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Answer:
Step-by-step explanation:
The shortest distance d, of a point (a, b, c) from a plane mx + ny + tz = r is given by:
--------------------(i)
From the question,
the point is (5, 0, -6)
the plane is x + y + z = 6
Therefore,
a = 5
b = 0
c = -6
m = 1
n = 1
t = 1
r = 6
Substitute these values into equation (i) as follows;
Therefore, the shortest distance from the point to the plane is
y=−3x+9 slope of this equation is -3 and y-intercept is 9
Plot the y -intercept first which is 9 and mark the points using the slope which is -3. Join the points. So blue line in the graph is y = -3x+9
y=−x−5 slope of this equation is -1 and y intercept is -5
Plot the y -intercept first which is -5 and mark the points using the slope which is -1 . Join the points. So pink line in the graph is y = -x-5
The intersection point of these two lines gives you the solution.
The coordinates of point of intersection (7,-12)
