Find the reflection of the point (x, y) in the line y=mx+c​

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Step-by-step explanation:

Consider the line that is perpendicular to and goes through .

Both and the reflection would be on this new line. Besides, the two points would be equidistant from the intersection of this new line and line .

Hence, if the vector between and that intersection could be found, adding twice that vector to would yield the coordinates of the reflection.

Since this new line is perpendicular to line , the slope of this new line would be .

Hence, would be a direction vector of this new line.

(a constant multiple of would also be a direction vector of this new line.)

Both and the aforementioned intersection are on this new line. Hence, their position vectors would differ only by a constant multiple of a direction vector of this new line.

In other words, for some constant , would be the position vector of the reflection of (the position vector of is .)

would be the coordinates of the intersection between the new line and . would be the vector between and that intersection.

Since that intersection is on the line , its coordinates should satisfy:

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Solve for :

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Hence, the vector between the position of and that of the intersection would be:

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Add twice the amount of this vector to position of to find the position of the reflection, .

-coordinate of the reflection:

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-coordinate of the reflection:

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y < -3

step-by-step explanation:

-3y - 2 > 7

-3y -2 + 2 > 7+2

-3y > 9

-3y/-3 < 9/-3   (change the inequality sign when multiplying or dividing by a negative number)

y < -3

d

step-by-step explanation:

i used the pitagoras theoreme