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Find the value at which lines intersect at a right angle.
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$begingroup$
I have a question I'm working on. I don't really understand it.
Find the value of $k$ for which the graphs of $2y + x + 3 = 0$ and $2y + kx + 2 = 0$ intersect at right angles.
I'm used to line equations with the formula $y = mx + b$. I don't really understand this. I also don't know exactly what is meant by "intersect at right angles." Can anyone provide some more insight to this? Thanks!
geometry
edited Mar 18 at 3:24
Robert Howard
2,2933935
asked Mar 18 at 3:16
scratchscratch
82
$endgroup$
add a comment |
$begingroup$
I have a question I'm working on. I don't really understand it.
Find the value of $k$ for which the graphs of $2y + x + 3 = 0$ and $2y + kx + 2 = 0$ intersect at right angles.
I'm used to line equations with the formula $y = mx + b$. I don't really understand this. I also don't know exactly what is meant by "intersect at right angles." Can anyone provide some more insight to this? Thanks!
geometry
edited Mar 18 at 3:24
Robert Howard
2,2933935
asked Mar 18 at 3:16
scratchscratch
82
$endgroup$
1
$begingroup$
If two lines intersect at a right angle the product of there slopes is -1 you can prove this using calculus or vectors anyone which is viable
$endgroup$
– Aditya Garg
Mar 18 at 3:23
$begingroup$
But what does intersecting at a right angle mean?
$endgroup$
– scratch
Mar 18 at 3:24
1
$begingroup$
They are perpendicular what else do you infer ?
$endgroup$
– Aditya Garg
Mar 18 at 3:24
$begingroup$
@AdityaGarg what ze needs to know is if the product of the slopes is -1 then the two lines intersect at right angles.
$endgroup$
– ErotemeObelus
Mar 18 at 3:39
$begingroup$
I can't understand what you want to say @ Tomislav Ostojich and can't understand what you want to know @ scratch
$endgroup$
– Aditya Garg
Mar 18 at 9:29
add a comment |
$begingroup$
I have a question I'm working on. I don't really understand it.
Find the value of $k$ for which the graphs of $2y + x + 3 = 0$ and $2y + kx + 2 = 0$ intersect at right angles.
I'm used to line equations with the formula $y = mx + b$. I don't really understand this. I also don't know exactly what is meant by "intersect at right angles." Can anyone provide some more insight to this? Thanks!
geometry
edited Mar 18 at 3:24
Robert Howard
2,2933935
asked Mar 18 at 3:16
scratchscratch
82
$endgroup$
I have a question I'm working on. I don't really understand it.
Find the value of $k$ for which the graphs of $2y + x + 3 = 0$ and $2y + kx + 2 = 0$ intersect at right angles.
I'm used to line equations with the formula $y = mx + b$. I don't really understand this. I also don't know exactly what is meant by "intersect at right angles." Can anyone provide some more insight to this? Thanks!
geometry
geometry
edited Mar 18 at 3:24
Robert Howard
2,2933935
asked Mar 18 at 3:16
scratchscratch
82
edited Mar 18 at 3:24
Robert Howard
2,2933935
asked Mar 18 at 3:16
scratchscratch
82
edited Mar 18 at 3:24
Robert Howard
2,2933935
edited Mar 18 at 3:24
Robert Howard
2,2933935
edited Mar 18 at 3:24
Robert Howard
2,2933935
2,2933935
asked Mar 18 at 3:16
scratchscratch
82
asked Mar 18 at 3:16
scratchscratch
82
asked Mar 18 at 3:16
scratchscratch
82
82
1
$begingroup$
If two lines intersect at a right angle the product of there slopes is -1 you can prove this using calculus or vectors anyone which is viable
$endgroup$
– Aditya Garg
Mar 18 at 3:23
$begingroup$
But what does intersecting at a right angle mean?
$endgroup$
– scratch
Mar 18 at 3:24
1
$begingroup$
They are perpendicular what else do you infer ?
$endgroup$
– Aditya Garg
Mar 18 at 3:24
$begingroup$
@AdityaGarg what ze needs to know is if the product of the slopes is -1 then the two lines intersect at right angles.
$endgroup$
– ErotemeObelus
Mar 18 at 3:39
$begingroup$
I can't understand what you want to say @ Tomislav Ostojich and can't understand what you want to know @ scratch
$endgroup$
– Aditya Garg
Mar 18 at 9:29
add a comment |
1
$begingroup$
If two lines intersect at a right angle the product of there slopes is -1 you can prove this using calculus or vectors anyone which is viable
$endgroup$
– Aditya Garg
Mar 18 at 3:23
$begingroup$
But what does intersecting at a right angle mean?
$endgroup$
– scratch
Mar 18 at 3:24
1
$begingroup$
They are perpendicular what else do you infer ?
$endgroup$
– Aditya Garg
Mar 18 at 3:24
$begingroup$
@AdityaGarg what ze needs to know is if the product of the slopes is -1 then the two lines intersect at right angles.
$endgroup$
– ErotemeObelus
Mar 18 at 3:39
$begingroup$
I can't understand what you want to say @ Tomislav Ostojich and can't understand what you want to know @ scratch
$endgroup$
– Aditya Garg
Mar 18 at 9:29
1
1
$begingroup$
If two lines intersect at a right angle the product of there slopes is -1 you can prove this using calculus or vectors anyone which is viable
$endgroup$
– Aditya Garg
Mar 18 at 3:23
$begingroup$
If two lines intersect at a right angle the product of there slopes is -1 you can prove this using calculus or vectors anyone which is viable
$endgroup$
– Aditya Garg
Mar 18 at 3:23
$begingroup$
But what does intersecting at a right angle mean?
$endgroup$
– scratch
Mar 18 at 3:24
$begingroup$
But what does intersecting at a right angle mean?
$endgroup$
– scratch
Mar 18 at 3:24
1
1
$begingroup$
They are perpendicular what else do you infer ?
$endgroup$
– Aditya Garg
Mar 18 at 3:24
$begingroup$
They are perpendicular what else do you infer ?
$endgroup$
– Aditya Garg
Mar 18 at 3:24
$begingroup$
@AdityaGarg what ze needs to know is if the product of the slopes is -1 then the two lines intersect at right angles.
$endgroup$
– ErotemeObelus
Mar 18 at 3:39
$begingroup$
@AdityaGarg what ze needs to know is if the product of the slopes is -1 then the two lines intersect at right angles.
$endgroup$
– ErotemeObelus
Mar 18 at 3:39
$begingroup$
I can't understand what you want to say @ Tomislav Ostojich and can't understand what you want to know @ scratch
$endgroup$
– Aditya Garg
Mar 18 at 9:29
$begingroup$
I can't understand what you want to say @ Tomislav Ostojich and can't understand what you want to know @ scratch
$endgroup$
– Aditya Garg
Mar 18 at 9:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If two lines intersect at right angles, they'll be perpendicular, and the slopes of two perpendicular lines are opposite reciprocals. That is, if a certain line has a slope of $m$, then a line that's perpendicular to that first line will have a slope of $-1/m$.
In order to find the value of $k$ such that those two lines are perpendicular, you'll first need to solve each equation for $y$ (that is, isolate $y$ on one side of each equation by moving all the other terms to the other side), and once you've done that, it shouldn't be too hard to figure out what value of $k$ will make the slope of one line the opposite reciprocal of the slope of the other line.
Here's an example of two perpendicular lines. In this case, the equation of the red line is $y=-2x+4$, and the equation of the blue line is $y=frac{1}{2}x+1$. Notice that $frac{-1}{text{either slope}}$ gives you the other slope.
answered Mar 18 at 3:23
Robert HowardRobert Howard
2,2933935
$endgroup$
add a comment |
$begingroup$
The condition for two lines to intersect at right angles is that the product of their gradients (slopes) is $-1$.
Put both equations in the standard form $y = mx+c$ where $m$ is the gradient.
$2y+x+3 = 0 implies y =-frac 12x - frac 32$
Gradient is $-frac 12$. Hence other line must have gradient of $frac{-1}{-frac 12}=2$.
$2y+kx+2=0 implies y=-frac k2 x - 1$
Gradient is $-frac k2$.
So $-frac k2 = 2 implies k=-4$.
Note that only the gradient is important here. The other term ($y$-intercept) is irrelevant.
answered Mar 18 at 3:46
DeepakDeepak
17.6k11539
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
If two lines intersect at right angles, they'll be perpendicular, and the slopes of two perpendicular lines are opposite reciprocals. That is, if a certain line has a slope of $m$, then a line that's perpendicular to that first line will have a slope of $-1/m$.
In order to find the value of $k$ such that those two lines are perpendicular, you'll first need to solve each equation for $y$ (that is, isolate $y$ on one side of each equation by moving all the other terms to the other side), and once you've done that, it shouldn't be too hard to figure out what value of $k$ will make the slope of one line the opposite reciprocal of the slope of the other line.
Here's an example of two perpendicular lines. In this case, the equation of the red line is $y=-2x+4$, and the equation of the blue line is $y=frac{1}{2}x+1$. Notice that $frac{-1}{text{either slope}}$ gives you the other slope.
answered Mar 18 at 3:23
Robert HowardRobert Howard
2,2933935
$endgroup$
add a comment |
$begingroup$
If two lines intersect at right angles, they'll be perpendicular, and the slopes of two perpendicular lines are opposite reciprocals. That is, if a certain line has a slope of $m$, then a line that's perpendicular to that first line will have a slope of $-1/m$.
In order to find the value of $k$ such that those two lines are perpendicular, you'll first need to solve each equation for $y$ (that is, isolate $y$ on one side of each equation by moving all the other terms to the other side), and once you've done that, it shouldn't be too hard to figure out what value of $k$ will make the slope of one line the opposite reciprocal of the slope of the other line.
Here's an example of two perpendicular lines. In this case, the equation of the red line is $y=-2x+4$, and the equation of the blue line is $y=frac{1}{2}x+1$. Notice that $frac{-1}{text{either slope}}$ gives you the other slope.
answered Mar 18 at 3:23
Robert HowardRobert Howard
2,2933935
$endgroup$
add a comment |
$begingroup$
If two lines intersect at right angles, they'll be perpendicular, and the slopes of two perpendicular lines are opposite reciprocals. That is, if a certain line has a slope of $m$, then a line that's perpendicular to that first line will have a slope of $-1/m$.
In order to find the value of $k$ such that those two lines are perpendicular, you'll first need to solve each equation for $y$ (that is, isolate $y$ on one side of each equation by moving all the other terms to the other side), and once you've done that, it shouldn't be too hard to figure out what value of $k$ will make the slope of one line the opposite reciprocal of the slope of the other line.
Here's an example of two perpendicular lines. In this case, the equation of the red line is $y=-2x+4$, and the equation of the blue line is $y=frac{1}{2}x+1$. Notice that $frac{-1}{text{either slope}}$ gives you the other slope.
answered Mar 18 at 3:23
Robert HowardRobert Howard
2,2933935
$endgroup$
If two lines intersect at right angles, they'll be perpendicular, and the slopes of two perpendicular lines are opposite reciprocals. That is, if a certain line has a slope of $m$, then a line that's perpendicular to that first line will have a slope of $-1/m$.
In order to find the value of $k$ such that those two lines are perpendicular, you'll first need to solve each equation for $y$ (that is, isolate $y$ on one side of each equation by moving all the other terms to the other side), and once you've done that, it shouldn't be too hard to figure out what value of $k$ will make the slope of one line the opposite reciprocal of the slope of the other line.
Here's an example of two perpendicular lines. In this case, the equation of the red line is $y=-2x+4$, and the equation of the blue line is $y=frac{1}{2}x+1$. Notice that $frac{-1}{text{either slope}}$ gives you the other slope.
answered Mar 18 at 3:23
Robert HowardRobert Howard
2,2933935
edited Mar 18 at 3:28
answered Mar 18 at 3:23
Robert HowardRobert Howard
2,2933935
answered Mar 18 at 3:23
Robert HowardRobert Howard
2,2933935
answered Mar 18 at 3:23
Robert HowardRobert Howard
2,2933935
2,2933935
add a comment |
add a comment |
$begingroup$
The condition for two lines to intersect at right angles is that the product of their gradients (slopes) is $-1$.
Put both equations in the standard form $y = mx+c$ where $m$ is the gradient.
$2y+x+3 = 0 implies y =-frac 12x - frac 32$
Gradient is $-frac 12$. Hence other line must have gradient of $frac{-1}{-frac 12}=2$.
$2y+kx+2=0 implies y=-frac k2 x - 1$
Gradient is $-frac k2$.
So $-frac k2 = 2 implies k=-4$.
Note that only the gradient is important here. The other term ($y$-intercept) is irrelevant.
answered Mar 18 at 3:46
DeepakDeepak
17.6k11539
$endgroup$
add a comment |
$begingroup$
The condition for two lines to intersect at right angles is that the product of their gradients (slopes) is $-1$.
Put both equations in the standard form $y = mx+c$ where $m$ is the gradient.
$2y+x+3 = 0 implies y =-frac 12x - frac 32$
Gradient is $-frac 12$. Hence other line must have gradient of $frac{-1}{-frac 12}=2$.
$2y+kx+2=0 implies y=-frac k2 x - 1$
Gradient is $-frac k2$.
So $-frac k2 = 2 implies k=-4$.
Note that only the gradient is important here. The other term ($y$-intercept) is irrelevant.
answered Mar 18 at 3:46
DeepakDeepak
17.6k11539
$endgroup$
add a comment |
$begingroup$
The condition for two lines to intersect at right angles is that the product of their gradients (slopes) is $-1$.
Put both equations in the standard form $y = mx+c$ where $m$ is the gradient.
$2y+x+3 = 0 implies y =-frac 12x - frac 32$
Gradient is $-frac 12$. Hence other line must have gradient of $frac{-1}{-frac 12}=2$.
$2y+kx+2=0 implies y=-frac k2 x - 1$
Gradient is $-frac k2$.
So $-frac k2 = 2 implies k=-4$.
Note that only the gradient is important here. The other term ($y$-intercept) is irrelevant.
answered Mar 18 at 3:46
DeepakDeepak
17.6k11539
$endgroup$
The condition for two lines to intersect at right angles is that the product of their gradients (slopes) is $-1$.
Put both equations in the standard form $y = mx+c$ where $m$ is the gradient.
$2y+x+3 = 0 implies y =-frac 12x - frac 32$
Gradient is $-frac 12$. Hence other line must have gradient of $frac{-1}{-frac 12}=2$.
$2y+kx+2=0 implies y=-frac k2 x - 1$
Gradient is $-frac k2$.
So $-frac k2 = 2 implies k=-4$.
Note that only the gradient is important here. The other term ($y$-intercept) is irrelevant.
answered Mar 18 at 3:46
DeepakDeepak
17.6k11539
answered Mar 18 at 3:46
DeepakDeepak
17.6k11539
answered Mar 18 at 3:46
DeepakDeepak
17.6k11539
answered Mar 18 at 3:46
DeepakDeepak
17.6k11539
17.6k11539
add a comment |
add a comment |
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1
$begingroup$
If two lines intersect at a right angle the product of there slopes is -1 you can prove this using calculus or vectors anyone which is viable
$endgroup$
– Aditya Garg
Mar 18 at 3:23
$begingroup$
But what does intersecting at a right angle mean?
$endgroup$
– scratch
Mar 18 at 3:24
1
$begingroup$
They are perpendicular what else do you infer ?
$endgroup$
– Aditya Garg
Mar 18 at 3:24
$begingroup$
@AdityaGarg what ze needs to know is if the product of the slopes is -1 then the two lines intersect at right angles.
$endgroup$
– ErotemeObelus
Mar 18 at 3:39
$begingroup$
I can't understand what you want to say @ Tomislav Ostojich and can't understand what you want to know @ scratch
$endgroup$
– Aditya Garg
Mar 18 at 9:29