From origin walk halfway to $(8,6)$, turn $90$ degrees left, then walk twice as far.
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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= $frac3x4$ - The equation for line PM is:
y= $frac-4x3$ + $frac253$ - 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
algebra-precalculus arithmetic coordinate-systems
edited Mar 2 at 15:48
Asaf Karagila♦
307k33439771
asked Mar 2 at 1:20
Sinestro 38Sinestro 38
251
$endgroup$
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= $frac3x4$ - The equation for line PM is:
y= $frac-4x3$ + $frac253$ - 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
algebra-precalculus arithmetic coordinate-systems
edited Mar 2 at 15:48
Asaf Karagila♦
307k33439771
asked Mar 2 at 1:20
Sinestro 38Sinestro 38
251
$endgroup$
1
$begingroup$
To solve this using brute force, just substitute $y = -frac43x + frac253$ 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
Mar 2 at 1:30
$begingroup$
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
Mar 2 at 1:37
$begingroup$
Thanks, this helped me so much!
$endgroup$
– Sinestro 38
Mar 2 at 3:51
2
$begingroup$
Please using subjective and superlative titles (in this case "incredibly hard"). Instead, provide an actual description of the question. Things like "question 23" are meaningless to almost everyone except you and those related to your class. You have 150 characters at your disposal, this is enough to provide a good title.
$endgroup$
– Asaf Karagila♦
Mar 2 at 12:55
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= $frac3x4$ - The equation for line PM is:
y= $frac-4x3$ + $frac253$ - 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
algebra-precalculus arithmetic coordinate-systems
edited Mar 2 at 15:48
Asaf Karagila♦
307k33439771
asked Mar 2 at 1:20
Sinestro 38Sinestro 38
251
$endgroup$
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= $frac3x4$ - The equation for line PM is:
y= $frac-4x3$ + $frac253$ - 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
algebra-precalculus arithmetic coordinate-systems
algebra-precalculus arithmetic coordinate-systems
edited Mar 2 at 15:48
Asaf Karagila♦
307k33439771
asked Mar 2 at 1:20
Sinestro 38Sinestro 38
251
edited Mar 2 at 15:48
Asaf Karagila♦
307k33439771
asked Mar 2 at 1:20
Sinestro 38Sinestro 38
251
edited Mar 2 at 15:48
Asaf Karagila♦
307k33439771
edited Mar 2 at 15:48
Asaf Karagila♦
307k33439771
edited Mar 2 at 15:48
Asaf Karagila♦
307k33439771
307k33439771
asked Mar 2 at 1:20
Sinestro 38Sinestro 38
251
asked Mar 2 at 1:20
Sinestro 38Sinestro 38
251
asked Mar 2 at 1:20
Sinestro 38Sinestro 38
251
251
1
$begingroup$
To solve this using brute force, just substitute $y = -frac43x + frac253$ 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
Mar 2 at 1:30
$begingroup$
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
Mar 2 at 1:37
$begingroup$
Thanks, this helped me so much!
$endgroup$
– Sinestro 38
Mar 2 at 3:51
2
$begingroup$
Please using subjective and superlative titles (in this case "incredibly hard"). Instead, provide an actual description of the question. Things like "question 23" are meaningless to almost everyone except you and those related to your class. You have 150 characters at your disposal, this is enough to provide a good title.
$endgroup$
– Asaf Karagila♦
Mar 2 at 12:55
add a comment |
1
$begingroup$
To solve this using brute force, just substitute $y = -frac43x + frac253$ 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
Mar 2 at 1:30
$begingroup$
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
Mar 2 at 1:37
$begingroup$
Thanks, this helped me so much!
$endgroup$
– Sinestro 38
Mar 2 at 3:51
2
$begingroup$
Please using subjective and superlative titles (in this case "incredibly hard"). Instead, provide an actual description of the question. Things like "question 23" are meaningless to almost everyone except you and those related to your class. You have 150 characters at your disposal, this is enough to provide a good title.
$endgroup$
– Asaf Karagila♦
Mar 2 at 12:55
1
1
$begingroup$
To solve this using brute force, just substitute $y = -frac43x + frac253$ 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
Mar 2 at 1:30
$begingroup$
To solve this using brute force, just substitute $y = -frac43x + frac253$ 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
Mar 2 at 1:30
$begingroup$
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
Mar 2 at 1:37
$begingroup$
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
Mar 2 at 1:37
$begingroup$
Thanks, this helped me so much!
$endgroup$
– Sinestro 38
Mar 2 at 3:51
$begingroup$
Thanks, this helped me so much!
$endgroup$
– Sinestro 38
Mar 2 at 3:51
2
2
$begingroup$
Please using subjective and superlative titles (in this case "incredibly hard"). Instead, provide an actual description of the question. Things like "question 23" are meaningless to almost everyone except you and those related to your class. You have 150 characters at your disposal, this is enough to provide a good title.
$endgroup$
– Asaf Karagila♦
Mar 2 at 12:55
$begingroup$
Please using subjective and superlative titles (in this case "incredibly hard"). Instead, provide an actual description of the question. Things like "question 23" are meaningless to almost everyone except you and those related to your class. You have 150 characters at your disposal, this is enough to provide a good title.
$endgroup$
– Asaf Karagila♦
Mar 2 at 12:55
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Equivalent to 4 steps east, 3 steps north, then 8 steps north, 6 steps west, resulting in (4-6, 3+8) or (-2, 11). After plotting, it's much easier to see the triangles you have to describe algebraically.
answered Mar 2 at 5:40
Witness Protection ID 44583292Witness Protection ID 44583292
23113
$endgroup$
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 Mar 2 at 1:33
DeepakDeepak
17.6k11539
$endgroup$
2
$begingroup$
He said he was in grade 10, so he probably doesn’t know complex numbers yet.
$endgroup$
– Bor Kari
Mar 2 at 1:40
$begingroup$
@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
Mar 2 at 1:48
1
$begingroup$
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
Mar 2 at 1:59
$begingroup$
@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
Mar 2 at 2:12
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 Mar 2 at 2:47
G Tony JacobsG Tony Jacobs
25.9k43686
$endgroup$
add a comment |
$begingroup$
It is quickly checked that an orthogonal vector to $binom43$ in the right direction is $binom-34$ and this vector has also length $5$.
So, the algebraic solution is
$$binom43 + 2binom-34 = binom-211$$
answered Mar 2 at 6:11
trancelocationtrancelocation
13.3k1827
$endgroup$
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 Mar 2 at 1:28
Bor KariBor Kari
3269
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Equivalent to 4 steps east, 3 steps north, then 8 steps north, 6 steps west, resulting in (4-6, 3+8) or (-2, 11). After plotting, it's much easier to see the triangles you have to describe algebraically.
answered Mar 2 at 5:40
Witness Protection ID 44583292Witness Protection ID 44583292
23113
$endgroup$
add a comment |
$begingroup$
Equivalent to 4 steps east, 3 steps north, then 8 steps north, 6 steps west, resulting in (4-6, 3+8) or (-2, 11). After plotting, it's much easier to see the triangles you have to describe algebraically.
answered Mar 2 at 5:40
Witness Protection ID 44583292Witness Protection ID 44583292
23113
$endgroup$
add a comment |
$begingroup$
Equivalent to 4 steps east, 3 steps north, then 8 steps north, 6 steps west, resulting in (4-6, 3+8) or (-2, 11). After plotting, it's much easier to see the triangles you have to describe algebraically.
answered Mar 2 at 5:40
Witness Protection ID 44583292Witness Protection ID 44583292
23113
$endgroup$
Equivalent to 4 steps east, 3 steps north, then 8 steps north, 6 steps west, resulting in (4-6, 3+8) or (-2, 11). After plotting, it's much easier to see the triangles you have to describe algebraically.
answered Mar 2 at 5:40
Witness Protection ID 44583292Witness Protection ID 44583292
23113
edited Mar 2 at 5:50
answered Mar 2 at 5:40
Witness Protection ID 44583292Witness Protection ID 44583292
23113
answered Mar 2 at 5:40
Witness Protection ID 44583292Witness Protection ID 44583292
23113
answered Mar 2 at 5:40
Witness Protection ID 44583292Witness Protection ID 44583292
23113
23113
add a comment |
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 Mar 2 at 1:33
DeepakDeepak
17.6k11539
$endgroup$
2
$begingroup$
He said he was in grade 10, so he probably doesn’t know complex numbers yet.
$endgroup$
– Bor Kari
Mar 2 at 1:40
$begingroup$
@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
Mar 2 at 1:48
1
$begingroup$
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
Mar 2 at 1:59
$begingroup$
@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
Mar 2 at 2:12
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 Mar 2 at 1:33
DeepakDeepak
17.6k11539
$endgroup$
2
$begingroup$
He said he was in grade 10, so he probably doesn’t know complex numbers yet.
$endgroup$
– Bor Kari
Mar 2 at 1:40
$begingroup$
@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
Mar 2 at 1:48
1
$begingroup$
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
Mar 2 at 1:59
$begingroup$
@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
Mar 2 at 2:12
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 Mar 2 at 1:33
DeepakDeepak
17.6k11539
$endgroup$
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 Mar 2 at 1:33
DeepakDeepak
17.6k11539
answered Mar 2 at 1:33
DeepakDeepak
17.6k11539
answered Mar 2 at 1:33
DeepakDeepak
17.6k11539
answered Mar 2 at 1:33
DeepakDeepak
17.6k11539
17.6k11539
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
Mar 2 at 1:40
$begingroup$
@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
Mar 2 at 1:48
1
$begingroup$
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
Mar 2 at 1:59
$begingroup$
@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
Mar 2 at 2:12
add a comment |
2
$begingroup$
He said he was in grade 10, so he probably doesn’t know complex numbers yet.
$endgroup$
– Bor Kari
Mar 2 at 1:40
$begingroup$
@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
Mar 2 at 1:48
1
$begingroup$
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
Mar 2 at 1:59
$begingroup$
@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
Mar 2 at 2:12
2
2
$begingroup$
He said he was in grade 10, so he probably doesn’t know complex numbers yet.
$endgroup$
– Bor Kari
Mar 2 at 1:40
$begingroup$
He said he was in grade 10, so he probably doesn’t know complex numbers yet.
$endgroup$
– Bor Kari
Mar 2 at 1:40
$begingroup$
@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
Mar 2 at 1:48
$begingroup$
@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
Mar 2 at 1:48
1
1
$begingroup$
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
Mar 2 at 1:59
$begingroup$
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
Mar 2 at 1:59
$begingroup$
@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
Mar 2 at 2:12
$begingroup$
@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
Mar 2 at 2:12
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 Mar 2 at 2:47
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 Mar 2 at 2:47
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 Mar 2 at 2:47
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 Mar 2 at 2:47
G Tony JacobsG Tony Jacobs
25.9k43686
answered Mar 2 at 2:47
G Tony JacobsG Tony Jacobs
25.9k43686
answered Mar 2 at 2:47
G Tony JacobsG Tony Jacobs
25.9k43686
answered Mar 2 at 2:47
G Tony JacobsG Tony Jacobs
25.9k43686
25.9k43686
add a comment |
add a comment |
$begingroup$
It is quickly checked that an orthogonal vector to $binom43$ in the right direction is $binom-34$ and this vector has also length $5$.
So, the algebraic solution is
$$binom43 + 2binom-34 = binom-211$$
answered Mar 2 at 6:11
trancelocationtrancelocation
13.3k1827
$endgroup$
add a comment |
$begingroup$
It is quickly checked that an orthogonal vector to $binom43$ in the right direction is $binom-34$ and this vector has also length $5$.
So, the algebraic solution is
$$binom43 + 2binom-34 = binom-211$$
answered Mar 2 at 6:11
trancelocationtrancelocation
13.3k1827
$endgroup$
add a comment |
$begingroup$
It is quickly checked that an orthogonal vector to $binom43$ in the right direction is $binom-34$ and this vector has also length $5$.
So, the algebraic solution is
$$binom43 + 2binom-34 = binom-211$$
answered Mar 2 at 6:11
trancelocationtrancelocation
13.3k1827
$endgroup$
It is quickly checked that an orthogonal vector to $binom43$ in the right direction is $binom-34$ and this vector has also length $5$.
So, the algebraic solution is
$$binom43 + 2binom-34 = binom-211$$
answered Mar 2 at 6:11
trancelocationtrancelocation
13.3k1827
answered Mar 2 at 6:11
trancelocationtrancelocation
13.3k1827
answered Mar 2 at 6:11
trancelocationtrancelocation
13.3k1827
answered Mar 2 at 6:11
trancelocationtrancelocation
13.3k1827
13.3k1827
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 Mar 2 at 1:28
Bor KariBor Kari
3269
$endgroup$
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 Mar 2 at 1:28
Bor KariBor Kari
3269
$endgroup$
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 Mar 2 at 1:28
Bor KariBor Kari
3269
$endgroup$
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 Mar 2 at 1:28
Bor KariBor Kari
3269
answered Mar 2 at 1:28
Bor KariBor Kari
3269
answered Mar 2 at 1:28
Bor KariBor Kari
3269
answered Mar 2 at 1:28
Bor KariBor Kari
3269
3269
add a comment |
add a comment |
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To solve this using brute force, just substitute $y = -frac43x + frac253$ 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
Mar 2 at 1:30
$begingroup$
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
Mar 2 at 1:37
$begingroup$
Thanks, this helped me so much!
$endgroup$
– Sinestro 38
Mar 2 at 3:51
2
$begingroup$
Please using subjective and superlative titles (in this case "incredibly hard"). Instead, provide an actual description of the question. Things like "question 23" are meaningless to almost everyone except you and those related to your class. You have 150 characters at your disposal, this is enough to provide a good title.
$endgroup$
– Asaf Karagila♦
Mar 2 at 12:55