Line through a a given point in the first quadrant of the coordinate plane to form a triangle [closed]
$begingroup$
Consider a straight line with negative gradient passing through the positive quadrant (where all co-ordinates are positive, or the first quadrant) of the co-ordinate plane and intercepting the $x$ and $y$ axes to form a triangle with them.
Which points in the positive quadrant can the line pass through such that the triangle always has area $2$?
I've considered the point $(1,1)$ and formed an equation in terms of the rectangle made by the point and the two triangles either side of it. My hypothesis is that anything right of $(1,1)$ doesn't work
geometry
edited Jan 7 at 1:41
Andrei
13.2k21230
asked Jan 7 at 1:30
user491842user491842
62
$endgroup$
closed as off-topic by Namaste, KReiser, Leucippus, Cesareo, José Carlos Santos Jan 7 at 8:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, KReiser, Leucippus, Cesareo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Consider a straight line with negative gradient passing through the positive quadrant (where all co-ordinates are positive, or the first quadrant) of the co-ordinate plane and intercepting the $x$ and $y$ axes to form a triangle with them.
Which points in the positive quadrant can the line pass through such that the triangle always has area $2$?
I've considered the point $(1,1)$ and formed an equation in terms of the rectangle made by the point and the two triangles either side of it. My hypothesis is that anything right of $(1,1)$ doesn't work
geometry
edited Jan 7 at 1:41
Andrei
13.2k21230
asked Jan 7 at 1:30
user491842user491842
62
$endgroup$
closed as off-topic by Namaste, KReiser, Leucippus, Cesareo, José Carlos Santos Jan 7 at 8:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, KReiser, Leucippus, Cesareo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
hi, what have you tried so far?
$endgroup$
– Siong Thye Goh
Jan 7 at 1:33
$begingroup$
Edited in a little
$endgroup$
– user491842
Jan 7 at 1:39
$begingroup$
any points (0,2x) and (x,0)
$endgroup$
– user29418
Jan 7 at 1:40
$begingroup$
A line going through $(1,1)$ and intersecting both the $x$ and $y$ axis in the first quadrant will have points on the right of $(1,1)$ (the point will just have lower $y$ coordinates)
$endgroup$
– Andrei
Jan 7 at 1:43
add a comment |
$begingroup$
Consider a straight line with negative gradient passing through the positive quadrant (where all co-ordinates are positive, or the first quadrant) of the co-ordinate plane and intercepting the $x$ and $y$ axes to form a triangle with them.
Which points in the positive quadrant can the line pass through such that the triangle always has area $2$?
I've considered the point $(1,1)$ and formed an equation in terms of the rectangle made by the point and the two triangles either side of it. My hypothesis is that anything right of $(1,1)$ doesn't work
geometry
edited Jan 7 at 1:41
Andrei
13.2k21230
asked Jan 7 at 1:30
user491842user491842
62
$endgroup$
Consider a straight line with negative gradient passing through the positive quadrant (where all co-ordinates are positive, or the first quadrant) of the co-ordinate plane and intercepting the $x$ and $y$ axes to form a triangle with them.
Which points in the positive quadrant can the line pass through such that the triangle always has area $2$?
I've considered the point $(1,1)$ and formed an equation in terms of the rectangle made by the point and the two triangles either side of it. My hypothesis is that anything right of $(1,1)$ doesn't work
geometry
geometry
edited Jan 7 at 1:41
Andrei
13.2k21230
asked Jan 7 at 1:30
user491842user491842
62
edited Jan 7 at 1:41
Andrei
13.2k21230
asked Jan 7 at 1:30
user491842user491842
62
edited Jan 7 at 1:41
Andrei
13.2k21230
edited Jan 7 at 1:41
Andrei
13.2k21230
edited Jan 7 at 1:41
Andrei
13.2k21230
13.2k21230
asked Jan 7 at 1:30
user491842user491842
62
asked Jan 7 at 1:30
user491842user491842
62
asked Jan 7 at 1:30
user491842user491842
62
62
closed as off-topic by Namaste, KReiser, Leucippus, Cesareo, José Carlos Santos Jan 7 at 8:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, KReiser, Leucippus, Cesareo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, KReiser, Leucippus, Cesareo, José Carlos Santos Jan 7 at 8:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, KReiser, Leucippus, Cesareo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
hi, what have you tried so far?
$endgroup$
– Siong Thye Goh
Jan 7 at 1:33
$begingroup$
Edited in a little
$endgroup$
– user491842
Jan 7 at 1:39
$begingroup$
any points (0,2x) and (x,0)
$endgroup$
– user29418
Jan 7 at 1:40
$begingroup$
A line going through $(1,1)$ and intersecting both the $x$ and $y$ axis in the first quadrant will have points on the right of $(1,1)$ (the point will just have lower $y$ coordinates)
$endgroup$
– Andrei
Jan 7 at 1:43
add a comment |
$begingroup$
hi, what have you tried so far?
$endgroup$
– Siong Thye Goh
Jan 7 at 1:33
$begingroup$
Edited in a little
$endgroup$
– user491842
Jan 7 at 1:39
$begingroup$
any points (0,2x) and (x,0)
$endgroup$
– user29418
Jan 7 at 1:40
$begingroup$
A line going through $(1,1)$ and intersecting both the $x$ and $y$ axis in the first quadrant will have points on the right of $(1,1)$ (the point will just have lower $y$ coordinates)
$endgroup$
– Andrei
Jan 7 at 1:43
$begingroup$
hi, what have you tried so far?
$endgroup$
– Siong Thye Goh
Jan 7 at 1:33
$begingroup$
hi, what have you tried so far?
$endgroup$
– Siong Thye Goh
Jan 7 at 1:33
$begingroup$
Edited in a little
$endgroup$
– user491842
Jan 7 at 1:39
$begingroup$
Edited in a little
$endgroup$
– user491842
Jan 7 at 1:39
$begingroup$
any points (0,2x) and (x,0)
$endgroup$
– user29418
Jan 7 at 1:40
$begingroup$
any points (0,2x) and (x,0)
$endgroup$
– user29418
Jan 7 at 1:40
$begingroup$
A line going through $(1,1)$ and intersecting both the $x$ and $y$ axis in the first quadrant will have points on the right of $(1,1)$ (the point will just have lower $y$ coordinates)
$endgroup$
– Andrei
Jan 7 at 1:43
$begingroup$
A line going through $(1,1)$ and intersecting both the $x$ and $y$ axis in the first quadrant will have points on the right of $(1,1)$ (the point will just have lower $y$ coordinates)
$endgroup$
– Andrei
Jan 7 at 1:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Guide:
Suppose the gradient is $m$ and the $x$-intercept is $x$, then the $y$ intercept is $-mx$.
That is we want to have $$frac12cdot xcdot (-mx) = 2$$
Hence solving for $x$ in terms of $m$ should fully determine the line.
answered Jan 7 at 1:42
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
The equation of a line in the plane, with slope $-m$ can be written as $$y=-mx+b$$To get the condition to have the area $2$, we need to calculate the intersection with the axes. If $x=0$, you get $y=b$. If $y=0$, you get $$x=frac bm$$
The area of the triangle is then $$2=frac 12 bfrac bm$$which yields $$b=2sqrt m$$
Going back to the original equation $$y=-mx+2sqrt m$$
Now let's suppose that I have a point $(x_0,y_0)$ on such line. What would happen if I keep $x_0$ constant, and I want to either increase or decrease $y_0$? Can I find a different $m$? Otherwise saying, given $x_0$, for what values of $y_0$ can I find a solution for the following equation: $$y_0=-mx_0+2sqrt m$$We can write this as a quadratic equation in $X=sqrt m$
$$x_0X^2-2X+y_0=0$$
To have a real solution, we need the discriminant to be positive, so $$1-x_0y_0ge 0$$ or $$x_0y_0le 1$$
What it means that any point that lies in the first quadrant below the hyperbola $xy=1$ is part of at least one line such that the intersections with the axis form a triangle with area $2$.
answered Jan 7 at 2:27
AndreiAndrei
13.2k21230
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Guide:
Suppose the gradient is $m$ and the $x$-intercept is $x$, then the $y$ intercept is $-mx$.
That is we want to have $$frac12cdot xcdot (-mx) = 2$$
Hence solving for $x$ in terms of $m$ should fully determine the line.
answered Jan 7 at 1:42
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
Guide:
Suppose the gradient is $m$ and the $x$-intercept is $x$, then the $y$ intercept is $-mx$.
That is we want to have $$frac12cdot xcdot (-mx) = 2$$
Hence solving for $x$ in terms of $m$ should fully determine the line.
answered Jan 7 at 1:42
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
Guide:
Suppose the gradient is $m$ and the $x$-intercept is $x$, then the $y$ intercept is $-mx$.
That is we want to have $$frac12cdot xcdot (-mx) = 2$$
Hence solving for $x$ in terms of $m$ should fully determine the line.
answered Jan 7 at 1:42
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
Guide:
Suppose the gradient is $m$ and the $x$-intercept is $x$, then the $y$ intercept is $-mx$.
That is we want to have $$frac12cdot xcdot (-mx) = 2$$
Hence solving for $x$ in terms of $m$ should fully determine the line.
answered Jan 7 at 1:42
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 7 at 1:42
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 7 at 1:42
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 7 at 1:42
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
add a comment |
add a comment |
$begingroup$
The equation of a line in the plane, with slope $-m$ can be written as $$y=-mx+b$$To get the condition to have the area $2$, we need to calculate the intersection with the axes. If $x=0$, you get $y=b$. If $y=0$, you get $$x=frac bm$$
The area of the triangle is then $$2=frac 12 bfrac bm$$which yields $$b=2sqrt m$$
Going back to the original equation $$y=-mx+2sqrt m$$
Now let's suppose that I have a point $(x_0,y_0)$ on such line. What would happen if I keep $x_0$ constant, and I want to either increase or decrease $y_0$? Can I find a different $m$? Otherwise saying, given $x_0$, for what values of $y_0$ can I find a solution for the following equation: $$y_0=-mx_0+2sqrt m$$We can write this as a quadratic equation in $X=sqrt m$
$$x_0X^2-2X+y_0=0$$
To have a real solution, we need the discriminant to be positive, so $$1-x_0y_0ge 0$$ or $$x_0y_0le 1$$
What it means that any point that lies in the first quadrant below the hyperbola $xy=1$ is part of at least one line such that the intersections with the axis form a triangle with area $2$.
answered Jan 7 at 2:27
AndreiAndrei
13.2k21230
$endgroup$
add a comment |
$begingroup$
The equation of a line in the plane, with slope $-m$ can be written as $$y=-mx+b$$To get the condition to have the area $2$, we need to calculate the intersection with the axes. If $x=0$, you get $y=b$. If $y=0$, you get $$x=frac bm$$
The area of the triangle is then $$2=frac 12 bfrac bm$$which yields $$b=2sqrt m$$
Going back to the original equation $$y=-mx+2sqrt m$$
Now let's suppose that I have a point $(x_0,y_0)$ on such line. What would happen if I keep $x_0$ constant, and I want to either increase or decrease $y_0$? Can I find a different $m$? Otherwise saying, given $x_0$, for what values of $y_0$ can I find a solution for the following equation: $$y_0=-mx_0+2sqrt m$$We can write this as a quadratic equation in $X=sqrt m$
$$x_0X^2-2X+y_0=0$$
To have a real solution, we need the discriminant to be positive, so $$1-x_0y_0ge 0$$ or $$x_0y_0le 1$$
What it means that any point that lies in the first quadrant below the hyperbola $xy=1$ is part of at least one line such that the intersections with the axis form a triangle with area $2$.
answered Jan 7 at 2:27
AndreiAndrei
13.2k21230
$endgroup$
add a comment |
$begingroup$
The equation of a line in the plane, with slope $-m$ can be written as $$y=-mx+b$$To get the condition to have the area $2$, we need to calculate the intersection with the axes. If $x=0$, you get $y=b$. If $y=0$, you get $$x=frac bm$$
The area of the triangle is then $$2=frac 12 bfrac bm$$which yields $$b=2sqrt m$$
Going back to the original equation $$y=-mx+2sqrt m$$
Now let's suppose that I have a point $(x_0,y_0)$ on such line. What would happen if I keep $x_0$ constant, and I want to either increase or decrease $y_0$? Can I find a different $m$? Otherwise saying, given $x_0$, for what values of $y_0$ can I find a solution for the following equation: $$y_0=-mx_0+2sqrt m$$We can write this as a quadratic equation in $X=sqrt m$
$$x_0X^2-2X+y_0=0$$
To have a real solution, we need the discriminant to be positive, so $$1-x_0y_0ge 0$$ or $$x_0y_0le 1$$
What it means that any point that lies in the first quadrant below the hyperbola $xy=1$ is part of at least one line such that the intersections with the axis form a triangle with area $2$.
answered Jan 7 at 2:27
AndreiAndrei
13.2k21230
$endgroup$
The equation of a line in the plane, with slope $-m$ can be written as $$y=-mx+b$$To get the condition to have the area $2$, we need to calculate the intersection with the axes. If $x=0$, you get $y=b$. If $y=0$, you get $$x=frac bm$$
The area of the triangle is then $$2=frac 12 bfrac bm$$which yields $$b=2sqrt m$$
Going back to the original equation $$y=-mx+2sqrt m$$
Now let's suppose that I have a point $(x_0,y_0)$ on such line. What would happen if I keep $x_0$ constant, and I want to either increase or decrease $y_0$? Can I find a different $m$? Otherwise saying, given $x_0$, for what values of $y_0$ can I find a solution for the following equation: $$y_0=-mx_0+2sqrt m$$We can write this as a quadratic equation in $X=sqrt m$
$$x_0X^2-2X+y_0=0$$
To have a real solution, we need the discriminant to be positive, so $$1-x_0y_0ge 0$$ or $$x_0y_0le 1$$
What it means that any point that lies in the first quadrant below the hyperbola $xy=1$ is part of at least one line such that the intersections with the axis form a triangle with area $2$.
answered Jan 7 at 2:27
AndreiAndrei
13.2k21230
answered Jan 7 at 2:27
AndreiAndrei
13.2k21230
answered Jan 7 at 2:27
AndreiAndrei
13.2k21230
answered Jan 7 at 2:27
AndreiAndrei
13.2k21230
13.2k21230
add a comment |
add a comment |
$begingroup$
hi, what have you tried so far?
$endgroup$
– Siong Thye Goh
Jan 7 at 1:33
$begingroup$
Edited in a little
$endgroup$
– user491842
Jan 7 at 1:39
$begingroup$
any points (0,2x) and (x,0)
$endgroup$
– user29418
Jan 7 at 1:40
$begingroup$
A line going through $(1,1)$ and intersecting both the $x$ and $y$ axis in the first quadrant will have points on the right of $(1,1)$ (the point will just have lower $y$ coordinates)
$endgroup$
– Andrei
Jan 7 at 1:43