Find all of the points on the line y = 2x + 2 which are 4 units from the point (−1, 4).

Question

I do not know what to do.

Answer 1

y = 2x +2 is an upward sloping straight line with slope = 2, y intercept = 2, and x intercept = -1

the point (-1,4) is 4 units from the line at the point (-1,0), the x intercept. Just graph the line and point, even just a rough sketch and you'll see it's 4 units away, directly above the line. the difference in y coordinates is 4.

another point is (2,6) directly to the right, horizontally to the line,but it's just 3 units away. Just look at the graph. to get 4 units, you need to move up higher on the line, a little larger than 2 and a little larger than 6. construct a right triangle with hoptenuse =4, base = 2+a, and height = 2a

(2+a)^2 + (2a)^2 = 4^2 by the Pythagorean theorem, with 2a=the altitude or height of the right triangle, 2+a = the base and 4 = the hypotenuse

4 + 4a + a^2 + 4a^2 = 16

5a^2 + 4a -12 = 0

(5a -6)(a+ 2) = 0

5a -6 = 0

a = 6/5 = 1.2

x = 1+a = 2.2

y = 4+2a = 6.4

point 4 units from (-1, 4) to the line is (2.2, 6.4)

check the answer

(-1-2.2)^2 + (6.4-4)^2

(-3.2)^2 + 2.4^2 = 4^2

10.24 + 5.76 = 16

16= 16

another method is:

find the closest point to the line from (-1,4) and then treat that as the midpoint between 2 endpoints with one endpoint as (-1, 0) and the other end point that's 4 units from (-1,4)

the line perpendicular to y=2x+2 has slope =-1/2 (take the x coefficient, 2, and find its negative inverse. change its sign and flip it upside down. change +2/1 to -1/2)

y=-x/2 +b. plug in (-1,4) to solve for b. b is the y intercept

4 =1/2 + b

b = 3 1/2 = 7/2

the perpendicular line is: y=-x/2 +7/2.

set -x/2 + 7/2 = 2x + 2 and solve for x, which is the x coordinate of the intersection of the 2 lines

5x/2 = 3/2

5x =3

x = 3/5 = 0.6

y= -3/10 + 7/2 = 35/10 -3/10 = 32/10 = 3.2

the point (.6, 3.2) is the point of intersection, the closest point to (-1,4), the midpoint between two endpoints where one endpoint is (-1,0). the other endpoint is (2.2, 6.4)

(2.2+1)^2 + (6.4-4)^2 = 3.2^2 + 2.4^2 = 10.24 + 5.76 = 16 = 4^2

there are only two points on the line that are 4 units from the given point.

(-1,0) and (2.2, 6.4)

************ BUT another and more direct general way to solve this problem is find the intersection points of the line y=2x+2 and the circle with center (-1,4) and radius 4.

(x+1)^2 + (y-4)^2 = 4^2 is the circle with radius 4 and center (-1,4) a straight line intersects a circle in at either 2 points, no points or is tangent at only 1 point.

general equation for a circle is (x-h)^2 + (y-k)^2 = r^2 where the center is (h,k) and radius = r

substitute 2x+2 for y

(x+1)^2 + (2x+2-4)^2 = 16

(x+1)^2 + (2x-2)^2 = 16

x^2 +2x + 1 + 4x^2 -8x + 4 = 16

5x^2 -6x -11 = 0

(5x -11)(x+1)= 0

5x -11 = 0

x = 11/5 = 2.2

y = 2(2.2)+2 = 6.4

x+1 = 0

x =-1

y = 2(-1) + 1 = 0

the two points are (2.2, 6.4) and (-1,0)

Answer 2

Hello Mariana,

The points that are 4 units away from the point (-1,4) belong to a circle of radius equal to 4. The equation of that circle would be: (x+1)² + (y-4)² = 16.

You could now look at this problem as "find all the points on the line y = 2x+2 which are ALSO on the circle" which means you're looking for where the circle intersects the line y= 2x+2.

How many times can a line intersect a circle? A maximum of 2 times.

Substitute y = 2x+2 in the circle equation ---> (x+1)² + (2x+2-4)² = 16.

---> (x+1)² + (2x-2)² = 16

---> (x² + 2x+1)+ (4x² -8x+4) = 16

---> 5x²-6x+5 = 16

---> 5x²-6x-11 = 0

you can use the quadratic formula or grouping to solve for x

Grouping: 5x²+5x-11x-11 = 0 --> 5x(x+1) - 11(x+1) = 0 ---> (x+1)(5x-11) = 0 --> x = -1 or x = 11/5

To find the y coordinates, you can plug each x value you found in the the equation y = 2x+2:

x = -1: y = 2(-1)+2 = -2+2 = 0 ---> the point is (-1,0)

x = 11/5: y = 2(11/5)+2 = 22/5 + 2 = 32/5 ---> the point is (11/5, 32/5) or (2.2, 6.4) if you prefer decimals

Cheers.