Question

NO LINKS!! Find a point with coordinates of the form (a, 2a) that is in the third quadrant and is a distance 5 from P(3,1).​

Answer 1

Answer:  The correct answer is (-1,-2)

Step-by-step explanation:

The distance 5  units from the point P(3,1)  to the third quadrant with a point of (x,y) = (3,1) is given by using the formula:

D = $\sqrt{(x-x1)^2 + (y-y1)^2}$

D = $\sqrt{(a-3)^2+(2a-1)^2}$

5 = $\sqrt{(a^2-6a+9)+(4a^2-4a+1)}$

5 = $\sqrt{5a^2-10a+10}$

Square both sides:

5^2=5a^2-10a+10

25=5a^2-10a+10

0=5a^2-10a-15

Divide both sides by 5:

0=a^2-2a-3

Using the zero product property, the zero values are:

a = -1

2a = -2

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Answer 2

Answer:

• The point is (- 1, - 2)

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Let the point be Q(a, 2a) and the other one is given P(3, 1).

The distance between them is 5 units and the Q is in the third quadrant.

Use the distance equation to express the length of PQ, then solve it for a:

• PQ² = (a - 3)² + (2a - 1)²

• (a - 3)² + (2a - 1)² = 5²
• a² - 6a + 9 + 4a² - 4a + 1 = 25
• 5a² - 10a + 10 = 25
• 5a² - 10a - 15 = 0
• a² - 2a - 3 = 0
• a² - 3a + a - 3 = 0
• a(a - 3) + (a + 3) = 0
• (a - 3)(a + 1) = 0

• a - 3 = 0 and a + 1 = 0
• a = 3 and a = - 1

The Q has coordinates of (3, 6) and (-1, - 2).

The (3, 6) is in the first quadrant and the (- 1, - 2) is in the third quadrant, hence the second pair of coordinates is our answer.