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Grade 10 Analytic Geometry Question 23- Incredibly hard

Analytic Geometry QuestionCoordinate Geometry Oblique Coordinates ProblemAnalytic geometry simple questionHelp with this coordinate geometry question involving cirlces and parabolas.Faster Alternative than Calculating Euclidian Distance to determine which Coordinate has Max Distance from a fixed coordinate (eg (0,0))Get coordinate origin from two pointsFind the radius of the circle for given conditionsAnalytical geometry - Finding the coordinates of point MRay-Casting algorithm in Ray-triangle intersectionDefine regular polygon while iterating through a two-dimensional array.

3

1

$begingroup$

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

share|cite|improve this question

asked 2 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
  2 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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, this helped me so much!
  $endgroup$
  – Sinestro 38
  16 mins ago
  

  
  
  
  

add a comment | 

3

1

$begingroup$

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

share|cite|improve this question

asked 2 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
  2 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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, this helped me so much!
  $endgroup$
  – Sinestro 38
  16 mins 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

share|cite|improve this question

asked 2 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.

$endgroup$

share|cite|improve this question

asked 2 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.

share|cite|improve this question

asked 2 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 2 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 2 hours ago

Sinestro 38Sinestro 38

211

3 Answers
3

active

oldest

votes

2

$begingroup$

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.

share|cite|improve this answer

answered 2 hours ago

DeepakDeepak

17.1k11536

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  He said he was in grade 10, so he probably doesn’t know complex numbers yet.
  $endgroup$
  – Bor Kari
  2 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
  2 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
  2 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
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered 2 hours ago

Bor KariBor Kari

1108

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered 1 hour ago

G Tony JacobsG Tony Jacobs

25.9k43686

$endgroup$

add a comment | 

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2

$begingroup$

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.

share|cite|improve this answer

answered 2 hours ago

DeepakDeepak

17.1k11536

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  He said he was in grade 10, so he probably doesn’t know complex numbers yet.
  $endgroup$
  – Bor Kari
  2 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
  2 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
  2 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
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

answered 2 hours ago

DeepakDeepak

17.1k11536

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  He said he was in grade 10, so he probably doesn’t know complex numbers yet.
  $endgroup$
  – Bor Kari
  2 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
  2 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
  2 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
  1 hour 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.

share|cite|improve this answer

answered 2 hours ago

DeepakDeepak

17.1k11536

$endgroup$

share|cite|improve this answer

answered 2 hours ago

DeepakDeepak

17.1k11536

answered 2 hours ago

DeepakDeepak

17.1k11536

answered 2 hours ago

DeepakDeepak

17.1k11536

1

$begingroup$

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.

share|cite|improve this answer

answered 2 hours ago

Bor KariBor Kari

1108

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered 2 hours ago

Bor KariBor Kari

1108

$endgroup$

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.

share|cite|improve this answer

answered 2 hours ago

Bor KariBor Kari

1108

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Bor KariBor Kari

1108

answered 2 hours ago

Bor KariBor Kari

1108

answered 2 hours ago

Bor KariBor Kari

1108

1

$begingroup$

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.

share|cite|improve this answer

answered 1 hour ago

G Tony JacobsG Tony Jacobs

25.9k43686

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered 1 hour 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.

share|cite|improve this answer

answered 1 hour ago

G Tony JacobsG Tony Jacobs

25.9k43686

$endgroup$

share|cite|improve this answer

answered 1 hour ago

G Tony JacobsG Tony Jacobs

25.9k43686

answered 1 hour ago

G Tony JacobsG Tony Jacobs

25.9k43686

answered 1 hour ago

G Tony JacobsG Tony Jacobs

25.9k43686