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ratelena [41]
2 years ago
14

What is the point-slope form of a line with slope -3 that contains the point (10, -1)?

Mathematics
1 answer:
Advocard [28]2 years ago
8 0

Answer:

y - (-1) = -3(x-10)

y+1= -3x+30

y= -3x+30-1

y= -3x+29

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The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2
maxonik [38]

Given:

The red figure dilated with a scale factor of s=\dfrac{1}{3} and the center of dilation is at the point (4,2) to get the green figure.

To find:

The coordinates of C' and A.

Solution:

If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

In given problem, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

4 0
3 years ago
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