5}{2}\right)$ .
Line $l$ splits the square's area in half.

Unknowns

The equation of line $l$ .

Constraints

One point on the line is known; a second point must come from the equal-area condition.

2 · Planchoose the strategy

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

Draw the square and notice it is symmetric about its center: turning it halfway around the center $(2, 2)$ lands it on itself. Any line through the center splits it into two matching halves, and a line that misses the center cannot split it evenly. So the center is the missing second point, and two points fix the line.

3 · Execute4 carry out the plan

1Find the center

#1 Draw a Diagram 6.G.A.1

The diagonals of the square meet at the midpoint of

\text{O}(0, 0)

and

\text{B}(4, 4)

.

\left(\frac{0+4}{2}, \frac{0+4}{2}\right)=(2, 2)

The center is the point every half-turn of the square keeps fixed.

2Why the line must pass through the center

#16 Count the Complement 6.G.A.1

A line through the center cuts the square into two pieces that are half-turns of each other, so they have equal area. If a line missed the center, the parallel line through the center would already split the area evenly, and our line would add or remove the strip between them from one side. So the bisecting line through

\text{P}

must also pass through

(2, 2)

.

Instead of measuring areas, ask which point every bisector must share.

3Slope through two points

#13 Convert to Algebra 8.F.B.4

Line

l

passes through

\text{P}\left(\frac{1}{2}, \frac{5}{2}\right)

and

(2, 2)

.

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

Rise over run between the two known points.

4Find the y-intercept

#13 Convert to Algebra 8.F.B.4

Write

y=-\frac{1}{3}x+b

and substitute

x=2

,

y=2

.

2=-\frac{1}{3}\times 2+b,\ b=\frac{8}{3}

One known point pins down the intercept once the slope is known.

Answer:

y=-\frac{1}{3}x+\frac{8}{3}

4 · Reviewdoes it hold up?

At $x=0$ the line is at $y=\frac{8}{3}$ (on side $\text{OC}$ ) and at $x=4$ it is at $y=\frac{4}{3}$ (on side $\text{AB}$ ). The two pieces are trapezoids with parallel sides $\frac{8}{3}$ and $\frac{4}{3}$ (bottom piece) and $\frac{4}{3}$ and $\frac{8}{3}$ (top piece), each of width $4$ , so each has area $\frac{1}{2}\times\left(\frac{8}{3}+\frac{4}{3}\right)\times 4=8$ , half of $16$ .

Another way: Algebra only: let the line meet $x=0$ at height $p$ and $x=4$ at height $q$ . The lower piece is a trapezoid of area $\frac{1}{2}(p+q)\times 4=8$ , so $p+q=4$ , which says the line's height at $x=2$ is $\frac{p+q}{2}=2$ -- the center again.

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Building the line's equation from two points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Comparing the areas of the two pieces of the square.

💡Takeaway. A shape with a center of symmetry is cut in half by exactly the lines through that center.

Variant 12 hard

As shown in the figure, square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(6, 0)$ , $\text{B}(6, 6)$ , and $\text{C}(0, 6)$ . A line $l$ passes through the point $\text{P}\left(\frac{3}{2}, \frac{5}{2}\right)$ inside the square and divides the area of square $\text{OABC}$ into two equal parts. Find the equation of line $l$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A line through the point $\text{P}\left(\frac{3}{2}, \frac{5}{2}\right)$ inside the square with corners $(0, 0)$ , $(6, 0)$ , $(6, 6)$ , $(0, 6)$ cuts the square into two pieces of equal area. We need the equation of that line.

Givens

Square $\text{OABC}$ has vertices $\text{O}(0, 0)$ , $\text{A}(6, 0)$ , $\text{B}(6, 6)$ , $\text{C}(0, 6)$ .
Line $l$ passes through $\text{P}\left(\frac{3}{2}, \frac{5}{2}\right)$ .
Line $l$ splits the square's area in half.

Unknowns

The equation of line $l$ .

Constraints

One point on the line is known; a second point must come from the equal-area condition.

2 · Planchoose the strategy

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

Draw the square and notice it is symmetric about its center: turning it halfway around the center $(3, 3)$ lands it on itself. Any line through the center splits it into two matching halves, and a line that misses the center cannot split it evenly. So the center is the missing second point, and two points fix the line.

3 · Execute4 carry out the plan

1Find the center

#1 Draw a Diagram 6.G.A.1

The diagonals of the square meet at the midpoint of

\text{O}(0, 0)

and

\text{B}(6, 6)

.

\left(\frac{0+6}{2}, \frac{0+6}{2}\right)=(3, 3)

The center is the point every half-turn of the square keeps fixed.

2Why the line must pass through the center

#16 Count the Complement 6.G.A.1

A line through the center cuts the square into two pieces that are half-turns of each other, so they have equal area. If a line missed the center, the parallel line through the center would already split the area evenly, and our line would add or remove the strip between them from one side. So the bisecting line through

\text{P}

must also pass through

(3, 3)

.

Instead of measuring areas, ask which point every bisector must share.

3Slope through two points

#13 Convert to Algebra 8.F.B.4

Line

l

passes through

\text{P}\left(\frac{3}{2}, \frac{5}{2}\right)

and

(3, 3)

.

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

Rise over run between the two known points.

4Find the y-intercept

#13 Convert to Algebra 8.F.B.4

Write

y=\frac{1}{3}x+b

and substitute

x=3

,

y=3

.

3=\frac{1}{3}\times 3+b,\ b=2

One known point pins down the intercept once the slope is known.

Answer:

y=\frac{1}{3}x+2

4 · Reviewdoes it hold up?

At $x=0$ the line is at $y=2$ (on side $\text{OC}$ ) and at $x=6$ it is at $y=4$ (on side $\text{AB}$ ). The two pieces are trapezoids with parallel sides $2$ and $4$ (bottom piece) and $4$ and $2$ (top piece), each of width $6$ , so each has area $\frac{1}{2}\times\left(2+4\right)\times 6=18$ , half of $36$ .

Another way: Algebra only: let the line meet $x=0$ at height $p$ and $x=6$ at height $q$ . The lower piece is a trapezoid of area $\frac{1}{2}(p+q)\times 6=18$ , so $p+q=6$ , which says the line's height at $x=3$ is $\frac{p+q}{2}=3$ -- the center again.

Standardsmin grade 8

8.F.B.4 Construct a function to model a linear relationship between two quantities — Building the line's equation from two points.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Comparing the areas of the two pieces of the square.

💡Takeaway. A shape with a center of symmetry is cut in half by exactly the lines through that center.

Area bisectors of a square pass through its center · 12 practice problems

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