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Both of the points P( 17, − 20) and Q( 25, t) are in the xy-
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10 Aug 2020, 10:01
10 Aug 2020, 10:01
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35% (02:05) correct
64% (01:53) wrong based on 101 sessions
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Both of the points P( 17, − 20) and Q( 25, t) are in the xy-plane. Which of the following statements alone give( s) sufficient additional information to determine whether \(t > − 18\) such sets of integers.?
Indicate all such statements.
A. The slope of the line that goes through the points P( 17, − 20) and Q( 25, t) is 3/4
B. The distance between the points P( 17, − 20) and Q( 25, t) is 10.
C. The point Q( 25, t) is the midpoint of the line segment whose endpoints are P( 17, − 20) and R( 33, 3t + 34).
Indicate all such statements.
A. The slope of the line that goes through the points P( 17, − 20) and Q( 25, t) is 3/4
B. The distance between the points P( 17, − 20) and Q( 25, t) is 10.
C. The point Q( 25, t) is the midpoint of the line segment whose endpoints are P( 17, − 20) and R( 33, 3t + 34).
Re: Both of the points P( 17, − 20) and Q( 25, t) are in the xy-
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18 Aug 2020, 12:50
18 Aug 2020, 12:50
2
A is a definite pick, because we can simply extend a line from P in the direction given by the slope, and find Q when it intersects x=25. At that point we can determine whether t > -18.
B shows only the distance from P, which could intersect the line x=25 in two places (positive and negative slope). Because that line is 8 units to the right from P, a distance of 10 would be the hypoteneuse of a right triangle, and the second leg would be 6. -20 +/- 6 won't provide enough information to determine whether t > -18.
C. Knowing that Q is the midpoint of line PR, we know that its y-component t is half of (3t+34) - (-20) (the vertical distance between P and R). So t = 1/2(3t+54), which we don't have to actually solve, but we know we can because we have one equation and one unknown, and that would allow us to determine whether t > -18.
The answer is A and C.
B shows only the distance from P, which could intersect the line x=25 in two places (positive and negative slope). Because that line is 8 units to the right from P, a distance of 10 would be the hypoteneuse of a right triangle, and the second leg would be 6. -20 +/- 6 won't provide enough information to determine whether t > -18.
C. Knowing that Q is the midpoint of line PR, we know that its y-component t is half of (3t+34) - (-20) (the vertical distance between P and R). So t = 1/2(3t+54), which we don't have to actually solve, but we know we can because we have one equation and one unknown, and that would allow us to determine whether t > -18.
The answer is A and C.
Carcass wrote:
Both of the points P( 17, − 20) and Q( 25, t) are in the xy-plane. Which of the following statements alone give( s) sufficient additional information to determine whether \(t > − 18\) such sets of integers.?
Indicate all such statements.
A. The slope of the line that goes through the points P( 17, − 20) and Q( 25, t) is 3/4
B. The distance between the points P( 17, − 20) and Q( 25, t) is 10.
C. The point Q( 25, t) is the midpoint of the line segment whose endpoints are P( 17, − 20) and R( 33, 3t + 34).
Indicate all such statements.
A. The slope of the line that goes through the points P( 17, − 20) and Q( 25, t) is 3/4
B. The distance between the points P( 17, − 20) and Q( 25, t) is 10.
C. The point Q( 25, t) is the midpoint of the line segment whose endpoints are P( 17, − 20) and R( 33, 3t + 34).
Re: Both of the points P( 17, − 20) and Q( 25, t) are in the xy-
[#permalink]
26 Aug 2020, 23:54
26 Aug 2020, 23:54
1
So, although it takes more than 2 min to solve it, let's do it quickly.
A. y = a +bx
let's solve a system of 2 eq.
a) -20 = a + (3/4) 17
b) t = a + (3/4) 25
subst a) - b)
-20 - t = (3/4)8
-t = 14 => -14 - Ok
B. Distance shows nothing - the line can have a negative slope => t <-18 - or it can have no slop - so t=-20 - but the distanse is still 10
C. Not sure - but it seems as being not correct.
Again we have a line
y = a+bx
-20 = a +17b
t = a + 25b
3t+34 = a +33b
New system
-20-t = -8b
-2t - 34 = -8b
And the last one
-3t - 54 = 0
t = -18
but it should be > -18 not equal. Not correct
A. y = a +bx
let's solve a system of 2 eq.
a) -20 = a + (3/4) 17
b) t = a + (3/4) 25
subst a) - b)
-20 - t = (3/4)8
-t = 14 => -14 - Ok
B. Distance shows nothing - the line can have a negative slope => t <-18 - or it can have no slop - so t=-20 - but the distanse is still 10
C. Not sure - but it seems as being not correct.
Again we have a line
y = a+bx
-20 = a +17b
t = a + 25b
3t+34 = a +33b
New system
-20-t = -8b
-2t - 34 = -8b
And the last one
-3t - 54 = 0
t = -18
but it should be > -18 not equal. Not correct
