Answer :
(a) To show that the point (3/5, 4/5) is on the unit circle, we need to prove that (3/5)² + (4/5)² = 1 .
(b) The distance between points (2,-5) and (-7,-5) is 9 units .
(c) The midpoint of the line segment that joins the given points is given by: (-5/2,-5) .
In the question ,
it is given that ,
the points given is (3/5 , 4/5)
(a) To show that the point (3/5, 4/5) is on the unit circle, we need to prove that (3/5)^2 + (4/5)^2 = 1
that is
9/25 + 16/25 = 1
25/25 = 1
1 = 1
hence proved that the point (3/5, 4/5) is on the unit circle .
(b) The distance between (2, -5) and (-7, -5) is calculated using the formula
[tex]=[/tex]√(x₂ - x₁)² [tex]+[/tex] (y₂ - y₁)²
= √(-7 - 2)² + (-5 -(-5))²
= √(-9)² + (-5 + 5)²
= √(-9)² = √81
= 9 units
(c) the mid point joining the points (2, -5) and (-7, -5) is (a,b) that is calculated by
a = (2-7)/2 , b = (-5-5)/2
a = -5/2 , b = -10/2
a = -5/2 and b = -5
the mid point is (-5/2,-5) .
Therefore , (a) To show that the point (3/5, 4/5) is on the unit circle, we need to prove that (3/5)² + (4/5)² = 1 .
(b) The distance between points (2, -5) and (-7, -5) is 9 units .
(c) The midpoint of the line segment that joins the given points is given by (-5/2,-5) .
The given question is incomplete , the complete question is
(a) To show that the point (3/5, 4/5) is on the unit circle, we need to prove that (3/5)² + (4/5)² = .
(b) The distance between (2, -5) and (-7, -5) is .
(c) The midpoint of the line segment that joins the given points is given by ( , ) .
Learn more about Distance Formula here
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