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3 years ago

6

Which is not an equation of the line that passes through the points (1, 1) and (5, 5)? y - x = 0 x - y = 0 -x - y = 0

2 answers:

slamgirl [31]3 years ago

4 0

Answer:

-x-y=0 is not the equation of the line

Step-by-step explanation:

Lets find equation of a line that passes through points (1, 1) and (5, 5)

Slope = $\frac{y2-y1}{x2-x1} =\frac{5-1}{5-1} =1$

Slope = 1 and use point (1,1)

y-y1= m(x-x1)

y-1= 1(x-1)

y-1 = x-1

add 1 on both sides

y=x

when we subtract x on both sides then equation becomes y-x=0

when we subtract y on both sides then equation becomes x-y=0

so -x-y=0 is not the equation of the line

dimulka [17.4K]3 years ago

3 0

The third option. you can always substitute the points into your equation to see if the equation is true or not.

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Let A be a 3×2 matrix, L its left inverse, and R its right inverse. L and R are then matrices such that LA = I₂ (the 2×2 identity matrix) and AR = I₃ (the 3×3 identity matrix). Clearly L must be 2×3 and R must be 3×2 in order for the matrix products to be defined.

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For the left inverse, we have

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$\left(L\left(AA^\top\right)\right)\left(AA^\top\right)^{-1} = A^\top \left(AA^\top\right)^{-1}$

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$LI = A^\top \left(AA^\top\right)^{-1}$

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We do the same thing for the right inverse, but take care with how we multiply both sides of AR = I₃.

$AR=I$

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$\left(A^\top A\right)R = A^\top$

$\left(A^\top A\right)^{-1} \left(\left(A^\top A\right)R\right) = \left(A^\top A\right)^{-1} A^\top$

$\left(\left(A^\top A\right)^{-1} \left(A^\top A\right)\right) R = \left(A^\top A\right)^{-1} A^\top$

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<h2>Answer </h2>

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<h2>Explanation </h2>

First we are going to find the length of JN; then we are subtracting from it the length of JU plus the length of CN.

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