Find both points on the line y = 3 that are 10 units from (2, -3)

Question

step by step answer please

Answer 1

The point ( 2, -3 ) is 6 units from the line  y = 3

There are 2 points on  y = 3  that are 10 units away.

Drawing a diagram shows these points to make a right triangle

with sides 6, 8, 10  w 10 as hypotenuse and 6 = length of the perpendicular from (2,-3) to y=3

6,8,10 is a Pythagorean Triple.

 

So the 2 points are + and -  8 units away from (2, -3) on the line y=3

They are

( -6, 3 )  and   ( 10, 3 )

 

 

Answer 2

The points in the plane that are 10 units from the point (2,-3) lie on a circle with center at (2,-3) with a radius of 10.

 

The equation for the circle is (x-2)² + (y+3)² = 100. So you want to find the points of intersection of that circle with the line y = 3. Use substitution, i.e. substitute 3 for y in the circle equation and solve for x. 

 

You will find two values for x: -6 and 10. So the points on the line y = 3 that are 10 units from (2,-3) are (-6,3) and (10,3).

 

See the following graph for a picture.

 

https://www.desmos.com/calculator/6vuodfa7rr

 

 

Answer 3

The line y = 3 is a horizontal line. The domain, x can be any real number but y can only ever be 3.

 

**Edit; Sorry Meredith, There are two points from the line y=3 that are exactly 10 units from (2,-3). The point (2,3), directly above it on the line is |3|+|-3|= 6 units. Every other point will make a right triangle with the hypotenuse as the distance between points, 6 units as one side length and the x-coordinate as the other. 10^2 = x^2 + 6^2