Equation parallel to y = 3x + 2 and passing through (1, 1)
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Answer:
Step-by-step explanation:
We have to see how the canonical vectors are transformed throught T. Lets first define T in any basis.
Since T is a reflection, then any element of the line y = x/2 if fixed by T. Therefore T(2,1) = (2,1).
On the other hand, any vector perpendicular to the line direction should be sent to its opposite value. We can take, for example, (-1,2) (note that the scalar product (2,1) * (-1,2) = -2+2 = 0). As a consecuence T(-1,2) = (1,-2). We have
- T(2,1) = (2,1)
- T(-1,2) = (1,-2)
By summing the first vector with the double of the second one we get, using linearity
T(0,5) = T( (2,1) + 2(-1,2)) = T(2,1) + 2T(-1,2) = (2,1) + 2(1,-2) = (4,-3)
Hence, T(0,1) = (4/5,-3/5)
Now, we take the second vector and substract it the double of the first one (to kill the second variable)
T(-3,0) = T( (-1,2) - 2*(2,1) ) = T(-1,2) -2T(2,1) = (1,-2)-2(2,1) = (-3,-4)
Therefore, T(1,0) = (1,4/3)
The matrix A induced by T has in its first column T(1,0) and in its second column T(0,1). We conclude that
Find the "difference" between the residents but division.
281421906/1000 = 281421.906
You always have to do the same to the top
8.7×281421.906 = 2,448,370.5822
There were 2,448,370.5822 deaths
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