Grade 10 Analytic Geometry Question 23- Incredibly hard












3












$begingroup$


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







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
    51 mins ago
















3












$begingroup$


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







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
    51 mins ago














3












3








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







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






share|cite|improve this question







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







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




share|cite|improve this question






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




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
    51 mins ago














  • 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
    51 mins 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
51 mins ago




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










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









$endgroup$









  • 1




    $begingroup$
    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
    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
    2 hours ago





















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.enter image description here






share|cite|improve this answer









$endgroup$





















    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









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });






      Sinestro 38 is a new contributor. Be nice, and check out our Code of Conduct.










      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3131979%2fgrade-10-analytic-geometry-question-23-incredibly-hard%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      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









      $endgroup$









      • 1




        $begingroup$
        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
        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
        2 hours ago


















      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









      $endgroup$









      • 1




        $begingroup$
        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
        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
        2 hours ago
















      2












      2








      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









      $endgroup$



      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.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 3 hours ago









      DeepakDeepak

      17.1k11536




      17.1k11536








      • 1




        $begingroup$
        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
        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
        2 hours ago
















      • 1




        $begingroup$
        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
        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
        2 hours ago










      1




      1




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




      $begingroup$
      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
      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




      $begingroup$
      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
      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
      2 hours ago













      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.enter image description here






      share|cite|improve this answer









      $endgroup$


















        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.enter image description here






        share|cite|improve this answer









        $endgroup$
















          1












          1








          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.enter image description here






          share|cite|improve this answer









          $endgroup$



          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.enter image description here







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          Bor KariBor Kari

          1108




          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









              $endgroup$


















                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









                $endgroup$
















                  1












                  1








                  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









                  $endgroup$



                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  G Tony JacobsG Tony Jacobs

                  25.9k43686




                  25.9k43686






















                      Sinestro 38 is a new contributor. Be nice, and check out our Code of Conduct.










                      draft saved

                      draft discarded


















                      Sinestro 38 is a new contributor. Be nice, and check out our Code of Conduct.













                      Sinestro 38 is a new contributor. Be nice, and check out our Code of Conduct.












                      Sinestro 38 is a new contributor. Be nice, and check out our Code of Conduct.
















                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3131979%2fgrade-10-analytic-geometry-question-23-incredibly-hard%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Fluorita

                      Image
                      Fluorita Fluorita (blau) amb pirita (daurat). Fórmula química CaF 2 Epònim fluor Localitat tipus Jáchymov Classificació Categoria Halurs Nickel-Strunz 10a ed. 3.AB.25 Nickel-Strunz 9a ed. 3.AB.25 Nickel-Strunz 8a ed. III/A.08 Dana 9.2.1.1 Heys 8.4.7 Propietats Sistema cristal·lí Cúbic Hàbit cristal·lí Sol presentar cubs; menys freqüentment dodecaedres. De vegades hexaoctaedres i tetrahexaedres. Per la combinació d'aquestes formes, les arestes dels cubs són sovint modificades. De vegades els cristalls poden presentar diferències de creixement entre les cares. Sovint s'observen cristalls compostos per sobrecreixement. Pot ser massiva, compacta, terrosa, columnar (poc freqüent), globular en agregats o botroïdal. Estructura cristal·lina a = 5.4626Å Simetria Classe (H-M): m 3 m (4/ m 3 2/ m ) - Hexoctaèdric; Grup espacial: Fm 3 m Color Blanc, groc, verd, violeta, vermell, rosa, blau o negre Mac...
                      Read more

                      Hulsita

                      Image
                      Hulsita Hulsita de la localitat tipus Fórmula química Fe 2+ 2 Fe 3+ O 2 BO 3 Epònim Alfred Hulse Brooks Localitat tipus mont Brooks, península de Seward, Nome Borough, Alaska, Estats Units Classificació Categoria borats Nickel-Strunz 10a ed. 6.AB.45 Nickel-Strunz 9a ed. 6.AB.45 Nickel-Strunz 8a ed. V/G.03 Dana 24.2.3.1 Heys 9.6.4 Propietats Sistema cristal·lí monoclínic Estructura cristal·lina a = 10,68Å; b = 3,09Å; c = 5,43Å; β = 94,15° Color negre Macles en [001]; un membre de la macla es troba girat 120° en relació a l'altre Exfoliació bona en {110} Duresa 3 Lluïssor vítria, submetàl·lica Densitat 4,28 g/cm 3 (mesurada); Més informació Estatus IMA mineral heretat (G) i mineral heretat (G) Any d'aprovació 1908 Referències [1] La hulsita és un mineral de la classe dels borats, que pertany al grup de la pinakiolita. Rep el seu nom en honor del nord-americà ...
                      Read more

                      Península de Txukotka

                      Image
                      Península de Txukotka Tipus Península Ubicació 66° N, 172° O  /  66°N , 172°O  / 66; -172 Mar mar dels Txuktxis La península de Txukotka a l'extrem oriental de Sibèria La península de Txukotka o península dels Txuktxis (en rus Чуко́тский полуо́стров , Txukotski poluóstrov ) és una península que es troba a l'extrem oriental d'Euràsia. Contingut 1 Descripció 2 Població 3 Vegeu també 4 Bibliografia Descripció L'estret de Bering, entre la península de Txukotka i la península de Seward a Alaska Foto de satèl·lit de la zona amb la península de Txukotka a l'esquerra El seu punt més extrem és el cap Dezhnev a la vora del nucli d'Uelen, un dels poquíssims llocs habitats de la zona. Al nord limita amb el mar dels Txuktxis i amb el mar de Bering al sud i amb l'estret de Bering a l'est. A la costa sud es troben la badia de Sant Lawrence i la badia de Kresta. La penínsu...
                      Read more