Solved on Jan 18, 2024

Translate a right triangle 4 left and 6 up. Show $\triangle$ transformation.

STEP 1

Assumptions
1. We have a right triangle in a coordinate plane.
2. We are translating the triangle 4 units to the left, which means in the negative x-direction.
3. We are translating the triangle 6 units up, which means in the positive y-direction.

STEP 2

Identify the coordinates of the vertices of the original right triangle. Let's assume the vertices are $A(x_1, y_1)$ , $B(x_2, y_2)$ , and $C(x_3, y_3)$ .

STEP 3

To translate a point to the left by 4 units, we subtract 4 from the x-coordinate of each vertex of the triangle.

STEP 4

To translate a point up by 6 units, we add 6 to the y-coordinate of each vertex of the triangle.

STEP 5

Apply the translation to vertex $A(x_1, y_1)$ .
$A'(x_1', y_1') = A(x_1 - 4, y_1 + 6)$

STEP 6

Apply the translation to vertex $B(x_2, y_2)$ .
$B'(x_2', y_2') = B(x_2 - 4, y_2 + 6)$

STEP 7

Apply the translation to vertex $C(x_3, y_3)$ .
$C'(x_3', y_3') = C(x_3 - 4, y_3 + 6)$

STEP 8

Now we have the new coordinates of the translated triangle's vertices $A'(x_1', y_1')$ , $B'(x_2', y_2')$ , and $C'(x_3', y_3')$ .

STEP 9

If we want to represent the translation algebraically, we can write the transformation rule as:
$(x, y) \rightarrow (x - 4, y + 6)$
This rule can be applied to any point on the original triangle to find its corresponding point on the translated triangle.

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Translate a right triangle 4 left and 6 up. Show $\triangle$ transformation.

STEP 1

Assumptions
1. We have a right triangle in a coordinate plane.
2. We are translating the triangle 4 units to the left, which means in the negative x-direction.
3. We are translating the triangle 6 units up, which means in the positive y-direction.

STEP 2

Identify the coordinates of the vertices of the original right triangle. Let's assume the vertices are $A(x_1, y_1)$ , $B(x_2, y_2)$ , and $C(x_3, y_3)$ .

STEP 3

To translate a point to the left by 4 units, we subtract 4 from the x-coordinate of each vertex of the triangle.

STEP 4

To translate a point up by 6 units, we add 6 to the y-coordinate of each vertex of the triangle.

STEP 5

Apply the translation to vertex $A(x_1, y_1)$ .
$A'(x_1', y_1') = A(x_1 - 4, y_1 + 6)$

STEP 6

Apply the translation to vertex $B(x_2, y_2)$ .
$B'(x_2', y_2') = B(x_2 - 4, y_2 + 6)$

STEP 7

Apply the translation to vertex $C(x_3, y_3)$ .
$C'(x_3', y_3') = C(x_3 - 4, y_3 + 6)$

STEP 8

Now we have the new coordinates of the translated triangle's vertices $A'(x_1', y_1')$ , $B'(x_2', y_2')$ , and $C'(x_3', y_3')$ .

STEP 9

If we want to represent the translation algebraically, we can write the transformation rule as:
$(x, y) \rightarrow (x - 4, y + 6)$
This rule can be applied to any point on the original triangle to find its corresponding point on the translated triangle.

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Translate a right triangle 4 left and 6 up. Show △\triangle△ transformation.

STEP 1

Assumptions1. We have a right triangle in a coordinate plane.2. We are translating the triangle 4 units to the left, which means in the negative x-direction.3. We are translating the triangle 6 units up, which means in the positive y-direction.

STEP 2

Identify the coordinates of the vertices of the original right triangle. Let's assume the vertices are A(x1,y1) A(x_1, y_1) A(x1​,y1​), B(x2,y2) B(x_2, y_2) B(x2​,y2​), and C(x3,y3) C(x_3, y_3) C(x3​,y3​).

STEP 3

To translate a point to the left by 4 units, we subtract 4 from the x-coordinate of each vertex of the triangle.

STEP 4

To translate a point up by 6 units, we add 6 to the y-coordinate of each vertex of the triangle.

STEP 5

Apply the translation to vertex A(x1,y1) A(x_1, y_1) A(x1​,y1​).A′(x1′,y1′)=A(x1−4,y1+6) A'(x_1', y_1') = A(x_1 - 4, y_1 + 6) A′(x1′​,y1′​)=A(x1​−4,y1​+6)

STEP 6

Apply the translation to vertex B(x2,y2) B(x_2, y_2) B(x2​,y2​).B′(x2′,y2′)=B(x2−4,y2+6) B'(x_2', y_2') = B(x_2 - 4, y_2 + 6) B′(x2′​,y2′​)=B(x2​−4,y2​+6)

STEP 7

Apply the translation to vertex C(x3,y3) C(x_3, y_3) C(x3​,y3​).C′(x3′,y3′)=C(x3−4,y3+6) C'(x_3', y_3') = C(x_3 - 4, y_3 + 6) C′(x3′​,y3′​)=C(x3​−4,y3​+6)

STEP 8

Now we have the new coordinates of the translated triangle's vertices A′(x1′,y1′) A'(x_1', y_1') A′(x1′​,y1′​), B′(x2′,y2′) B'(x_2', y_2') B′(x2′​,y2′​), and C′(x3′,y3′) C'(x_3', y_3') C′(x3′​,y3′​).

STEP 9

If we want to represent the translation algebraically, we can write the transformation rule as:(x,y)→(x−4,y+6) (x, y) \rightarrow (x - 4, y + 6) (x,y)→(x−4,y+6)This rule can be applied to any point on the original triangle to find its corresponding point on the translated triangle.

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Translate a right triangle 4 left and 6 up. Show $\triangle$ transformation.

Assumptions
1. We have a right triangle in a coordinate plane.
2. We are translating the triangle 4 units to the left, which means in the negative x-direction.
3. We are translating the triangle 6 units up, which means in the positive y-direction.

Identify the coordinates of the vertices of the original right triangle. Let's assume the vertices are $A(x_1, y_1)$ , $B(x_2, y_2)$ , and $C(x_3, y_3)$ .

Apply the translation to vertex $A(x_1, y_1)$ .
$A'(x_1', y_1') = A(x_1 - 4, y_1 + 6)$

Apply the translation to vertex $B(x_2, y_2)$ .
$B'(x_2', y_2') = B(x_2 - 4, y_2 + 6)$

Apply the translation to vertex $C(x_3, y_3)$ .
$C'(x_3', y_3') = C(x_3 - 4, y_3 + 6)$

Now we have the new coordinates of the translated triangle's vertices $A'(x_1', y_1')$ , $B'(x_2', y_2')$ , and $C'(x_3', y_3')$ .

If we want to represent the translation algebraically, we can write the transformation rule as:
$(x, y) \rightarrow (x - 4, y + 6)$
This rule can be applied to any point on the original triangle to find its corresponding point on the translated triangle.