Maths MC 2題 ....5分

1.AB=√[(a-5a)2+(-a-2a)2]
=√(16a2+9a2)
=√(25a2)
=5a
ANS=A

2.Slope of AB=[6-(-2)]/(5-3)
=8/2
=4
∴slope of straight line L =-1/4
ANS=B

1.Find the distance between (a,-a) and (5a,2a)
√((5a - a)2 + (2a - -a)2)
= √(25a2)
= 5a (A)

A.5a B.√17a C.√37a D.√45a

2.Given two points A(5,6) and B(3,-2),find the slope of straight line L which is perpendicular to AB .
slope of AB
= (6 - -2)/(5 - 3)
= 4
slope of the line
= -1/(slope of AB)
= -1/4 (B)

A.-1/2 B.-1/4 C.1/4 D.1/2

1. A : distance = √(2a--a)^2 (5a-a)^2=√9a 16a=5a

2. B : slope of AB = [6-(-2)]/(5-3) = 8/2 = 4
since slope of line L * slope of AB = -1
so slope of line L = -1/4