Answer:
the slope of both lines are the same.
Step-by-step explanation:
Given the following segment of the Quadrilateral EFGH on a coordinate Segment FG is on the line 3x − y = −2,
segment EH is on the 3x − y = −6.
To determine their relationship, we can find the slope of the lines
For line FG: 3x - y = -2
Rewrite in standard form y = mx+c
-y = -3x - 2
Multiply through by-1
y = 3x + 2
Compare
mx = 3x
m = 3
The slope of the line segment FG is 3
For line EH: 3x - y = -6
Rewrite in standard form y = mx+c
-y = -3x - 6
Multiply through by-1
y = 3x + 6
Compare
mx = 3x
m = 3
The slope of the line segment EH is 3
Hence the statement that proves their relationship is that the slope of both lines are the same.
I hope this helps, I wrote the new equation each time you get a new number and where it should be placed
The properties that apply true are:
1–as it implies the Pythagorean theorem; a^2+b^2=c^2
3–as it implies the basic rules of visualized geometry.
4–as the greater side(hypotenuse) is opposite the right angle.
NOT 2, as the accrue angles ARE complimentary.
Well if it is a number like .933333333 you will have to round it, So now it is .93 and .93 in a fraction is 93/100
Answer:
x -2y -2=0
Step-by-step explanation:
using the formula, m= ( Y2 - Y1)/( X2-X1)
the gradient is found. the coordinates chosen are (2,0) and ( 0,-1)
the equation of the line is obtained using the formula
y -y1= m(x -x1)
