- #1

singleton

- 121

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Hey,

Usually I have a general idea how to tackle a problem when posting on here.

This time I am pretty darn lost :(

When asked to find the equation of the line that passes through a point (1,1) and maps onto itself under the translation vector a = (3,2).

I start with Ax + By + C = 0

The equation of the image is

A(x - 3) + B(y - 2) + C = 0

Ax - 3A + By - 2B + C = 0

The point (1, 1) satisfies the equation since the translation "maps onto itself" (I'm taking this as it means a parallel and coincident line? not heard that terminology before..."maps onto itself") So the point is on both lines...

Basically I know that (1, 1) satisfies the equation of the image, and that when I substitute I end up with

Ax - 3A + By - 2B + C = 0

A(1) - 3A + B(1) - 2B + C = 0

-2A -B + C = 0

I almost wanted to say the normal vector for the translation was (-2, -1) but that is incorrect, as those numbers (-2 and -1) are the coefficients I have for A and B AFTER I substitute not BEFORE...

Just a small hint or suggestion how to tackle this Q would be great :)

I know (rather, I think) that I should be exploring the equation of the image (the translation) but I do not know what more I can do to find the normal vector (so I can write the Cartesian equation of the unknown line)

Usually I have a general idea how to tackle a problem when posting on here.

This time I am pretty darn lost :(

When asked to find the equation of the line that passes through a point (1,1) and maps onto itself under the translation vector a = (3,2).

I start with Ax + By + C = 0

The equation of the image is

A(x - 3) + B(y - 2) + C = 0

Ax - 3A + By - 2B + C = 0

The point (1, 1) satisfies the equation since the translation "maps onto itself" (I'm taking this as it means a parallel and coincident line? not heard that terminology before..."maps onto itself") So the point is on both lines...

Basically I know that (1, 1) satisfies the equation of the image, and that when I substitute I end up with

Ax - 3A + By - 2B + C = 0

A(1) - 3A + B(1) - 2B + C = 0

-2A -B + C = 0

I almost wanted to say the normal vector for the translation was (-2, -1) but that is incorrect, as those numbers (-2 and -1) are the coefficients I have for A and B AFTER I substitute not BEFORE...

Just a small hint or suggestion how to tackle this Q would be great :)

I know (rather, I think) that I should be exploring the equation of the image (the translation) but I do not know what more I can do to find the normal vector (so I can write the Cartesian equation of the unknown line)

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