Grade 10 Analytic Geometry Question 23- Incredibly hard
$begingroup$
So, I've spent hours on this question and it's frustrating me way too much so I created an account for StackExchange just to understand how you solve this problem.
Let me start by sharing the question: "Sally has hidden her brother's birthday present somewhere in the backyard. When writing instructions for finding the present, she used a coordinate system with each unit on the grid representing 1 m. The positive y-axis of this grid points north. The instructions read "Start at the origin, walk halfway to (8,6), turn 90 degrees left, and then walk twice as far." Where is the present?" By the way, you have to answer the question algebraically and the answer in the textbook is (-2,11)
What I Know
Let P represent the coordinate of the present
Let M represent the midpoint between (0,0) and (8,6)
- M is at (4,3) which is solved using the midpoint formula.
- The distance from P to M is 10 m.
- The equation for the line from the origin to the midpoint is:
y= $frac{3x}{4}$
- The equation for line PM is:
y= $frac{-4x}{3}$ + $frac{25}{3}$ - Using the distance formula for two coordinates I know that the distance from M to P can be represented in this equation: $100=(4-x)^2 + (3-y)^2$ The x and y in the equation represents the x,y coordinates of the point P. I don't know how to simplify this equation further.
My troubles
I come to this point where I don't know what to do anymore. Please explain how you solved this problem.
Thank you so much
analytic-geometry
asked 3 hours ago
Sinestro 38Sinestro 38
211
New contributor
Sinestro 38 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
So, I've spent hours on this question and it's frustrating me way too much so I created an account for StackExchange just to understand how you solve this problem.
Let me start by sharing the question: "Sally has hidden her brother's birthday present somewhere in the backyard. When writing instructions for finding the present, she used a coordinate system with each unit on the grid representing 1 m. The positive y-axis of this grid points north. The instructions read "Start at the origin, walk halfway to (8,6), turn 90 degrees left, and then walk twice as far." Where is the present?" By the way, you have to answer the question algebraically and the answer in the textbook is (-2,11)
What I Know
Let P represent the coordinate of the present
Let M represent the midpoint between (0,0) and (8,6)
- M is at (4,3) which is solved using the midpoint formula.
- The distance from P to M is 10 m.
- The equation for the line from the origin to the midpoint is:
y= $frac{3x}{4}$
- The equation for line PM is:
y= $frac{-4x}{3}$ + $frac{25}{3}$ - Using the distance formula for two coordinates I know that the distance from M to P can be represented in this equation: $100=(4-x)^2 + (3-y)^2$ The x and y in the equation represents the x,y coordinates of the point P. I don't know how to simplify this equation further.
My troubles
I come to this point where I don't know what to do anymore. Please explain how you solved this problem.
Thank you so much
analytic-geometry
asked 3 hours ago
Sinestro 38Sinestro 38
211
New contributor
Sinestro 38 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
To solve this using brute force, just substitute $y = -frac{4}{3}x + frac{25}{3}$ into $100 = (4-x)^2 + (3-y)^4$ to get a quadratic equation in $x$. Solve that using factorization (or the quadratic formula if you're not comfortable with factoring) and you'll end up with two points. These correspond to turning left and turning right. Pick the one with the smaller $x$ coordinate to indicate that you turned left. You can also draw this out with the origin moved to the point (4,3). You get a right triangle with horizontal side $3a$, vertical side $4a$ and hypotenuse $10$. Solve for $a$.
$endgroup$
– forgottenarrow
3 hours ago
$begingroup$
The above comment gives a great "algebraic" answer to your problem. You will find two points (two points that come from the intersection of a circle and a line). If the x coordinate gets larger, you are moving to the right. If it gets smaller, you are moving to the left. As for the triangle solution... you could just also draw a 2D Plot and use the information given, but I don't think that was the point of the exercise
$endgroup$
– Jonathan Perales
3 hours ago
$begingroup$
Thanks, this helped me so much!
$endgroup$
– Sinestro 38
1 hour ago
add a comment |
$begingroup$
So, I've spent hours on this question and it's frustrating me way too much so I created an account for StackExchange just to understand how you solve this problem.
Let me start by sharing the question: "Sally has hidden her brother's birthday present somewhere in the backyard. When writing instructions for finding the present, she used a coordinate system with each unit on the grid representing 1 m. The positive y-axis of this grid points north. The instructions read "Start at the origin, walk halfway to (8,6), turn 90 degrees left, and then walk twice as far." Where is the present?" By the way, you have to answer the question algebraically and the answer in the textbook is (-2,11)
What I Know
Let P represent the coordinate of the present
Let M represent the midpoint between (0,0) and (8,6)
- M is at (4,3) which is solved using the midpoint formula.
- The distance from P to M is 10 m.
- The equation for the line from the origin to the midpoint is:
y= $frac{3x}{4}$
- The equation for line PM is:
y= $frac{-4x}{3}$ + $frac{25}{3}$ - Using the distance formula for two coordinates I know that the distance from M to P can be represented in this equation: $100=(4-x)^2 + (3-y)^2$ The x and y in the equation represents the x,y coordinates of the point P. I don't know how to simplify this equation further.
My troubles
I come to this point where I don't know what to do anymore. Please explain how you solved this problem.
Thank you so much
analytic-geometry
asked 3 hours ago
Sinestro 38Sinestro 38
211
New contributor
Sinestro 38 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
So, I've spent hours on this question and it's frustrating me way too much so I created an account for StackExchange just to understand how you solve this problem.
Let me start by sharing the question: "Sally has hidden her brother's birthday present somewhere in the backyard. When writing instructions for finding the present, she used a coordinate system with each unit on the grid representing 1 m. The positive y-axis of this grid points north. The instructions read "Start at the origin, walk halfway to (8,6), turn 90 degrees left, and then walk twice as far." Where is the present?" By the way, you have to answer the question algebraically and the answer in the textbook is (-2,11)
What I Know
Let P represent the coordinate of the present
Let M represent the midpoint between (0,0) and (8,6)
- M is at (4,3) which is solved using the midpoint formula.
- The distance from P to M is 10 m.
- The equation for the line from the origin to the midpoint is:
y= $frac{3x}{4}$
- The equation for line PM is:
y= $frac{-4x}{3}$ + $frac{25}{3}$ - Using the distance formula for two coordinates I know that the distance from M to P can be represented in this equation: $100=(4-x)^2 + (3-y)^2$ The x and y in the equation represents the x,y coordinates of the point P. I don't know how to simplify this equation further.
My troubles
I come to this point where I don't know what to do anymore. Please explain how you solved this problem.
Thank you so much
analytic-geometry
analytic-geometry
asked 3 hours ago
Sinestro 38Sinestro 38
211
New contributor
Sinestro 38 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Sinestro 38Sinestro 38
211
New contributor
Sinestro 38 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Sinestro 38Sinestro 38
211
New contributor
Sinestro 38 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Sinestro 38Sinestro 38
211
asked 3 hours ago
Sinestro 38Sinestro 38
211
211
New contributor
Sinestro 38 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sinestro 38 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sinestro 38 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
To solve this using brute force, just substitute $y = -frac{4}{3}x + frac{25}{3}$ into $100 = (4-x)^2 + (3-y)^4$ to get a quadratic equation in $x$. Solve that using factorization (or the quadratic formula if you're not comfortable with factoring) and you'll end up with two points. These correspond to turning left and turning right. Pick the one with the smaller $x$ coordinate to indicate that you turned left. You can also draw this out with the origin moved to the point (4,3). You get a right triangle with horizontal side $3a$, vertical side $4a$ and hypotenuse $10$. Solve for $a$.
$endgroup$
– forgottenarrow
3 hours ago
$begingroup$
The above comment gives a great "algebraic" answer to your problem. You will find two points (two points that come from the intersection of a circle and a line). If the x coordinate gets larger, you are moving to the right. If it gets smaller, you are moving to the left. As for the triangle solution... you could just also draw a 2D Plot and use the information given, but I don't think that was the point of the exercise
$endgroup$
– Jonathan Perales
3 hours ago
$begingroup$
Thanks, this helped me so much!
$endgroup$
– Sinestro 38
1 hour ago
add a comment |
1
$begingroup$
To solve this using brute force, just substitute $y = -frac{4}{3}x + frac{25}{3}$ into $100 = (4-x)^2 + (3-y)^4$ to get a quadratic equation in $x$. Solve that using factorization (or the quadratic formula if you're not comfortable with factoring) and you'll end up with two points. These correspond to turning left and turning right. Pick the one with the smaller $x$ coordinate to indicate that you turned left. You can also draw this out with the origin moved to the point (4,3). You get a right triangle with horizontal side $3a$, vertical side $4a$ and hypotenuse $10$. Solve for $a$.
$endgroup$
– forgottenarrow
3 hours ago
$begingroup$
The above comment gives a great "algebraic" answer to your problem. You will find two points (two points that come from the intersection of a circle and a line). If the x coordinate gets larger, you are moving to the right. If it gets smaller, you are moving to the left. As for the triangle solution... you could just also draw a 2D Plot and use the information given, but I don't think that was the point of the exercise
$endgroup$
– Jonathan Perales
3 hours ago
$begingroup$
Thanks, this helped me so much!
$endgroup$
– Sinestro 38
1 hour ago
1
1
$begingroup$
To solve this using brute force, just substitute $y = -frac{4}{3}x + frac{25}{3}$ into $100 = (4-x)^2 + (3-y)^4$ to get a quadratic equation in $x$. Solve that using factorization (or the quadratic formula if you're not comfortable with factoring) and you'll end up with two points. These correspond to turning left and turning right. Pick the one with the smaller $x$ coordinate to indicate that you turned left. You can also draw this out with the origin moved to the point (4,3). You get a right triangle with horizontal side $3a$, vertical side $4a$ and hypotenuse $10$. Solve for $a$.
$endgroup$
– forgottenarrow
3 hours ago
$begingroup$
To solve this using brute force, just substitute $y = -frac{4}{3}x + frac{25}{3}$ into $100 = (4-x)^2 + (3-y)^4$ to get a quadratic equation in $x$. Solve that using factorization (or the quadratic formula if you're not comfortable with factoring) and you'll end up with two points. These correspond to turning left and turning right. Pick the one with the smaller $x$ coordinate to indicate that you turned left. You can also draw this out with the origin moved to the point (4,3). You get a right triangle with horizontal side $3a$, vertical side $4a$ and hypotenuse $10$. Solve for $a$.
$endgroup$
– forgottenarrow
3 hours ago
$begingroup$
The above comment gives a great "algebraic" answer to your problem. You will find two points (two points that come from the intersection of a circle and a line). If the x coordinate gets larger, you are moving to the right. If it gets smaller, you are moving to the left. As for the triangle solution... you could just also draw a 2D Plot and use the information given, but I don't think that was the point of the exercise
$endgroup$
– Jonathan Perales
3 hours ago
$begingroup$
The above comment gives a great "algebraic" answer to your problem. You will find two points (two points that come from the intersection of a circle and a line). If the x coordinate gets larger, you are moving to the right. If it gets smaller, you are moving to the left. As for the triangle solution... you could just also draw a 2D Plot and use the information given, but I don't think that was the point of the exercise
$endgroup$
– Jonathan Perales
3 hours ago
$begingroup$
Thanks, this helped me so much!
$endgroup$
– Sinestro 38
1 hour ago
$begingroup$
Thanks, this helped me so much!
$endgroup$
– Sinestro 38
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If you know complex numbers, the problem becomes almost trivial to solve. Complex numbers lend themselves very naturally to mapping to vectors and points on the 2-d plane.
So you start at the origin (point $0+0i$) and walk halfway to $8+6i$, which means you end up at $4+3i$. At this point you turn ninety degrees to your left (which is counter-clockwise), and the vector for the new direction can be found by multiplying the previous one by $i$. And since you're walking twice as far, this is equivalent to again multiplying the result by two. So you're now taking the path $2i(4+3i) = -6 + 8i$. The final position is simply the sum of $4+3i$ and $-6+8i$, i.e. $4+3i -6+8i = -2 + 11i$. This is the point $(-2,11)$ as required.
answered 3 hours ago
DeepakDeepak
17.1k11536
$endgroup$
2
$begingroup$
He said he was in grade 10, so he probably doesn’t know complex numbers yet.
$endgroup$
– Bor Kari
3 hours ago
$begingroup$
@BorKari You have a point there. I'm not familiar with the US school system (I'm from Singapore, and have taken the UK-based GCE exams) - but even here, I think complex numbers are only "officially" covered starting from age 17 or so. Still, I had self-taught myself math from a young age. I think complex numbers are not that hard to learn, and they are so very useful in many fields. I am hoping to encourage the asker (and others) to pick up a new technique they may be unfamiliar with, or perhaps not have considered. :)
$endgroup$
– Deepak
3 hours ago
$begingroup$
Could be done in a similar way but instead of using complex numbers, using 2D Coordinates...If the vector is 4i+3j, the perpendicular vector would be -3i+4j [i is vector notation, not imaginary, perpendicular turning to the left], That new vector has a lenght of 5, so if you need it to double the lenght, you just multiply it by 2 and work it the same way as in complex numbers, you'll end up with the same answer.
$endgroup$
– Jonathan Perales
3 hours ago
$begingroup$
@JonathanPerales Yes, I realise that. I wanted to post the complex numbers method because a) it's my first line as it somehow feels more natural to me (it's basically just arithmetic) and b) a lot of people are unfamiliar with the versatility of the method. But thanks for the comment, that provides an alternative formulation that others may wish to use.
$endgroup$
– Deepak
3 hours ago
add a comment |
$begingroup$
Go to (4,3), and then walk in the direction of slope -4/3, i.e., left 3 and up 4, twice as far, i.e., left 6 and up 8. So, 4-6=-2, and 3+8=11. There’s a 3,4,5 right triangle and a 6,8,10 right triangle involved.
Solving quadratic equations is serious overkill.
answered 2 hours ago
G Tony JacobsG Tony Jacobs
25.9k43686
$endgroup$
add a comment |
$begingroup$
Hint:
You can make a second equation by considering the equation of the line because you know that the point has to be on that line.
answered 3 hours ago
Bor KariBor Kari
1088
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you know complex numbers, the problem becomes almost trivial to solve. Complex numbers lend themselves very naturally to mapping to vectors and points on the 2-d plane.
So you start at the origin (point $0+0i$) and walk halfway to $8+6i$, which means you end up at $4+3i$. At this point you turn ninety degrees to your left (which is counter-clockwise), and the vector for the new direction can be found by multiplying the previous one by $i$. And since you're walking twice as far, this is equivalent to again multiplying the result by two. So you're now taking the path $2i(4+3i) = -6 + 8i$. The final position is simply the sum of $4+3i$ and $-6+8i$, i.e. $4+3i -6+8i = -2 + 11i$. This is the point $(-2,11)$ as required.
answered 3 hours ago
DeepakDeepak
17.1k11536
$endgroup$
2
$begingroup$
He said he was in grade 10, so he probably doesn’t know complex numbers yet.
$endgroup$
– Bor Kari
3 hours ago
$begingroup$
@BorKari You have a point there. I'm not familiar with the US school system (I'm from Singapore, and have taken the UK-based GCE exams) - but even here, I think complex numbers are only "officially" covered starting from age 17 or so. Still, I had self-taught myself math from a young age. I think complex numbers are not that hard to learn, and they are so very useful in many fields. I am hoping to encourage the asker (and others) to pick up a new technique they may be unfamiliar with, or perhaps not have considered. :)
$endgroup$
– Deepak
3 hours ago
$begingroup$
Could be done in a similar way but instead of using complex numbers, using 2D Coordinates...If the vector is 4i+3j, the perpendicular vector would be -3i+4j [i is vector notation, not imaginary, perpendicular turning to the left], That new vector has a lenght of 5, so if you need it to double the lenght, you just multiply it by 2 and work it the same way as in complex numbers, you'll end up with the same answer.
$endgroup$
– Jonathan Perales
3 hours ago
$begingroup$
@JonathanPerales Yes, I realise that. I wanted to post the complex numbers method because a) it's my first line as it somehow feels more natural to me (it's basically just arithmetic) and b) a lot of people are unfamiliar with the versatility of the method. But thanks for the comment, that provides an alternative formulation that others may wish to use.
$endgroup$
– Deepak
3 hours ago
add a comment |
$begingroup$
If you know complex numbers, the problem becomes almost trivial to solve. Complex numbers lend themselves very naturally to mapping to vectors and points on the 2-d plane.
So you start at the origin (point $0+0i$) and walk halfway to $8+6i$, which means you end up at $4+3i$. At this point you turn ninety degrees to your left (which is counter-clockwise), and the vector for the new direction can be found by multiplying the previous one by $i$. And since you're walking twice as far, this is equivalent to again multiplying the result by two. So you're now taking the path $2i(4+3i) = -6 + 8i$. The final position is simply the sum of $4+3i$ and $-6+8i$, i.e. $4+3i -6+8i = -2 + 11i$. This is the point $(-2,11)$ as required.
answered 3 hours ago
DeepakDeepak
17.1k11536
$endgroup$
2
$begingroup$
He said he was in grade 10, so he probably doesn’t know complex numbers yet.
$endgroup$
– Bor Kari
3 hours ago
$begingroup$
@BorKari You have a point there. I'm not familiar with the US school system (I'm from Singapore, and have taken the UK-based GCE exams) - but even here, I think complex numbers are only "officially" covered starting from age 17 or so. Still, I had self-taught myself math from a young age. I think complex numbers are not that hard to learn, and they are so very useful in many fields. I am hoping to encourage the asker (and others) to pick up a new technique they may be unfamiliar with, or perhaps not have considered. :)
$endgroup$
– Deepak
3 hours ago
$begingroup$
Could be done in a similar way but instead of using complex numbers, using 2D Coordinates...If the vector is 4i+3j, the perpendicular vector would be -3i+4j [i is vector notation, not imaginary, perpendicular turning to the left], That new vector has a lenght of 5, so if you need it to double the lenght, you just multiply it by 2 and work it the same way as in complex numbers, you'll end up with the same answer.
$endgroup$
– Jonathan Perales
3 hours ago
$begingroup$
@JonathanPerales Yes, I realise that. I wanted to post the complex numbers method because a) it's my first line as it somehow feels more natural to me (it's basically just arithmetic) and b) a lot of people are unfamiliar with the versatility of the method. But thanks for the comment, that provides an alternative formulation that others may wish to use.
$endgroup$
– Deepak
3 hours ago
add a comment |
$begingroup$
If you know complex numbers, the problem becomes almost trivial to solve. Complex numbers lend themselves very naturally to mapping to vectors and points on the 2-d plane.
So you start at the origin (point $0+0i$) and walk halfway to $8+6i$, which means you end up at $4+3i$. At this point you turn ninety degrees to your left (which is counter-clockwise), and the vector for the new direction can be found by multiplying the previous one by $i$. And since you're walking twice as far, this is equivalent to again multiplying the result by two. So you're now taking the path $2i(4+3i) = -6 + 8i$. The final position is simply the sum of $4+3i$ and $-6+8i$, i.e. $4+3i -6+8i = -2 + 11i$. This is the point $(-2,11)$ as required.
answered 3 hours ago
DeepakDeepak
17.1k11536
$endgroup$
If you know complex numbers, the problem becomes almost trivial to solve. Complex numbers lend themselves very naturally to mapping to vectors and points on the 2-d plane.
So you start at the origin (point $0+0i$) and walk halfway to $8+6i$, which means you end up at $4+3i$. At this point you turn ninety degrees to your left (which is counter-clockwise), and the vector for the new direction can be found by multiplying the previous one by $i$. And since you're walking twice as far, this is equivalent to again multiplying the result by two. So you're now taking the path $2i(4+3i) = -6 + 8i$. The final position is simply the sum of $4+3i$ and $-6+8i$, i.e. $4+3i -6+8i = -2 + 11i$. This is the point $(-2,11)$ as required.
answered 3 hours ago
DeepakDeepak
17.1k11536
answered 3 hours ago
DeepakDeepak
17.1k11536
answered 3 hours ago
DeepakDeepak
17.1k11536
answered 3 hours ago
DeepakDeepak
17.1k11536
17.1k11536
2
$begingroup$
He said he was in grade 10, so he probably doesn’t know complex numbers yet.
$endgroup$
– Bor Kari
3 hours ago
$begingroup$
@BorKari You have a point there. I'm not familiar with the US school system (I'm from Singapore, and have taken the UK-based GCE exams) - but even here, I think complex numbers are only "officially" covered starting from age 17 or so. Still, I had self-taught myself math from a young age. I think complex numbers are not that hard to learn, and they are so very useful in many fields. I am hoping to encourage the asker (and others) to pick up a new technique they may be unfamiliar with, or perhaps not have considered. :)
$endgroup$
– Deepak
3 hours ago
$begingroup$
Could be done in a similar way but instead of using complex numbers, using 2D Coordinates...If the vector is 4i+3j, the perpendicular vector would be -3i+4j [i is vector notation, not imaginary, perpendicular turning to the left], That new vector has a lenght of 5, so if you need it to double the lenght, you just multiply it by 2 and work it the same way as in complex numbers, you'll end up with the same answer.
$endgroup$
– Jonathan Perales
3 hours ago
$begingroup$
@JonathanPerales Yes, I realise that. I wanted to post the complex numbers method because a) it's my first line as it somehow feels more natural to me (it's basically just arithmetic) and b) a lot of people are unfamiliar with the versatility of the method. But thanks for the comment, that provides an alternative formulation that others may wish to use.
$endgroup$
– Deepak
3 hours ago
add a comment |
2
$begingroup$
He said he was in grade 10, so he probably doesn’t know complex numbers yet.
$endgroup$
– Bor Kari
3 hours ago
$begingroup$
@BorKari You have a point there. I'm not familiar with the US school system (I'm from Singapore, and have taken the UK-based GCE exams) - but even here, I think complex numbers are only "officially" covered starting from age 17 or so. Still, I had self-taught myself math from a young age. I think complex numbers are not that hard to learn, and they are so very useful in many fields. I am hoping to encourage the asker (and others) to pick up a new technique they may be unfamiliar with, or perhaps not have considered. :)
$endgroup$
– Deepak
3 hours ago
$begingroup$
Could be done in a similar way but instead of using complex numbers, using 2D Coordinates...If the vector is 4i+3j, the perpendicular vector would be -3i+4j [i is vector notation, not imaginary, perpendicular turning to the left], That new vector has a lenght of 5, so if you need it to double the lenght, you just multiply it by 2 and work it the same way as in complex numbers, you'll end up with the same answer.
$endgroup$
– Jonathan Perales
3 hours ago
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@JonathanPerales Yes, I realise that. I wanted to post the complex numbers method because a) it's my first line as it somehow feels more natural to me (it's basically just arithmetic) and b) a lot of people are unfamiliar with the versatility of the method. But thanks for the comment, that provides an alternative formulation that others may wish to use.
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– Deepak
3 hours ago
2
2
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He said he was in grade 10, so he probably doesn’t know complex numbers yet.
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– Bor Kari
3 hours ago
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He said he was in grade 10, so he probably doesn’t know complex numbers yet.
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– Bor Kari
3 hours ago
$begingroup$
@BorKari You have a point there. I'm not familiar with the US school system (I'm from Singapore, and have taken the UK-based GCE exams) - but even here, I think complex numbers are only "officially" covered starting from age 17 or so. Still, I had self-taught myself math from a young age. I think complex numbers are not that hard to learn, and they are so very useful in many fields. I am hoping to encourage the asker (and others) to pick up a new technique they may be unfamiliar with, or perhaps not have considered. :)
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– Deepak
3 hours ago
$begingroup$
@BorKari You have a point there. I'm not familiar with the US school system (I'm from Singapore, and have taken the UK-based GCE exams) - but even here, I think complex numbers are only "officially" covered starting from age 17 or so. Still, I had self-taught myself math from a young age. I think complex numbers are not that hard to learn, and they are so very useful in many fields. I am hoping to encourage the asker (and others) to pick up a new technique they may be unfamiliar with, or perhaps not have considered. :)
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– Deepak
3 hours ago
$begingroup$
Could be done in a similar way but instead of using complex numbers, using 2D Coordinates...If the vector is 4i+3j, the perpendicular vector would be -3i+4j [i is vector notation, not imaginary, perpendicular turning to the left], That new vector has a lenght of 5, so if you need it to double the lenght, you just multiply it by 2 and work it the same way as in complex numbers, you'll end up with the same answer.
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– Jonathan Perales
3 hours ago
$begingroup$
Could be done in a similar way but instead of using complex numbers, using 2D Coordinates...If the vector is 4i+3j, the perpendicular vector would be -3i+4j [i is vector notation, not imaginary, perpendicular turning to the left], That new vector has a lenght of 5, so if you need it to double the lenght, you just multiply it by 2 and work it the same way as in complex numbers, you'll end up with the same answer.
$endgroup$
– Jonathan Perales
3 hours ago
$begingroup$
@JonathanPerales Yes, I realise that. I wanted to post the complex numbers method because a) it's my first line as it somehow feels more natural to me (it's basically just arithmetic) and b) a lot of people are unfamiliar with the versatility of the method. But thanks for the comment, that provides an alternative formulation that others may wish to use.
$endgroup$
– Deepak
3 hours ago
$begingroup$
@JonathanPerales Yes, I realise that. I wanted to post the complex numbers method because a) it's my first line as it somehow feels more natural to me (it's basically just arithmetic) and b) a lot of people are unfamiliar with the versatility of the method. But thanks for the comment, that provides an alternative formulation that others may wish to use.
$endgroup$
– Deepak
3 hours ago
add a comment |
$begingroup$
Go to (4,3), and then walk in the direction of slope -4/3, i.e., left 3 and up 4, twice as far, i.e., left 6 and up 8. So, 4-6=-2, and 3+8=11. There’s a 3,4,5 right triangle and a 6,8,10 right triangle involved.
Solving quadratic equations is serious overkill.
answered 2 hours ago
G Tony JacobsG Tony Jacobs
25.9k43686
$endgroup$
add a comment |
$begingroup$
Go to (4,3), and then walk in the direction of slope -4/3, i.e., left 3 and up 4, twice as far, i.e., left 6 and up 8. So, 4-6=-2, and 3+8=11. There’s a 3,4,5 right triangle and a 6,8,10 right triangle involved.
Solving quadratic equations is serious overkill.
answered 2 hours ago
G Tony JacobsG Tony Jacobs
25.9k43686
$endgroup$
add a comment |
$begingroup$
Go to (4,3), and then walk in the direction of slope -4/3, i.e., left 3 and up 4, twice as far, i.e., left 6 and up 8. So, 4-6=-2, and 3+8=11. There’s a 3,4,5 right triangle and a 6,8,10 right triangle involved.
Solving quadratic equations is serious overkill.
answered 2 hours ago
G Tony JacobsG Tony Jacobs
25.9k43686
$endgroup$
Go to (4,3), and then walk in the direction of slope -4/3, i.e., left 3 and up 4, twice as far, i.e., left 6 and up 8. So, 4-6=-2, and 3+8=11. There’s a 3,4,5 right triangle and a 6,8,10 right triangle involved.
Solving quadratic equations is serious overkill.
answered 2 hours ago
G Tony JacobsG Tony Jacobs
25.9k43686
answered 2 hours ago
G Tony JacobsG Tony Jacobs
25.9k43686
answered 2 hours ago
G Tony JacobsG Tony Jacobs
25.9k43686
answered 2 hours ago
G Tony JacobsG Tony Jacobs
25.9k43686
25.9k43686
add a comment |
add a comment |
$begingroup$
Hint:
You can make a second equation by considering the equation of the line because you know that the point has to be on that line.
answered 3 hours ago
Bor KariBor Kari
1088
$endgroup$
add a comment |
$begingroup$
Hint:
You can make a second equation by considering the equation of the line because you know that the point has to be on that line.
answered 3 hours ago
Bor KariBor Kari
1088
$endgroup$
add a comment |
$begingroup$
Hint:
You can make a second equation by considering the equation of the line because you know that the point has to be on that line.
answered 3 hours ago
Bor KariBor Kari
1088
$endgroup$
Hint:
You can make a second equation by considering the equation of the line because you know that the point has to be on that line.
answered 3 hours ago
Bor KariBor Kari
1088
answered 3 hours ago
Bor KariBor Kari
1088
answered 3 hours ago
Bor KariBor Kari
1088
answered 3 hours ago
Bor KariBor Kari
1088
1088
add a comment |
add a comment |
Sinestro 38 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
To solve this using brute force, just substitute $y = -frac{4}{3}x + frac{25}{3}$ into $100 = (4-x)^2 + (3-y)^4$ to get a quadratic equation in $x$. Solve that using factorization (or the quadratic formula if you're not comfortable with factoring) and you'll end up with two points. These correspond to turning left and turning right. Pick the one with the smaller $x$ coordinate to indicate that you turned left. You can also draw this out with the origin moved to the point (4,3). You get a right triangle with horizontal side $3a$, vertical side $4a$ and hypotenuse $10$. Solve for $a$.
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– forgottenarrow
3 hours ago
$begingroup$
The above comment gives a great "algebraic" answer to your problem. You will find two points (two points that come from the intersection of a circle and a line). If the x coordinate gets larger, you are moving to the right. If it gets smaller, you are moving to the left. As for the triangle solution... you could just also draw a 2D Plot and use the information given, but I don't think that was the point of the exercise
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– Jonathan Perales
3 hours ago
$begingroup$
Thanks, this helped me so much!
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– Sinestro 38
1 hour ago