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Nikolay [14]
3 years ago
15

Reflect line AB over the x- axis.

Mathematics
2 answers:
LenKa [72]3 years ago
8 0

a would be at (1,-2) and b would be at (5,-4)

balu736 [363]3 years ago
4 0

1. Think of a x axis as a mirror line

2. So point A is currently two squares away from the mirror line so if your reflecting line AB over the x axis you would have to count two squares downwards to reflect point A over the line so by looking at the graph the precise coordinate of A(reflected) would be (1,-2)

3. Now do the exact same with point B and finally join the points together.

4. Furthermore if this question asked to reflect AB over the y axis all thats different is that you are treating the y axis as the mirror line.

I hope this made sense sorry if I failed to explain it well .

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Find the distance between the lines<br> Y=3x+10<br> Y=3x-20<br> AND <br> Y=3/2x+3/2<br> Y=3/2x-5
rosijanka [135]
At first we need to know how to find the distance between two parallel lines
The parallel lines which have the same slope 
If we have two parallel lines ⇒ y₁ = mx + c₁   and    y₂ = mx + c₂
So, the distance between y₁ and y₂ = d = \frac{|c_{1} - c_{2}|}{\sqrt{1 + m^{2}}}
========================================================

Part (1):
=======

<span>The distance between the lines
</span>
<span> Y₁ = 3x + 10    and       Y₂ = 3x - 20</span>
∴ m = 3  , c₁ = 10  and  c₂ = -20
∴ d = \frac{| 10 - (-20)|}{\sqrt{1 + 3^{2}}}= \framebox{9.5}

========================================================

Part (2):
=======

<span>The distance between the lines
</span>
Y₁ = <span>3/2 x + 3/2</span>    and       Y₂ = <span>3/2 x - 5</span>
∴ m = 3/2 = 1.5  , c₁ = 3/2 = 1.5   and  c₂ = -5
∴ d = \frac{| 1.5- (-5)|}{\sqrt{1 + 1.5^{2}}}= \framebox{3.6}

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Step-by-step explanation:

5 0
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Read 2 more answers
what is the point-slope form of the equation of a line that passes through the point (3,5) and has the slope of 2? A)y+3=2(x+5)
lara31 [8.8K]
Point-slope form
y-y₀=m.(x-x₀)
P(3,5)
m=2

y-5=2(x-3)


3 0
3 years ago
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