# Finding Intersections 9897 . 136

Finding Intersections

• Equation of arc 1:

y

 

.

0070

x

2

1 .

2349

x

• Equation of arc 2:

y

 

.

0254

x

2

4 .

4856

x

136 .

9897

• To find intersections, use substitution:

» EQ ARC 1 = EQ ARC 2

.

0070

x

2

Finding Intersections

1 .

2349

x

 

.

0254

x

2

4 .

4856

x

136 .

9897

.

0070

x

2

1 .

2349

x

.

0070

x

2

1 .

2349

x

0

 

.

0184

x

2

3 .

2507

x

136 .

9897

• I solved the system of equations so that one side equaled 0. Now, I will use the quadratic formula to find my x-values.

0

Finding Intersections

.

0184

x

2

3 .

2507

x

136 .

9897

x x

x

b

2

a

3 .

2507

2 (

.

0184 )

3 .

2507

.

0368

b

2

4

ac

2

a

3 .

2507

2

10 .

4 (

.

0184 )(

136 .

9897 )

5671

.

2 (

.

0184 )

10 .

0824

0368

Finding Intersections

x x

.

0368

3 .

2507

3

.

.

2507

0368

.

.

.

4846

0368

6961

.

0368

x

88 .

3342

18 .

9168

x

107 .

2510

x

69 .

4174

Finding Intersections

• Now that we have found the x-values of the intersections, substitute into an equation for an arc to find the y-value.

• I find the non-translated one to be just a bit easier.

y

Finding Intersections

.

0070

x

2

1 .

2349

x x x

107 .

2510

69 .

4174

y

 

.

0070 ( 107 .

2510 )

2

1 .

2349 ( 107 .

2510 )

y

 

.

0070 ( 11502 .

7770 )

1 .

2349 ( 107 .

2510 )

y y

80 .

5194

51 .

9249

132 .

4443

Finding Intersections

x

107 .

2510

y

 

.

0070

x

2

1 .

2349

x x

69 .

4174

y y

 

.

0070 ( 69 .

4174 )

2

.

0070 ( 4818 .

7754 )

1 .

2349 ( 69 .

4174 )

1 .

2349 ( 69 .

4174 )

y y

33 .

7314

51 .

9921

85 .

7235

I know that the y-values should be exactly the same since I have the same line of symmetry. These are basically the same, where the rounding I have done caused the difference. I’ll label the y-value as the midpoint between the two solutions I got – 59.9585

Finding Intersections