Question

Question:-Obtain coordinates of points on the line x/2=y/2=z/1 which are at 6 unit distance from the origin.(Fill in the blanks) ​

Answer 1

Answer:

Let O be the origin and Q be the foot of the perpendicular dropped from P onto the x axis.

So ΔOPQ is right-angled at Q.

By definition of coordinates:

OQ=x coordinate of P= distance of P from y axis =∣x∣

Similarly, QP=∣y∣.

Thus, by using Pythagoras theorem on ΔOPQ, we get OP= OQ 2+QP 2= x 2 +y 2

Thus the distance of the point P(x,y) from the origin O(0,0) is x 2+y 2

Explanation:

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Answer 2

Answer:

So the point be (4,4,2) or(-4,-4,-2).

Explanation:

GIVEN

$\frac{x}{2} =\frac{y}{2} =\frac{z}{1} =k\\\\from this we get:\\x=2k;y=2k;z=k\\distance of the points from origin(000)=\sqrt{(2k-0)^{2}+(2k-0)^{2}+(k-0)^{2 } }\\ =6 units\\\\\sqrt{(2k)^{2}+(2k)^{2}+k^{2} } =6\\therefore,\\\\\sqrt{9k^{2} } =6;$

squaring both sides we get,

9k²=36

k²=4

k=±2.

If k=2, so the points become

x=2k =4;

y=2k= 4;

z=k=2

point(4,4,2)

second case k=-2

x=2k=-4

y=2k=-4

z=k=-2

The point becomes(-4,-4,-2)

So the point be (4,4,2) or(-4,-4,-2).