MAth 一問

Let L: y = 3

10a )
Since the straight line perpendicular to the line y = 3.
Slope of L = 0
Slope of the line Slope of L = –1
Slope of the line 0 = –1
Slope of the line = –1/0
Slope of the line = ∞

Equation of the line is:
y – 6 = ∞[x – (–2)]
(y – 6)/ ∞ = x + 2
0 = x + 2
∴ x = –2


10b)
Since the straight line parallel to the line y = 3.
Slope of the line = Slope of L = 0
Slope of the line = 0

Equation of the line is:
y – 6 = 0[x – (–2)]
y – 6 = 0
∴ y = 6



希望幫到您！

(a)y = 3 is a horizontal line, that means any vertical line will be perpendicular to it, that is x = a is perpendicular to y= 3 for any values of a. Now you want the line to pass through (-2,6), that means x = -2 is the required line.
(b) Similarly, y = b is any line that will be parallel to y = 3 for any value of b. Now you want it to pass through (-2,6) as well, so y = 6 is the required line.