Question

(100 points boys) Just this one question for my math.

Answer 1

Answer: Its C
`x^(2) x 6 = 24 x \alpha + thE 9 =23- 518 x_(123)`

Explanation:

Answer 2

`A^(')` = (1,
`(5)/(2)`),
`B^(')` = (
`(3)/(2)`,-1) and
`C^(') = (2,2)`

OA =
`\sqrt[]{29}` units, OB =
`\sqrt[]{13}` units and OC =
`\sqrt[]{32}` units

O
`A^(')` =
`(√(29) )/(2)` units,

O
`B^(')` =
`(√(13) )/(2)` units and

O
`C^(')` =
`(√(32) )/(2)` units

Explanation:

Given vertices are A(2,5), B(3,-2), C(4,4) and the new vertices after dilation are
`A^('), B^('), C^(')`

a.

Triangle needs to be dilated about the origin.

⇒The centre of dilation is the origin O(0,0).

scaling factor is
`(1)/(2)`

`(OA^('))/(OA)` =
`(OB^('))/(OB)` =
`(OC^('))/(OC)` =
`(1)/(2)`

⇒
`A^('), B^(') and C^(')` divide OA, OB and OC in the ratio 1:1 respectively.

Using the formula for the coordinates of a point dividing two points (0,0), (x,y) in the ratio 1:1 is (
`(x)/(2)`,
`(y)/(2)`) :

`A^(')` = (1,
`(5)/(2)`),
`B^(')` = (
`(3)/(2)`,-1) and
`C^(') = (2,2)`

b.

Using the formula for the distance between the points (0,0) and (x,y) is
`\sqrt[]{x^(2 )+ y^(2) }` :

OA =
`\sqrt[]{2^(2) +5^(2)}` =
`\sqrt[]{29}` units

OB =
`\sqrt[]{3^(2) +-2^(2)}` =
`\sqrt[]{13}` units

OC =
`\sqrt[]{4^(2) +4^(2)}` =
`\sqrt[]{32}` units

c.

From the property of dilation :

`(OA^('))/(OA)` =
`(OB^('))/(OB)` =
`(OC^('))/(OC)` =
`(1)/(2)`

⇒O
`A^(')` =
`(OA)/(2)` =
`(√(29) )/(2)` units

O
`B^(')` =
`(OB)/(2)` =
`(√(13) )/(2)` units

O
`C^(')` =
`(OC)/(2)` =
`(√(32) )/(2)` units

d.

Using the formula for the distance between two points :

O
`A^(')` =
`\sqrt{1^(2) + ((5)/(2))^(2) }` =
`(√(29) )/(2)` units

O
`B^(')` =
`\sqrt{((3)/(2))^(2) + -1^(2)}` =
`(√(13) )/(2)` units

O
`C^(')` =
`\sqrt[]{2^(2) +2^(2)}` =
`(√(32) )/(2)` units

which are the same as we predicted in part c.