Question

A point P lies on the line with equation y= 4-3X. The point P is a distance √34 from the origin. Find the two possible positions Of point P

Answer 1

The two possible positions of point P are:

(3, -5) and (0.6, 5.8)

How to find the two possible positions of point P?

We know that point P lies on the line:

y = 4 - 3*x

And that the distance between P and the origin is √34, then if the coordinates of point P are (x, y), we have that:

$\sqrt{34} = \sqrt{x^2 + y^2}$

Now, we can replace "y" in the distance equation by the linear equation, and also remove the square roots:

$34 = x^2 + (4 - 3x)^2$

Now we can solve the quadratic equation for x:

$34 = x^2 + 9x^2 + 16 - 24x\\\\10x^2 - 24x - 18 = 0$

The solutions are:

$x = \frac{24 \pm \sqrt{(-24)^2 - 4*10*(-18)} }{2*10} \\\\x = \frac{24 \pm36}{20}$

So the two solutions are:

x = (24 + 36)/20 = 3

x = (24 - 36)/20 = -0.6

To get the points, we need to evaluate y on these values:

y = 4 - 3*3 = -5   So we have P = (3, -5)

y = 4 - 3*(-0.6) = 2.2 So we have the point (0.6, 5.8)

There are the two possible positions of point P.

If you want to learn more about quadratic equations:

brainly.com/question/1214333

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