Answer:
D. d = sqrt( ( x2-x1)^2 + (y2-y1)^2)
C. d = sqrt( ( x1-x2)^2 + (y1-y2)^2)
B. d = sqrt( (|x2-x1|^2 + |y2-y1|^2)
Step-by-step explanation:
Given two points (x1, y1) and (x2,y2) we can find the distance using
d = sqrt( ( x2-x1)^2 + (y2-y1)^2)
The order of the terms inside the square doesn't matter
d = sqrt( ( x1-x2)^2 + (y1-y2)^2)
When we are squaring are term, we can take the absolute value before we square and it does not change the value
d = sqrt( (|x2-x1|^2 + |y2-y1|^2)
comparing the relation given
thus y= -2x +1 to the general equation of a line
where y= mx + c
where m is the gradient and c is the intercept on the y-axis.
from the question, the gradient is -2 and since the line is perpendicular, the gradient is given as
so substituting the value of m in to the equation
=
as the gradient
equation of line is given by
y - y1 = m( x-x1)
from the question y1= -2 and x1=1
substitute them
y-(-2) = 1/2 (x -1)
y+2 =1/2 (x-1)
multiplying through by 2
2 (y+2) = x-1
2y +4 = x-1
2y =x-1-4
2y =x-5
x -2y -5 = 0
therefore the equation for the line is x-2y -5=0
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