Answer:
[tex]P_{1}(-14, 12) , P_{2}(5,12)[/tex] are 19 units apart
Step-by-step explanation:
Answer:
We are given four pair of points and we have to find which points are 19 units apart:
[tex]P_{1}(x_{1},y_{1}) , P_{2}(x_{2},y_{2})[/tex]
[tex]P_{1}(-14, 12) , P_{2}(5,12)\\\\P_{1}(-5, 19) , P_{2}(-12, 19)\\\\P_{1}(-20, 5) , P_{2}(1, 5)\\\\P_{1}(14, 12) , P_{2}(5,12)[/tex]
Using distance formula to find how far the points are from each other
[tex]d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}[/tex]
For first pair:
[tex]P_{1}(x_{1},y_{1}) , P_{2}(x_{2},y_{2}) = P_{1}(-14, 12) , P_{2}(5,12)\\\\d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\\\\d = \sqrt{(-14 - 5)^{2} + (12-12)^{2}} = 19[/tex]
For second pair:
[tex]P_{1}(x_{1},y_{1}) , P_{2}(x_{2},y_{2}) = P_{1}(-5, 19) , P_{2}(-12,19)\\\\d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\\\\d = \sqrt{(-5 - 12)^{2} + (19 - 19)^{2}} = 17[/tex]
For third pair:
[tex]P_{1}(x_{1},y_{1}) , P_{2}(x_{2},y_{2}) = P_{1}(-20, 15) , P_{2}(1, 5)\\\\d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\\\\d = \sqrt{(-20 - 1)^{2} + (15 - 5)^{2}} = 23.26[/tex]
For 4th pair:
[tex]P_{1}(x_{1},y_{1}) , P_{2}(x_{2},y_{2}) = P_{1}(14, 12) , P_{2}(5,12)\\\\d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\\\\d = \sqrt{(14 - 5)^{2} + (12 - 12)^{2}} = 9[/tex]
Answer:
Choice a is correct answer.
Step-by-step explanation:
We have given four set of points.
(a) (-14,12) and (5,12)
(b) (-5,19) and (-12,19)
(c) (-20,5) and (1,5)
(d) (14,12) and (5,12)
We have to find the distance between each set of points.
The formula for distance is:
d = √(x₂-x₁)²+(y₂-y₁)²
The set of points which have 19 units distance is our answer.
For (a):
d =√(5-(-14)²+(12-12)²
d = √(5+14)²+(0)²
d = 19 units
for (b) :
d = √(-12-(-5))²+(19-19)²
d = √(-12+5)²+(0)²
d = √(-7)²
d = 7 units
for (c):
d = √(1-(-20)²+(5-5)²
d = √(1+20)²+(0)²
d = 21 units
for (d):
d =√(5-14)²+(12-12)²
d = √(-9)²+(0)²
d = 9 units
Hence , correct choice is (a).
