Answer: C) 10.3 units
This value is approximate. This also means that segment JK is roughly 10.3 units.
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Explanation:
The "5 units right and 9 units up" means we have a right triangle with legs 5 and 9. The hypotenuse is unknown, which I'll call c.
Apply the pythagorean theorem. Solve for c.
a^2 + b^2 = c^2
5^2 + 9^2 = c^2
25 + 81 = c^2
106 = c^2
c^2 = 106
c = sqrt(106)
c = 10.2956301 approximately
c = 10.3 approximately
The distance from J to K is roughly 10.3 units
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A slightly different method:
Start at point J (-2, -5) and move 5 units to the right and 9 units up to arrive at K(3, 4)
Let's use the distance formula to find the distance from (-2,-5) to (3,4)
Define the following
- (x1,y1) = (-2,-5) which means x1 = -2 and y1 = -5
- (x2,y2) = (3,4) which means x2 = 3 and y2 = 4
Those values are then plugged into the distance formula
[tex]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-2-3)^2 + (-5-4)^2}\\\\d = \sqrt{(-5)^2 + (-9)^2}\\\\d = \sqrt{25 + 81}\\\\d = \sqrt{106}\\\\d \approx 10.2956301\\\\d \approx 10.3\\\\[/tex]
The distance from J to K is roughly 10.3 units
One thing to point out, as you probably already noticed, is that the distance formula and pythagorean theorem effectively are saying the same thing when computing distances like this. The distance formula is based off the pythagorean theorem. Note the x1-x2 portion which represents the horizontal distance from J to K. This is the "move 5 units right" portion. Then the "y1-y2" is the vertical change in distance, which is the "move 9 units up". Also, we can see that the square root of 106 shows up again; a lot of elements are repeated (just phrased in a slightly different way) from the last section.
