Rectangular Coordinate Systems
$begingroup$
Let $A(0,2)$ and $B(6,0).$ Find all points $P$ on the line $y = 4$ such that $APperp BP.$
(a) by slope
(b) by Pythagorean Theorem
I tried getting the midpoint which is $M=(3,1)$ and then I looked for the slope $m= -1/3$ and the inverse of slope $3$ to find the equation of the line perpendicular to the points $A$ and $B$ which is $y = 3x -8.$
algebra-precalculus trigonometry slope
edited Dec 10 '18 at 15:36
user376343
3,7883828
asked Sep 15 '16 at 3:45
nomad09nomad09
61
$endgroup$
add a comment |
$begingroup$
Let $A(0,2)$ and $B(6,0).$ Find all points $P$ on the line $y = 4$ such that $APperp BP.$
(a) by slope
(b) by Pythagorean Theorem
I tried getting the midpoint which is $M=(3,1)$ and then I looked for the slope $m= -1/3$ and the inverse of slope $3$ to find the equation of the line perpendicular to the points $A$ and $B$ which is $y = 3x -8.$
algebra-precalculus trigonometry slope
edited Dec 10 '18 at 15:36
user376343
3,7883828
asked Sep 15 '16 at 3:45
nomad09nomad09
61
$endgroup$
$begingroup$
I tried getting the midpoint which is M=(3,1) and then I looked for the slope m= -1/3 and the inverse of slope 3 to find the equation of the line perpendicular to the points A and B which is y = 3x -8
$endgroup$
– nomad09
Sep 15 '16 at 3:49
$begingroup$
There is no need for midpoint. The perpendicular need not be to the middle
$endgroup$
– N.S.JOHN
Sep 15 '16 at 3:51
$begingroup$
That's where I get lost. I tried using the coordinates (0,4) and the inverse slope 3.
$endgroup$
– nomad09
Sep 15 '16 at 3:53
add a comment |
$begingroup$
Let $A(0,2)$ and $B(6,0).$ Find all points $P$ on the line $y = 4$ such that $APperp BP.$
(a) by slope
(b) by Pythagorean Theorem
I tried getting the midpoint which is $M=(3,1)$ and then I looked for the slope $m= -1/3$ and the inverse of slope $3$ to find the equation of the line perpendicular to the points $A$ and $B$ which is $y = 3x -8.$
algebra-precalculus trigonometry slope
edited Dec 10 '18 at 15:36
user376343
3,7883828
asked Sep 15 '16 at 3:45
nomad09nomad09
61
$endgroup$
Let $A(0,2)$ and $B(6,0).$ Find all points $P$ on the line $y = 4$ such that $APperp BP.$
(a) by slope
(b) by Pythagorean Theorem
I tried getting the midpoint which is $M=(3,1)$ and then I looked for the slope $m= -1/3$ and the inverse of slope $3$ to find the equation of the line perpendicular to the points $A$ and $B$ which is $y = 3x -8.$
algebra-precalculus trigonometry slope
algebra-precalculus trigonometry slope
edited Dec 10 '18 at 15:36
user376343
3,7883828
asked Sep 15 '16 at 3:45
nomad09nomad09
61
edited Dec 10 '18 at 15:36
user376343
3,7883828
asked Sep 15 '16 at 3:45
nomad09nomad09
61
edited Dec 10 '18 at 15:36
user376343
3,7883828
edited Dec 10 '18 at 15:36
user376343
3,7883828
edited Dec 10 '18 at 15:36
user376343
3,7883828
3,7883828
asked Sep 15 '16 at 3:45
nomad09nomad09
61
asked Sep 15 '16 at 3:45
nomad09nomad09
61
asked Sep 15 '16 at 3:45
nomad09nomad09
61
61
$begingroup$
I tried getting the midpoint which is M=(3,1) and then I looked for the slope m= -1/3 and the inverse of slope 3 to find the equation of the line perpendicular to the points A and B which is y = 3x -8
$endgroup$
– nomad09
Sep 15 '16 at 3:49
$begingroup$
There is no need for midpoint. The perpendicular need not be to the middle
$endgroup$
– N.S.JOHN
Sep 15 '16 at 3:51
$begingroup$
That's where I get lost. I tried using the coordinates (0,4) and the inverse slope 3.
$endgroup$
– nomad09
Sep 15 '16 at 3:53
add a comment |
$begingroup$
I tried getting the midpoint which is M=(3,1) and then I looked for the slope m= -1/3 and the inverse of slope 3 to find the equation of the line perpendicular to the points A and B which is y = 3x -8
$endgroup$
– nomad09
Sep 15 '16 at 3:49
$begingroup$
There is no need for midpoint. The perpendicular need not be to the middle
$endgroup$
– N.S.JOHN
Sep 15 '16 at 3:51
$begingroup$
That's where I get lost. I tried using the coordinates (0,4) and the inverse slope 3.
$endgroup$
– nomad09
Sep 15 '16 at 3:53
$begingroup$
I tried getting the midpoint which is M=(3,1) and then I looked for the slope m= -1/3 and the inverse of slope 3 to find the equation of the line perpendicular to the points A and B which is y = 3x -8
$endgroup$
– nomad09
Sep 15 '16 at 3:49
$begingroup$
I tried getting the midpoint which is M=(3,1) and then I looked for the slope m= -1/3 and the inverse of slope 3 to find the equation of the line perpendicular to the points A and B which is y = 3x -8
$endgroup$
– nomad09
Sep 15 '16 at 3:49
$begingroup$
There is no need for midpoint. The perpendicular need not be to the middle
$endgroup$
– N.S.JOHN
Sep 15 '16 at 3:51
$begingroup$
There is no need for midpoint. The perpendicular need not be to the middle
$endgroup$
– N.S.JOHN
Sep 15 '16 at 3:51
$begingroup$
That's where I get lost. I tried using the coordinates (0,4) and the inverse slope 3.
$endgroup$
– nomad09
Sep 15 '16 at 3:53
$begingroup$
That's where I get lost. I tried using the coordinates (0,4) and the inverse slope 3.
$endgroup$
– nomad09
Sep 15 '16 at 3:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Take P $(x,4)$ Find slope of AP and BP in terms of $x $.
Now use the relation between perpendicular slopes and equate and solve for $x $
answered Sep 15 '16 at 3:57
N.S.JOHNN.S.JOHN
1,177620
$endgroup$
add a comment |
$begingroup$
If $AP$ and $BP$ are perpendicular, that means they form a right angle and $AB$ is the hypotenuse of a right triangle. So you can use the distance formula to find the distance between $(0, 2)$ and $(p, 4)$ and the distance between $(6, 0)$ and $(p, 4)$. Plug these distances in to the Pythagorean Theorem to see if they make a right triangle.
So we have these distances:
$AB=sqrt{left(6-0right)^2+left(0-2right)^2}=sqrt{36+4}=sqrt{40}$
$AP=sqrt{left(p-0right)^2+left(4-2right)^2}=sqrt{p^2+4}$
$BP=sqrt{left(6-pright)^2+left(0-4right)^2}=sqrt{36-12p+p^2+16}=sqrt{p^2-12p+52}$
Now we use the converse of the Pythagorean Theorem to see which values of $p$ (if any) make a right triangle. (This is the fun part because we make the square root symbols go away.)
$left(sqrt{p^2-12p+52}right)^2+left(sqrt{p^2+4}right)^2=left(sqrt{40}right)^2$
$p^2-12p+52+p^2+4=40$
$2p^2-12p+56=40$
$2p^2-12p+16=0$
This quadratic equation is easy to factor:
$2left(p^2-6p+8right)=0$
$2left(p-2right)left(p-4right)=0$
Thus we have $p=2$ or $p=4$. In the context of the problem, this means the points that make a right triangle (where $AP$ and $BP$ are perpendicular) are $(2,4)$ and $(4,4)$.
answered Sep 16 '16 at 0:18
scottscott
1,459511
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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votes
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oldest
votes
$begingroup$
Take P $(x,4)$ Find slope of AP and BP in terms of $x $.
Now use the relation between perpendicular slopes and equate and solve for $x $
answered Sep 15 '16 at 3:57
N.S.JOHNN.S.JOHN
1,177620
$endgroup$
add a comment |
$begingroup$
Take P $(x,4)$ Find slope of AP and BP in terms of $x $.
Now use the relation between perpendicular slopes and equate and solve for $x $
answered Sep 15 '16 at 3:57
N.S.JOHNN.S.JOHN
1,177620
$endgroup$
add a comment |
$begingroup$
Take P $(x,4)$ Find slope of AP and BP in terms of $x $.
Now use the relation between perpendicular slopes and equate and solve for $x $
answered Sep 15 '16 at 3:57
N.S.JOHNN.S.JOHN
1,177620
$endgroup$
Take P $(x,4)$ Find slope of AP and BP in terms of $x $.
Now use the relation between perpendicular slopes and equate and solve for $x $
answered Sep 15 '16 at 3:57
N.S.JOHNN.S.JOHN
1,177620
answered Sep 15 '16 at 3:57
N.S.JOHNN.S.JOHN
1,177620
answered Sep 15 '16 at 3:57
N.S.JOHNN.S.JOHN
1,177620
answered Sep 15 '16 at 3:57
N.S.JOHNN.S.JOHN
1,177620
1,177620
add a comment |
add a comment |
$begingroup$
If $AP$ and $BP$ are perpendicular, that means they form a right angle and $AB$ is the hypotenuse of a right triangle. So you can use the distance formula to find the distance between $(0, 2)$ and $(p, 4)$ and the distance between $(6, 0)$ and $(p, 4)$. Plug these distances in to the Pythagorean Theorem to see if they make a right triangle.
So we have these distances:
$AB=sqrt{left(6-0right)^2+left(0-2right)^2}=sqrt{36+4}=sqrt{40}$
$AP=sqrt{left(p-0right)^2+left(4-2right)^2}=sqrt{p^2+4}$
$BP=sqrt{left(6-pright)^2+left(0-4right)^2}=sqrt{36-12p+p^2+16}=sqrt{p^2-12p+52}$
Now we use the converse of the Pythagorean Theorem to see which values of $p$ (if any) make a right triangle. (This is the fun part because we make the square root symbols go away.)
$left(sqrt{p^2-12p+52}right)^2+left(sqrt{p^2+4}right)^2=left(sqrt{40}right)^2$
$p^2-12p+52+p^2+4=40$
$2p^2-12p+56=40$
$2p^2-12p+16=0$
This quadratic equation is easy to factor:
$2left(p^2-6p+8right)=0$
$2left(p-2right)left(p-4right)=0$
Thus we have $p=2$ or $p=4$. In the context of the problem, this means the points that make a right triangle (where $AP$ and $BP$ are perpendicular) are $(2,4)$ and $(4,4)$.
answered Sep 16 '16 at 0:18
scottscott
1,459511
$endgroup$
add a comment |
$begingroup$
If $AP$ and $BP$ are perpendicular, that means they form a right angle and $AB$ is the hypotenuse of a right triangle. So you can use the distance formula to find the distance between $(0, 2)$ and $(p, 4)$ and the distance between $(6, 0)$ and $(p, 4)$. Plug these distances in to the Pythagorean Theorem to see if they make a right triangle.
So we have these distances:
$AB=sqrt{left(6-0right)^2+left(0-2right)^2}=sqrt{36+4}=sqrt{40}$
$AP=sqrt{left(p-0right)^2+left(4-2right)^2}=sqrt{p^2+4}$
$BP=sqrt{left(6-pright)^2+left(0-4right)^2}=sqrt{36-12p+p^2+16}=sqrt{p^2-12p+52}$
Now we use the converse of the Pythagorean Theorem to see which values of $p$ (if any) make a right triangle. (This is the fun part because we make the square root symbols go away.)
$left(sqrt{p^2-12p+52}right)^2+left(sqrt{p^2+4}right)^2=left(sqrt{40}right)^2$
$p^2-12p+52+p^2+4=40$
$2p^2-12p+56=40$
$2p^2-12p+16=0$
This quadratic equation is easy to factor:
$2left(p^2-6p+8right)=0$
$2left(p-2right)left(p-4right)=0$
Thus we have $p=2$ or $p=4$. In the context of the problem, this means the points that make a right triangle (where $AP$ and $BP$ are perpendicular) are $(2,4)$ and $(4,4)$.
answered Sep 16 '16 at 0:18
scottscott
1,459511
$endgroup$
add a comment |
$begingroup$
If $AP$ and $BP$ are perpendicular, that means they form a right angle and $AB$ is the hypotenuse of a right triangle. So you can use the distance formula to find the distance between $(0, 2)$ and $(p, 4)$ and the distance between $(6, 0)$ and $(p, 4)$. Plug these distances in to the Pythagorean Theorem to see if they make a right triangle.
So we have these distances:
$AB=sqrt{left(6-0right)^2+left(0-2right)^2}=sqrt{36+4}=sqrt{40}$
$AP=sqrt{left(p-0right)^2+left(4-2right)^2}=sqrt{p^2+4}$
$BP=sqrt{left(6-pright)^2+left(0-4right)^2}=sqrt{36-12p+p^2+16}=sqrt{p^2-12p+52}$
Now we use the converse of the Pythagorean Theorem to see which values of $p$ (if any) make a right triangle. (This is the fun part because we make the square root symbols go away.)
$left(sqrt{p^2-12p+52}right)^2+left(sqrt{p^2+4}right)^2=left(sqrt{40}right)^2$
$p^2-12p+52+p^2+4=40$
$2p^2-12p+56=40$
$2p^2-12p+16=0$
This quadratic equation is easy to factor:
$2left(p^2-6p+8right)=0$
$2left(p-2right)left(p-4right)=0$
Thus we have $p=2$ or $p=4$. In the context of the problem, this means the points that make a right triangle (where $AP$ and $BP$ are perpendicular) are $(2,4)$ and $(4,4)$.
answered Sep 16 '16 at 0:18
scottscott
1,459511
$endgroup$
If $AP$ and $BP$ are perpendicular, that means they form a right angle and $AB$ is the hypotenuse of a right triangle. So you can use the distance formula to find the distance between $(0, 2)$ and $(p, 4)$ and the distance between $(6, 0)$ and $(p, 4)$. Plug these distances in to the Pythagorean Theorem to see if they make a right triangle.
So we have these distances:
$AB=sqrt{left(6-0right)^2+left(0-2right)^2}=sqrt{36+4}=sqrt{40}$
$AP=sqrt{left(p-0right)^2+left(4-2right)^2}=sqrt{p^2+4}$
$BP=sqrt{left(6-pright)^2+left(0-4right)^2}=sqrt{36-12p+p^2+16}=sqrt{p^2-12p+52}$
Now we use the converse of the Pythagorean Theorem to see which values of $p$ (if any) make a right triangle. (This is the fun part because we make the square root symbols go away.)
$left(sqrt{p^2-12p+52}right)^2+left(sqrt{p^2+4}right)^2=left(sqrt{40}right)^2$
$p^2-12p+52+p^2+4=40$
$2p^2-12p+56=40$
$2p^2-12p+16=0$
This quadratic equation is easy to factor:
$2left(p^2-6p+8right)=0$
$2left(p-2right)left(p-4right)=0$
Thus we have $p=2$ or $p=4$. In the context of the problem, this means the points that make a right triangle (where $AP$ and $BP$ are perpendicular) are $(2,4)$ and $(4,4)$.
answered Sep 16 '16 at 0:18
scottscott
1,459511
answered Sep 16 '16 at 0:18
scottscott
1,459511
answered Sep 16 '16 at 0:18
scottscott
1,459511
answered Sep 16 '16 at 0:18
scottscott
1,459511
1,459511
add a comment |
add a comment |
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$begingroup$
I tried getting the midpoint which is M=(3,1) and then I looked for the slope m= -1/3 and the inverse of slope 3 to find the equation of the line perpendicular to the points A and B which is y = 3x -8
$endgroup$
– nomad09
Sep 15 '16 at 3:49
$begingroup$
There is no need for midpoint. The perpendicular need not be to the middle
$endgroup$
– N.S.JOHN
Sep 15 '16 at 3:51
$begingroup$
That's where I get lost. I tried using the coordinates (0,4) and the inverse slope 3.
$endgroup$
– nomad09
Sep 15 '16 at 3:53