What is the linear equation that passes through point a (1,2) and has a slope of half?

What is the linear equation that passes through point a (1,2) and has a slope of half?

Y-2 = 1 / 2 (x-1), that is: x-2y + 3 = 0

The equation of a line passing through the origin and perpendicular to a line with a slope of minus half is

Let the linear equation be y = KX + B
Because the line goes through the origin, B = 0
Because the straight line is perpendicular to the straight line with the slope of negative half, the formula K1 * K2 = - 1 (the slope of two straight lines multiplied by = - 1) deduces - 1 / 2 * K2 = - 1, and K2 = 2
So the linear equation is y = 2x

Find the relationship between the line L1 passing through two points a (2,3), B (- 1,0) and the line L2 passing through point P (1,0) with slope 1

1、 Formula method
1. From the two-point equation of the straight line: (Y-Y1) / (x-x1) = (y2-y1) / (x2-x1), it is obtained that:
L1：(y-3)/(x-2)=(0-3)/(-1-2)
Sorting: X-Y + 1 = 0
2. From the point oblique equation of the straight line: Y-Y1 = K (x-x1), it is obtained that:
L2：y－0=(x－1)
Sorting: x-y-1 = 0
The slope of L1 and L2 is k = 1
Conclusion L1, L1 parallel
2、 Analytical method
L1 slope K1 = (y2-y1) / (x2-x1) = (0-3) / (- 1-2) = 1
L2 slope K2 = 1
k1=k2
Conclusion L1 is parallel to L2

Given that the line L1: y = 2x + 3, L2 and L1 are symmetric with respect to the line y = - x, and the line L3 ⊥ L2, then the slope of L3 is______ ．

∵ line L1: y = 2x + 3, L2 and L1 are symmetric with respect to the line y = - x, the equation of L2 is - x = 2 (- y) + 3, that is, x-2y + 3 = 0, the slope of L2 is 12, from the line L3 ⊥ L2: the slope of L3 is - 2, so the answer is - 2