Solve the linear equation of point a (- 3,2) and the distance from the origin is 3

Solve the linear equation of point a (- 3,2) and the distance from the origin is 3

① The slope of the line does not exist
Then x = - 3
The distance from the origin to the straight line is d = 3, which is consistent with
So the line is x = - 3
② If the slope of the line exists, set it to K
Then Y-2 = K (x + 3)
That is kx-y + 3K + 2 = 0
So | 3K + 2 | / √ (K & # 178; + 1) = 3
So (3K + 2) ² = 9 (K & #178; + 1)
That is, k = 5 / 12
So the straight line is (5 / 12) X-Y + 3 * 5 / 12 + 2 = 0
That is, 5x-12y + 39 = 0
To sum up, the straight line is x = - 3 or 5x-12y + 39 = 0
If you don't understand, I wish you a happy study!

Find the trajectory equation of a (1,1) and the point with the same distance to the straight line x + 2Y = 3

Set to (x, y)
Then √ [(x-1) & sup2; + (Y-1) & sup2;] = | x + 2y-3 | / √ (1 & sup2; + 2 & sup2;)
square
(x²+y²-2x-2y+2)=(x²+4y²+9+4xy-6x-12y)/5
5x²+5y²-10x-10y+10=x²+4y²+9+4xy-6x-12y
4x²-4xy+y²-4x+2y+1=0

How to find the trajectory equation of the point with equal distance to two straight lines X, y = 0 and X-Y = 0 by distance formula
It's x + y = 0 and X-Y = 0

Let the distance between the point (x, y) and the two straight lines be equal,
The distance formula is
|x+y|/√2 =|x-y|/√2
The trajectory equations of the simplified points are x = 0 and y = 0, that is, X axis and Y axis