Step-by-step explanation:
12a. y-y1=m(x-x1)
y-5=(2-5)/(1--3)(x--3)
y-5=-3/4(x+3)
4y-20=-3x-9
4y=-3x-9+20
4y=-3x+11
y=-¾x+11/4
12b.
Perpendicular lines have the product of their gradient equal to -1
For L1,the gradient is 3
For L2,we express in slope-intercept form
6y=1-2x
y=1/6-⅓x
The slope is -⅓
Now product of both slopes 3×-⅓=-1,thus the lines are perpendicular
