Answer:
Step-by-step explanation:
The required plane Π contains the line
L: (-1,1,2)+t(7,6,2)
means that Π is perpendicular to the direction vector of the line L, namely
vl=<7,6,2>
It is also required that Π be perpendicular to the plane
Π 1 : 5y-7z+8=0
means that Π is also perpendicular to the normal vector of the given plane, vp=<0,5,-7>.
Thus the normal vector of the required plane, Π can be obtained by the cross product of vl and vp, or vl x vp:
i j k
7 6 2
0 5 -7
=<-42-10, 0+49, 35-0>
=<-52, 49, 35>
which is the normal vector of Π
Since Π has to contain the line, it must pass through the point (-1,1,2), so the equation of the plane is
Π : -52(x-(-1))+49(y-1)+35(z-2)=0
=>
Π : -52x+49y+35z = 171
Check that normal vector of plane is orthogonal to line direction vector
<-52,49,35>.<7,6,2>
=-364+294+70
=0 ok
Answers:
- slope = 1
- equation is y = x-4
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Explanation:
Two points on this diagonal line are (0,-4) and (1,-3)
Let's apply the slope formula to get the following
m = (y2-y1)/(x2-x1)
m = (-3-(-4))/(1-0)
m = (-3+4)/(1-0)
m = 1/1
m = 1
The slope is 1. We can think of it as 1/1 to indicate "go up 1, and over to the right 1". In other words, slope = rise/run = 1/1.
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Note how the graph crosses the y axis at -4, so this is the y intercept.
Since the y intercept is -4, this means b = -4
So with m = 1 and b = -4, we go from y = mx+b to y = 1x+(-4) which simplifies to y = x-4
Answer:
I'm just guessing but I'm 1000000000%sure out of 10000000000000000000000000000
Answer:
i think the answer is B
Step-by-step explanation:
formula is A^2+B^2=(A+B)(A-B)
