Hii happy holidays !! any help is appreciated consider the following graph
1 answer:
9514 1404 393
Answer:
A, M, N, F
Step-by-step explanation:
I find it easier to look at the graph, rather than mess with the coordinate transformations. Each image point is the same distance from the line of reflection that its pre-image point is. The line joining the two points is perpendicular to the line of reflection.
See attached for the reflected points.
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The red and turquoise dashed lines are the lines y=x and y=-x, respectively. The same-colored arrows show the reflection of the relevant point.
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The transformations of interest are ...
(x, y) ⇒ (y, x) . . . . reflection over y = x
(x, y) ⇒ (-x, y) . . . . reflection over y-axis
(x, y) ⇒ (x, -y) . . . . reflection over x-axis
(x, y) ⇒ (-y, -x) . . . . reflection over y = -x
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Given:
A plane is normal to the vector = -2i+5j+k
It contains the point (-10,7,5).
To find:
The component equation of the plane.
Solution:
The equation of plane is
Where, is the point on the plane and
is normal vector.
Normal vector is -2i+5j+k and plane passes through (-10,7,5). So, the equation of the plane is
Therefore, the equation of the plane is .