Practice · Constant xy fixes the triangle's area · Sensim Math

Variant 1 easy

The graph of an inverse proportion, shown in the figure, passes through the point P $(-2, -1)$ . Point A is on the branch in the first quadrant and point B is on the branch in the third quadrant, and segment AB passes through the origin O. Point C is chosen so that AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis. Find the area of triangle ABC.

Your answer

Show solution

1 · Understandwhat's really being asked

Find the inverse proportion from P, then let A $(p, q)$ be any point on the first-quadrant branch. Because AB passes through O, B is A reflected through the origin, and the right triangle ABC has legs we can write with $p$ and $q$ . Its area depends only on $pq$ .

Givens

The inverse proportion passes through P $(-2, -1)$ .
A is on the first-quadrant branch, B on the third-quadrant branch, and AB passes through O.
AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis.

Unknowns

The area of triangle ABC.

Constraints

The position of A is not given, so the area must be the same for every A.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#17 Visualize Spatial Relationships

Instead of hunting for A, change focus to the product $xy$ , which is the same for every point on the graph. Point symmetry about O turns the legs of the triangle into $2p$ and $2q$ .

3 · Execute4 carry out the plan

1Find the constant

#13 Convert to Algebra 7.NS.A.2

For

y=\frac{k}{x}

, the product

xy

is the constant

k

. Use P.

k = (-2) \times (-1) = 2 \Rightarrow y = \frac{2}{x}

Every point on this graph has

xy = 2

.

2Name the points

#17 Visualize Spatial Relationships 6.NS.C.8

Let A

=(p, q)

with

pq=2

. The line through O and A is

y=\frac{q}{p}x

, and it meets the curve again at the reflection of A through O.

B = (-p, -q),\quad C = (p, -q)

(-p)(-q) = pq = 2

, so

(-p, -q)

is on the graph and on the line through O and A.

3Leg lengths

#17 Visualize Spatial Relationships 6.NS.C.8

AC is vertical from

q

down to

-q

; BC is horizontal from

-p

to

p

.

AC = q - (-q) = 2q,\quad BC = p - (-p) = 2p

Distance along a grid line is the difference of coordinates.

4Area

#16 Count the Complement 7.G.B.6

Triangle ABC is right-angled at C.

\frac{1}{2} \times 2p \times 2q = 2pq = 2 \times 2 = 4

The unknown

p

and

q

appear only as the product

pq

, which we know.

Answer:

4

4 · Reviewdoes it hold up?

Try a concrete A: A $(1, 2)$ gives B $(-1, -2)$ , C $(1, -2)$ , legs $2$ and $4$ , area $\frac{1}{2} \times 2 \times 4 = 4$ . A $(2, 1)$ gives legs $4$ and $2$ , again $4$ .

Another way: Let D be where AC crosses the $x$ -axis. Triangle AOD has area $\frac{1}{2}pq = 1$ , and triangle ABC is the same shape scaled by $2$ (O is the midpoint of AB), so its area is $4 \times 1 = 4$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying the signed coordinates of P to get the constant $(-2)\times(-1)=2$.
6.NS.C.8 Solve problems by graphing points in four quadrants — Reflecting through the origin and finding lengths from coordinates.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Area of the right triangle.

💡Takeaway. When a point on

y=\frac{k}{x}

is not pinned down, the answer usually only needs

xy = k

.

Variant 2 easy

The graph of an inverse proportion, shown in the figure, passes through the point P $(-1, -2)$ . Point A is on the branch in the first quadrant and point B is on the branch in the third quadrant, and segment AB passes through the origin O. Point C is chosen so that AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis. Find the area of triangle ABC.

Your answer

Show solution

1 · Understandwhat's really being asked

Find the inverse proportion from P, then let A $(p, q)$ be any point on the first-quadrant branch. Because AB passes through O, B is A reflected through the origin, and the right triangle ABC has legs we can write with $p$ and $q$ . Its area depends only on $pq$ .

Givens

The inverse proportion passes through P $(-1, -2)$ .
A is on the first-quadrant branch, B on the third-quadrant branch, and AB passes through O.
AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis.

Unknowns

The area of triangle ABC.

Constraints

The position of A is not given, so the area must be the same for every A.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#17 Visualize Spatial Relationships

Instead of hunting for A, change focus to the product $xy$ , which is the same for every point on the graph. Point symmetry about O turns the legs of the triangle into $2p$ and $2q$ .

3 · Execute4 carry out the plan

1Find the constant

#13 Convert to Algebra 7.NS.A.2

For

y=\frac{k}{x}

, the product

xy

is the constant

k

. Use P.

k = (-1) \times (-2) = 2 \Rightarrow y = \frac{2}{x}

Every point on this graph has

xy = 2

.

2Name the points

#17 Visualize Spatial Relationships 6.NS.C.8

Let A

=(p, q)

with

pq=2

. The line through O and A is

y=\frac{q}{p}x

, and it meets the curve again at the reflection of A through O.

B = (-p, -q),\quad C = (p, -q)

(-p)(-q) = pq = 2

, so

(-p, -q)

is on the graph and on the line through O and A.

3Leg lengths

#17 Visualize Spatial Relationships 6.NS.C.8

AC is vertical from

q

down to

-q

; BC is horizontal from

-p

to

p

.

AC = q - (-q) = 2q,\quad BC = p - (-p) = 2p

Distance along a grid line is the difference of coordinates.

4Area

#16 Count the Complement 7.G.B.6

Triangle ABC is right-angled at C.

\frac{1}{2} \times 2p \times 2q = 2pq = 2 \times 2 = 4

The unknown

p

and

q

appear only as the product

pq

, which we know.

Answer:

4

4 · Reviewdoes it hold up?

Try a concrete A: A $(1, 2)$ gives B $(-1, -2)$ , C $(1, -2)$ , legs $2$ and $4$ , area $\frac{1}{2} \times 2 \times 4 = 4$ . A $(2, 1)$ gives legs $4$ and $2$ , again $4$ .

Another way: Let D be where AC crosses the $x$ -axis. Triangle AOD has area $\frac{1}{2}pq = 1$ , and triangle ABC is the same shape scaled by $2$ (O is the midpoint of AB), so its area is $4 \times 1 = 4$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying the signed coordinates of P to get the constant $(-1)\times(-2)=2$.
6.NS.C.8 Solve problems by graphing points in four quadrants — Reflecting through the origin and finding lengths from coordinates.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Area of the right triangle.

💡Takeaway. When a point on

y=\frac{k}{x}

is not pinned down, the answer usually only needs

xy = k

.

Variant 3 easy

The graph of an inverse proportion, shown in the figure, passes through the point P $(-2, -2)$ . Point A is on the branch in the first quadrant and point B is on the branch in the third quadrant, and segment AB passes through the origin O. Point C is chosen so that AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis. Find the area of triangle ABC.

Your answer

Show solution

1 · Understandwhat's really being asked

Find the inverse proportion from P, then let A $(p, q)$ be any point on the first-quadrant branch. Because AB passes through O, B is A reflected through the origin, and the right triangle ABC has legs we can write with $p$ and $q$ . Its area depends only on $pq$ .

Givens

The inverse proportion passes through P $(-2, -2)$ .
A is on the first-quadrant branch, B on the third-quadrant branch, and AB passes through O.
AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis.

Unknowns

The area of triangle ABC.

Constraints

The position of A is not given, so the area must be the same for every A.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#17 Visualize Spatial Relationships

Instead of hunting for A, change focus to the product $xy$ , which is the same for every point on the graph. Point symmetry about O turns the legs of the triangle into $2p$ and $2q$ .

3 · Execute4 carry out the plan

1Find the constant

#13 Convert to Algebra 7.NS.A.2

For

y=\frac{k}{x}

, the product

xy

is the constant

k

. Use P.

k = (-2) \times (-2) = 4 \Rightarrow y = \frac{4}{x}

Every point on this graph has

xy = 4

.

2Name the points

#17 Visualize Spatial Relationships 6.NS.C.8

Let A

=(p, q)

with

pq=4

. The line through O and A is

y=\frac{q}{p}x

, and it meets the curve again at the reflection of A through O.

B = (-p, -q),\quad C = (p, -q)

(-p)(-q) = pq = 4

, so

(-p, -q)

is on the graph and on the line through O and A.

3Leg lengths

#17 Visualize Spatial Relationships 6.NS.C.8

AC is vertical from

q

down to

-q

; BC is horizontal from

-p

to

p

.

AC = q - (-q) = 2q,\quad BC = p - (-p) = 2p

Distance along a grid line is the difference of coordinates.

4Area

#16 Count the Complement 7.G.B.6

Triangle ABC is right-angled at C.

\frac{1}{2} \times 2p \times 2q = 2pq = 2 \times 4 = 8

The unknown

p

and

q

appear only as the product

pq

, which we know.

Answer:

8

4 · Reviewdoes it hold up?

Try a concrete A: A $(1, 4)$ gives B $(-1, -4)$ , C $(1, -4)$ , legs $2$ and $8$ , area $\frac{1}{2} \times 2 \times 8 = 8$ . A $(4, 1)$ gives legs $8$ and $2$ , again $8$ .

Another way: Let D be where AC crosses the $x$ -axis. Triangle AOD has area $\frac{1}{2}pq = 2$ , and triangle ABC is the same shape scaled by $2$ (O is the midpoint of AB), so its area is $4 \times 2 = 8$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying the signed coordinates of P to get the constant $(-2)\times(-2)=4$.
6.NS.C.8 Solve problems by graphing points in four quadrants — Reflecting through the origin and finding lengths from coordinates.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Area of the right triangle.

💡Takeaway. When a point on

y=\frac{k}{x}

is not pinned down, the answer usually only needs

xy = k

.

Variant 4 medium

The graph of an inverse proportion, shown in the figure, passes through the point P $(-2, -3)$ . Point A is on the branch in the first quadrant and point B is on the branch in the third quadrant, and segment AB passes through the origin O. Point C is chosen so that AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis. Find the area of triangle ABC.

Your answer

Show solution

1 · Understandwhat's really being asked

Find the inverse proportion from P, then let A $(p, q)$ be any point on the first-quadrant branch. Because AB passes through O, B is A reflected through the origin, and the right triangle ABC has legs we can write with $p$ and $q$ . Its area depends only on $pq$ .

Givens

The inverse proportion passes through P $(-2, -3)$ .
A is on the first-quadrant branch, B on the third-quadrant branch, and AB passes through O.
AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis.

Unknowns

The area of triangle ABC.

Constraints

The position of A is not given, so the area must be the same for every A.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#17 Visualize Spatial Relationships

Instead of hunting for A, change focus to the product $xy$ , which is the same for every point on the graph. Point symmetry about O turns the legs of the triangle into $2p$ and $2q$ .

3 · Execute4 carry out the plan

1Find the constant

#13 Convert to Algebra 7.NS.A.2

For

y=\frac{k}{x}

, the product

xy

is the constant

k

. Use P.

k = (-2) \times (-3) = 6 \Rightarrow y = \frac{6}{x}

Every point on this graph has

xy = 6

.

2Name the points

#17 Visualize Spatial Relationships 6.NS.C.8

Let A

=(p, q)

with

pq=6

. The line through O and A is

y=\frac{q}{p}x

, and it meets the curve again at the reflection of A through O.

B = (-p, -q),\quad C = (p, -q)

(-p)(-q) = pq = 6

, so

(-p, -q)

is on the graph and on the line through O and A.

3Leg lengths

#17 Visualize Spatial Relationships 6.NS.C.8

AC is vertical from

q

down to

-q

; BC is horizontal from

-p

to

p

.

AC = q - (-q) = 2q,\quad BC = p - (-p) = 2p

Distance along a grid line is the difference of coordinates.

4Area

#16 Count the Complement 7.G.B.6

Triangle ABC is right-angled at C.

\frac{1}{2} \times 2p \times 2q = 2pq = 2 \times 6 = 12

The unknown

p

and

q

appear only as the product

pq

, which we know.

Answer:

12

4 · Reviewdoes it hold up?

Try a concrete A: A $(1, 6)$ gives B $(-1, -6)$ , C $(1, -6)$ , legs $2$ and $12$ , area $\frac{1}{2} \times 2 \times 12 = 12$ . A $(6, 1)$ gives legs $12$ and $2$ , again $12$ .

Another way: Let D be where AC crosses the $x$ -axis. Triangle AOD has area $\frac{1}{2}pq = 3$ , and triangle ABC is the same shape scaled by $2$ (O is the midpoint of AB), so its area is $4 \times 3 = 12$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying the signed coordinates of P to get the constant $(-2)\times(-3)=6$.
6.NS.C.8 Solve problems by graphing points in four quadrants — Reflecting through the origin and finding lengths from coordinates.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Area of the right triangle.

💡Takeaway. When a point on

y=\frac{k}{x}

is not pinned down, the answer usually only needs

xy = k

.

Variant 5 easy

The graph of an inverse proportion, shown in the figure, passes through the point P $(-3, -1)$ . Point A is on the branch in the first quadrant and point B is on the branch in the third quadrant, and segment AB passes through the origin O. Point C is chosen so that AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis. Find the area of triangle ABC.

Your answer

Show solution

1 · Understandwhat's really being asked

Find the inverse proportion from P, then let A $(p, q)$ be any point on the first-quadrant branch. Because AB passes through O, B is A reflected through the origin, and the right triangle ABC has legs we can write with $p$ and $q$ . Its area depends only on $pq$ .

Givens

The inverse proportion passes through P $(-3, -1)$ .
A is on the first-quadrant branch, B on the third-quadrant branch, and AB passes through O.
AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis.

Unknowns

The area of triangle ABC.

Constraints

The position of A is not given, so the area must be the same for every A.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#17 Visualize Spatial Relationships

Instead of hunting for A, change focus to the product $xy$ , which is the same for every point on the graph. Point symmetry about O turns the legs of the triangle into $2p$ and $2q$ .

3 · Execute4 carry out the plan

1Find the constant

#13 Convert to Algebra 7.NS.A.2

For

y=\frac{k}{x}

, the product

xy

is the constant

k

. Use P.

k = (-3) \times (-1) = 3 \Rightarrow y = \frac{3}{x}

Every point on this graph has

xy = 3

.

2Name the points

#17 Visualize Spatial Relationships 6.NS.C.8

Let A

=(p, q)

with

pq=3

. The line through O and A is

y=\frac{q}{p}x

, and it meets the curve again at the reflection of A through O.

B = (-p, -q),\quad C = (p, -q)

(-p)(-q) = pq = 3

, so

(-p, -q)

is on the graph and on the line through O and A.

3Leg lengths

#17 Visualize Spatial Relationships 6.NS.C.8

AC is vertical from

q

down to

-q

; BC is horizontal from

-p

to

p

.

AC = q - (-q) = 2q,\quad BC = p - (-p) = 2p

Distance along a grid line is the difference of coordinates.

4Area

#16 Count the Complement 7.G.B.6

Triangle ABC is right-angled at C.

\frac{1}{2} \times 2p \times 2q = 2pq = 2 \times 3 = 6

The unknown

p

and

q

appear only as the product

pq

, which we know.

Answer:

6

4 · Reviewdoes it hold up?

Try a concrete A: A $(1, 3)$ gives B $(-1, -3)$ , C $(1, -3)$ , legs $2$ and $6$ , area $\frac{1}{2} \times 2 \times 6 = 6$ . A $(3, 1)$ gives legs $6$ and $2$ , again $6$ .

Another way: Let D be where AC crosses the $x$ -axis. Triangle AOD has area $\frac{1}{2}pq = \frac{3}{2}$ , and triangle ABC is the same shape scaled by $2$ (O is the midpoint of AB), so its area is $4 \times \frac{3}{2} = 6$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying the signed coordinates of P to get the constant $(-3)\times(-1)=3$.
6.NS.C.8 Solve problems by graphing points in four quadrants — Reflecting through the origin and finding lengths from coordinates.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Area of the right triangle.

💡Takeaway. When a point on

y=\frac{k}{x}

is not pinned down, the answer usually only needs

xy = k

.

Variant 6 medium

The graph of an inverse proportion, shown in the figure, passes through the point P $(-3, -3)$ . Point A is on the branch in the first quadrant and point B is on the branch in the third quadrant, and segment AB passes through the origin O. Point C is chosen so that AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis. Find the area of triangle ABC.

Your answer

Show solution

1 · Understandwhat's really being asked

Find the inverse proportion from P, then let A $(p, q)$ be any point on the first-quadrant branch. Because AB passes through O, B is A reflected through the origin, and the right triangle ABC has legs we can write with $p$ and $q$ . Its area depends only on $pq$ .

Givens

The inverse proportion passes through P $(-3, -3)$ .
A is on the first-quadrant branch, B on the third-quadrant branch, and AB passes through O.
AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis.

Unknowns

The area of triangle ABC.

Constraints

The position of A is not given, so the area must be the same for every A.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#17 Visualize Spatial Relationships

Instead of hunting for A, change focus to the product $xy$ , which is the same for every point on the graph. Point symmetry about O turns the legs of the triangle into $2p$ and $2q$ .

3 · Execute4 carry out the plan

1Find the constant

#13 Convert to Algebra 7.NS.A.2

For

y=\frac{k}{x}

, the product

xy

is the constant

k

. Use P.

k = (-3) \times (-3) = 9 \Rightarrow y = \frac{9}{x}

Every point on this graph has

xy = 9

.

2Name the points

#17 Visualize Spatial Relationships 6.NS.C.8

Let A

=(p, q)

with

pq=9

. The line through O and A is

y=\frac{q}{p}x

, and it meets the curve again at the reflection of A through O.

B = (-p, -q),\quad C = (p, -q)

(-p)(-q) = pq = 9

, so

(-p, -q)

is on the graph and on the line through O and A.

3Leg lengths

#17 Visualize Spatial Relationships 6.NS.C.8

AC is vertical from

q

down to

-q

; BC is horizontal from

-p

to

p

.

AC = q - (-q) = 2q,\quad BC = p - (-p) = 2p

Distance along a grid line is the difference of coordinates.

4Area

#16 Count the Complement 7.G.B.6

Triangle ABC is right-angled at C.

\frac{1}{2} \times 2p \times 2q = 2pq = 2 \times 9 = 18

The unknown

p

and

q

appear only as the product

pq

, which we know.

Answer:

18

4 · Reviewdoes it hold up?

Try a concrete A: A $(1, 9)$ gives B $(-1, -9)$ , C $(1, -9)$ , legs $2$ and $18$ , area $\frac{1}{2} \times 2 \times 18 = 18$ . A $(9, 1)$ gives legs $18$ and $2$ , again $18$ .

Another way: Let D be where AC crosses the $x$ -axis. Triangle AOD has area $\frac{1}{2}pq = \frac{9}{2}$ , and triangle ABC is the same shape scaled by $2$ (O is the midpoint of AB), so its area is $4 \times \frac{9}{2} = 18$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying the signed coordinates of P to get the constant $(-3)\times(-3)=9$.
6.NS.C.8 Solve problems by graphing points in four quadrants — Reflecting through the origin and finding lengths from coordinates.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Area of the right triangle.

💡Takeaway. When a point on

y=\frac{k}{x}

is not pinned down, the answer usually only needs

xy = k

.

Variant 7 hard

The graph of an inverse proportion, shown in the figure, passes through the point P $(-4, -3)$ . Point A is on the branch in the first quadrant and point B is on the branch in the third quadrant, and segment AB passes through the origin O. Point C is chosen so that AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis. Find the area of triangle ABC.

Your answer

Show solution

1 · Understandwhat's really being asked

Find the inverse proportion from P, then let A $(p, q)$ be any point on the first-quadrant branch. Because AB passes through O, B is A reflected through the origin, and the right triangle ABC has legs we can write with $p$ and $q$ . Its area depends only on $pq$ .

Givens

The inverse proportion passes through P $(-4, -3)$ .
A is on the first-quadrant branch, B on the third-quadrant branch, and AB passes through O.
AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis.

Unknowns

The area of triangle ABC.

Constraints

The position of A is not given, so the area must be the same for every A.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#17 Visualize Spatial Relationships

Instead of hunting for A, change focus to the product $xy$ , which is the same for every point on the graph. Point symmetry about O turns the legs of the triangle into $2p$ and $2q$ .

3 · Execute4 carry out the plan

1Find the constant

#13 Convert to Algebra 7.NS.A.2

For

y=\frac{k}{x}

, the product

xy

is the constant

k

. Use P.

k = (-4) \times (-3) = 12 \Rightarrow y = \frac{12}{x}

Every point on this graph has

xy = 12

.

2Name the points

#17 Visualize Spatial Relationships 6.NS.C.8

Let A

=(p, q)

with

pq=12

. The line through O and A is

y=\frac{q}{p}x

, and it meets the curve again at the reflection of A through O.

B = (-p, -q),\quad C = (p, -q)

(-p)(-q) = pq = 12

, so

(-p, -q)

is on the graph and on the line through O and A.

3Leg lengths

#17 Visualize Spatial Relationships 6.NS.C.8

AC is vertical from

q

down to

-q

; BC is horizontal from

-p

to

p

.

AC = q - (-q) = 2q,\quad BC = p - (-p) = 2p

Distance along a grid line is the difference of coordinates.

4Area

#16 Count the Complement 7.G.B.6

Triangle ABC is right-angled at C.

\frac{1}{2} \times 2p \times 2q = 2pq = 2 \times 12 = 24

The unknown

p

and

q

appear only as the product

pq

, which we know.

Answer:

24

4 · Reviewdoes it hold up?

Try a concrete A: A $(1, 12)$ gives B $(-1, -12)$ , C $(1, -12)$ , legs $2$ and $24$ , area $\frac{1}{2} \times 2 \times 24 = 24$ . A $(12, 1)$ gives legs $24$ and $2$ , again $24$ .

Another way: Let D be where AC crosses the $x$ -axis. Triangle AOD has area $\frac{1}{2}pq = 6$ , and triangle ABC is the same shape scaled by $2$ (O is the midpoint of AB), so its area is $4 \times 6 = 24$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying the signed coordinates of P to get the constant $(-4)\times(-3)=12$.
6.NS.C.8 Solve problems by graphing points in four quadrants — Reflecting through the origin and finding lengths from coordinates.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Area of the right triangle.

💡Takeaway. When a point on

y=\frac{k}{x}

is not pinned down, the answer usually only needs

xy = k

.

Variant 8 medium

The graph of an inverse proportion, shown in the figure, passes through the point P $(-4, -2)$ . Point A is on the branch in the first quadrant and point B is on the branch in the third quadrant, and segment AB passes through the origin O. Point C is chosen so that AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis. Find the area of triangle ABC.

Your answer

Show solution

1 · Understandwhat's really being asked

Find the inverse proportion from P, then let A $(p, q)$ be any point on the first-quadrant branch. Because AB passes through O, B is A reflected through the origin, and the right triangle ABC has legs we can write with $p$ and $q$ . Its area depends only on $pq$ .

Givens

The inverse proportion passes through P $(-4, -2)$ .
A is on the first-quadrant branch, B on the third-quadrant branch, and AB passes through O.
AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis.

Unknowns

The area of triangle ABC.

Constraints

The position of A is not given, so the area must be the same for every A.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#17 Visualize Spatial Relationships

Instead of hunting for A, change focus to the product $xy$ , which is the same for every point on the graph. Point symmetry about O turns the legs of the triangle into $2p$ and $2q$ .

3 · Execute4 carry out the plan

1Find the constant

#13 Convert to Algebra 7.NS.A.2

For

y=\frac{k}{x}

, the product

xy

is the constant

k

. Use P.

k = (-4) \times (-2) = 8 \Rightarrow y = \frac{8}{x}

Every point on this graph has

xy = 8

.

2Name the points

#17 Visualize Spatial Relationships 6.NS.C.8

Let A

=(p, q)

with

pq=8

. The line through O and A is

y=\frac{q}{p}x

, and it meets the curve again at the reflection of A through O.

B = (-p, -q),\quad C = (p, -q)

(-p)(-q) = pq = 8

, so

(-p, -q)

is on the graph and on the line through O and A.

3Leg lengths

#17 Visualize Spatial Relationships 6.NS.C.8

AC is vertical from

q

down to

-q

; BC is horizontal from

-p

to

p

.

AC = q - (-q) = 2q,\quad BC = p - (-p) = 2p

Distance along a grid line is the difference of coordinates.

4Area

#16 Count the Complement 7.G.B.6

Triangle ABC is right-angled at C.

\frac{1}{2} \times 2p \times 2q = 2pq = 2 \times 8 = 16

The unknown

p

and

q

appear only as the product

pq

, which we know.

Answer:

16

4 · Reviewdoes it hold up?

Try a concrete A: A $(1, 8)$ gives B $(-1, -8)$ , C $(1, -8)$ , legs $2$ and $16$ , area $\frac{1}{2} \times 2 \times 16 = 16$ . A $(8, 1)$ gives legs $16$ and $2$ , again $16$ .

Another way: Let D be where AC crosses the $x$ -axis. Triangle AOD has area $\frac{1}{2}pq = 4$ , and triangle ABC is the same shape scaled by $2$ (O is the midpoint of AB), so its area is $4 \times 4 = 16$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying the signed coordinates of P to get the constant $(-4)\times(-2)=8$.
6.NS.C.8 Solve problems by graphing points in four quadrants — Reflecting through the origin and finding lengths from coordinates.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Area of the right triangle.

💡Takeaway. When a point on

y=\frac{k}{x}

is not pinned down, the answer usually only needs

xy = k

.

Variant 9 medium

The graph of an inverse proportion, shown in the figure, passes through the point P $(-4, -1)$ . Point A is on the branch in the first quadrant and point B is on the branch in the third quadrant, and segment AB passes through the origin O. Point C is chosen so that AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis. Find the area of triangle ABC.

Your answer

Show solution

1 · Understandwhat's really being asked

Find the inverse proportion from P, then let A $(p, q)$ be any point on the first-quadrant branch. Because AB passes through O, B is A reflected through the origin, and the right triangle ABC has legs we can write with $p$ and $q$ . Its area depends only on $pq$ .

Givens

The inverse proportion passes through P $(-4, -1)$ .
A is on the first-quadrant branch, B on the third-quadrant branch, and AB passes through O.
AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis.

Unknowns

The area of triangle ABC.

Constraints

The position of A is not given, so the area must be the same for every A.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#17 Visualize Spatial Relationships

Instead of hunting for A, change focus to the product $xy$ , which is the same for every point on the graph. Point symmetry about O turns the legs of the triangle into $2p$ and $2q$ .

3 · Execute4 carry out the plan

1Find the constant

#13 Convert to Algebra 7.NS.A.2

For

y=\frac{k}{x}

, the product

xy

is the constant

k

. Use P.

k = (-4) \times (-1) = 4 \Rightarrow y = \frac{4}{x}

Every point on this graph has

xy = 4

.

2Name the points

#17 Visualize Spatial Relationships 6.NS.C.8

Let A

=(p, q)

with

pq=4

. The line through O and A is

y=\frac{q}{p}x

, and it meets the curve again at the reflection of A through O.

B = (-p, -q),\quad C = (p, -q)

(-p)(-q) = pq = 4

, so

(-p, -q)

is on the graph and on the line through O and A.

3Leg lengths

#17 Visualize Spatial Relationships 6.NS.C.8

AC is vertical from

q

down to

-q

; BC is horizontal from

-p

to

p

.

AC = q - (-q) = 2q,\quad BC = p - (-p) = 2p

Distance along a grid line is the difference of coordinates.

4Area

#16 Count the Complement 7.G.B.6

Triangle ABC is right-angled at C.

\frac{1}{2} \times 2p \times 2q = 2pq = 2 \times 4 = 8

The unknown

p

and

q

appear only as the product

pq

, which we know.

Answer:

8

4 · Reviewdoes it hold up?

Try a concrete A: A $(1, 4)$ gives B $(-1, -4)$ , C $(1, -4)$ , legs $2$ and $8$ , area $\frac{1}{2} \times 2 \times 8 = 8$ . A $(4, 1)$ gives legs $8$ and $2$ , again $8$ .

Another way: Let D be where AC crosses the $x$ -axis. Triangle AOD has area $\frac{1}{2}pq = 2$ , and triangle ABC is the same shape scaled by $2$ (O is the midpoint of AB), so its area is $4 \times 2 = 8$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying the signed coordinates of P to get the constant $(-4)\times(-1)=4$.
6.NS.C.8 Solve problems by graphing points in four quadrants — Reflecting through the origin and finding lengths from coordinates.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Area of the right triangle.

💡Takeaway. When a point on

y=\frac{k}{x}

is not pinned down, the answer usually only needs

xy = k

.

Variant 10 hard

The graph of an inverse proportion, shown in the figure, passes through the point P $(-5, -1)$ . Point A is on the branch in the first quadrant and point B is on the branch in the third quadrant, and segment AB passes through the origin O. Point C is chosen so that AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis. Find the area of triangle ABC.

Your answer

Show solution

1 · Understandwhat's really being asked

Find the inverse proportion from P, then let A $(p, q)$ be any point on the first-quadrant branch. Because AB passes through O, B is A reflected through the origin, and the right triangle ABC has legs we can write with $p$ and $q$ . Its area depends only on $pq$ .

Givens

The inverse proportion passes through P $(-5, -1)$ .
A is on the first-quadrant branch, B on the third-quadrant branch, and AB passes through O.
AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis.

Unknowns

The area of triangle ABC.

Constraints

The position of A is not given, so the area must be the same for every A.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#17 Visualize Spatial Relationships

Instead of hunting for A, change focus to the product $xy$ , which is the same for every point on the graph. Point symmetry about O turns the legs of the triangle into $2p$ and $2q$ .

3 · Execute4 carry out the plan

1Find the constant

#13 Convert to Algebra 7.NS.A.2

For

y=\frac{k}{x}

, the product

xy

is the constant

k

. Use P.

k = (-5) \times (-1) = 5 \Rightarrow y = \frac{5}{x}

Every point on this graph has

xy = 5

.

2Name the points

#17 Visualize Spatial Relationships 6.NS.C.8

Let A

=(p, q)

with

pq=5

. The line through O and A is

y=\frac{q}{p}x

, and it meets the curve again at the reflection of A through O.

B = (-p, -q),\quad C = (p, -q)

(-p)(-q) = pq = 5

, so

(-p, -q)

is on the graph and on the line through O and A.

3Leg lengths

#17 Visualize Spatial Relationships 6.NS.C.8

AC is vertical from

q

down to

-q

; BC is horizontal from

-p

to

p

.

AC = q - (-q) = 2q,\quad BC = p - (-p) = 2p

Distance along a grid line is the difference of coordinates.

4Area

#16 Count the Complement 7.G.B.6

Triangle ABC is right-angled at C.

\frac{1}{2} \times 2p \times 2q = 2pq = 2 \times 5 = 10

The unknown

p

and

q

appear only as the product

pq

, which we know.

Answer:

10

4 · Reviewdoes it hold up?

Try a concrete A: A $(1, 5)$ gives B $(-1, -5)$ , C $(1, -5)$ , legs $2$ and $10$ , area $\frac{1}{2} \times 2 \times 10 = 10$ . A $(5, 1)$ gives legs $10$ and $2$ , again $10$ .

Another way: Let D be where AC crosses the $x$ -axis. Triangle AOD has area $\frac{1}{2}pq = \frac{5}{2}$ , and triangle ABC is the same shape scaled by $2$ (O is the midpoint of AB), so its area is $4 \times \frac{5}{2} = 10$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying the signed coordinates of P to get the constant $(-5)\times(-1)=5$.
6.NS.C.8 Solve problems by graphing points in four quadrants — Reflecting through the origin and finding lengths from coordinates.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Area of the right triangle.

💡Takeaway. When a point on

y=\frac{k}{x}

is not pinned down, the answer usually only needs

xy = k

.

Variant 11 hard

The graph of an inverse proportion, shown in the figure, passes through the point P $(-2, -5)$ . Point A is on the branch in the first quadrant and point B is on the branch in the third quadrant, and segment AB passes through the origin O. Point C is chosen so that AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis. Find the area of triangle ABC.

Your answer

Show solution

1 · Understandwhat's really being asked

Find the inverse proportion from P, then let A $(p, q)$ be any point on the first-quadrant branch. Because AB passes through O, B is A reflected through the origin, and the right triangle ABC has legs we can write with $p$ and $q$ . Its area depends only on $pq$ .

Givens

The inverse proportion passes through P $(-2, -5)$ .
A is on the first-quadrant branch, B on the third-quadrant branch, and AB passes through O.
AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis.

Unknowns

The area of triangle ABC.

Constraints

The position of A is not given, so the area must be the same for every A.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#17 Visualize Spatial Relationships

Instead of hunting for A, change focus to the product $xy$ , which is the same for every point on the graph. Point symmetry about O turns the legs of the triangle into $2p$ and $2q$ .

3 · Execute4 carry out the plan

1Find the constant

#13 Convert to Algebra 7.NS.A.2

For

y=\frac{k}{x}

, the product

xy

is the constant

k

. Use P.

k = (-2) \times (-5) = 10 \Rightarrow y = \frac{10}{x}

Every point on this graph has

xy = 10

.

2Name the points

#17 Visualize Spatial Relationships 6.NS.C.8

Let A

=(p, q)

with

pq=10

. The line through O and A is

y=\frac{q}{p}x

, and it meets the curve again at the reflection of A through O.

B = (-p, -q),\quad C = (p, -q)

(-p)(-q) = pq = 10

, so

(-p, -q)

is on the graph and on the line through O and A.

3Leg lengths

#17 Visualize Spatial Relationships 6.NS.C.8

AC is vertical from

q

down to

-q

; BC is horizontal from

-p

to

p

.

AC = q - (-q) = 2q,\quad BC = p - (-p) = 2p

Distance along a grid line is the difference of coordinates.

4Area

#16 Count the Complement 7.G.B.6

Triangle ABC is right-angled at C.

\frac{1}{2} \times 2p \times 2q = 2pq = 2 \times 10 = 20

The unknown

p

and

q

appear only as the product

pq

, which we know.

Answer:

20

4 · Reviewdoes it hold up?

Try a concrete A: A $(1, 10)$ gives B $(-1, -10)$ , C $(1, -10)$ , legs $2$ and $20$ , area $\frac{1}{2} \times 2 \times 20 = 20$ . A $(10, 1)$ gives legs $20$ and $2$ , again $20$ .

Another way: Let D be where AC crosses the $x$ -axis. Triangle AOD has area $\frac{1}{2}pq = 5$ , and triangle ABC is the same shape scaled by $2$ (O is the midpoint of AB), so its area is $4 \times 5 = 20$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying the signed coordinates of P to get the constant $(-2)\times(-5)=10$.
6.NS.C.8 Solve problems by graphing points in four quadrants — Reflecting through the origin and finding lengths from coordinates.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Area of the right triangle.

💡Takeaway. When a point on

y=\frac{k}{x}

is not pinned down, the answer usually only needs

xy = k

.

Variant 12 hard

The graph of an inverse proportion, shown in the figure, passes through the point P $(-1, -6)$ . Point A is on the branch in the first quadrant and point B is on the branch in the third quadrant, and segment AB passes through the origin O. Point C is chosen so that AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis. Find the area of triangle ABC.

Your answer

Show solution

1 · Understandwhat's really being asked

Find the inverse proportion from P, then let A $(p, q)$ be any point on the first-quadrant branch. Because AB passes through O, B is A reflected through the origin, and the right triangle ABC has legs we can write with $p$ and $q$ . Its area depends only on $pq$ .

Givens

The inverse proportion passes through P $(-1, -6)$ .
A is on the first-quadrant branch, B on the third-quadrant branch, and AB passes through O.
AC is parallel to the $y$ -axis and BC is parallel to the $x$ -axis.

Unknowns

The area of triangle ABC.

Constraints

The position of A is not given, so the area must be the same for every A.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#17 Visualize Spatial Relationships

Instead of hunting for A, change focus to the product $xy$ , which is the same for every point on the graph. Point symmetry about O turns the legs of the triangle into $2p$ and $2q$ .

3 · Execute4 carry out the plan

1Find the constant

#13 Convert to Algebra 7.NS.A.2

For

y=\frac{k}{x}

, the product

xy

is the constant

k

. Use P.

k = (-1) \times (-6) = 6 \Rightarrow y = \frac{6}{x}

Every point on this graph has

xy = 6

.

2Name the points

#17 Visualize Spatial Relationships 6.NS.C.8

Let A

=(p, q)

with

pq=6

. The line through O and A is

y=\frac{q}{p}x

, and it meets the curve again at the reflection of A through O.

B = (-p, -q),\quad C = (p, -q)

(-p)(-q) = pq = 6

, so

(-p, -q)

is on the graph and on the line through O and A.

3Leg lengths

#17 Visualize Spatial Relationships 6.NS.C.8

AC is vertical from

q

down to

-q

; BC is horizontal from

-p

to

p

.

AC = q - (-q) = 2q,\quad BC = p - (-p) = 2p

Distance along a grid line is the difference of coordinates.

4Area

#16 Count the Complement 7.G.B.6

Triangle ABC is right-angled at C.

\frac{1}{2} \times 2p \times 2q = 2pq = 2 \times 6 = 12

The unknown

p

and

q

appear only as the product

pq

, which we know.

Answer:

12

4 · Reviewdoes it hold up?

Try a concrete A: A $(1, 6)$ gives B $(-1, -6)$ , C $(1, -6)$ , legs $2$ and $12$ , area $\frac{1}{2} \times 2 \times 12 = 12$ . A $(6, 1)$ gives legs $12$ and $2$ , again $12$ .

Another way: Let D be where AC crosses the $x$ -axis. Triangle AOD has area $\frac{1}{2}pq = 3$ , and triangle ABC is the same shape scaled by $2$ (O is the midpoint of AB), so its area is $4 \times 3 = 12$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Multiplying the signed coordinates of P to get the constant $(-1)\times(-6)=6$.
6.NS.C.8 Solve problems by graphing points in four quadrants — Reflecting through the origin and finding lengths from coordinates.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Area of the right triangle.

💡Takeaway. When a point on

y=\frac{k}{x}

is not pinned down, the answer usually only needs

xy = k

.

Constant xy fixes the triangle's area · 12 practice problems

Generated variants — 12

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Where next?