What is the equation of the line that passes through the point (3,0) and is perpendicular to the line 2x-y=5?

Slope of the line (2x - y = 5)
= -2/(-1)
= 2

Due to perpendicular :
(Slope of the required line) × 2 = -1
Slope of the required line = -1/2

Point-slope form of the required line :
y - 0 = (-1/2)(x - 3)
2y = -(x - 3)
2y = -x + 3
x + 2y = 3

...... The answer is : B. x + 2y = 3

The answer is B since the slope of x+2y=3 is
-1/2 and the slope os 2x-y=6 is 2 and
(-1/2)(2)=-1. Also (3,0) lies on line B.

2x - y = 5
-y = -2x + 5
y = 2x - 5

given slope is 2
perp slope is -1/2

y - 0 = (-1/2)(x - 3)
y = (-x + 3) / 2
2y = -x + 3
x + 2y = 3