Find the value of k, so that the following lines intersect at the same point [closed]
$begingroup$
Find the value of k, so that the following lines intersect at the same point:
$$3x + y - 2 = 0$$
$$kx + 2y - 3 = 0$$
$$2x - y + 3 = 0$$
How can I resolve this?
thanks
I was able to find that $(-frac15,frac{13}5)$ is the intersection of the first and the third line.
analytic-geometry
edited Dec 10 '18 at 11:46
Martin Sleziak
44.9k10122275
asked Dec 10 '18 at 5:38
englishworkvgsenglishworkvgs
32
$endgroup$
closed as off-topic by Isaac Browne, Shailesh, Gibbs, John B, Brahadeesh Dec 10 '18 at 12:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Isaac Browne, Shailesh, Gibbs, John B, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 2 more comments
$begingroup$
Find the value of k, so that the following lines intersect at the same point:
$$3x + y - 2 = 0$$
$$kx + 2y - 3 = 0$$
$$2x - y + 3 = 0$$
How can I resolve this?
thanks
I was able to find that $(-frac15,frac{13}5)$ is the intersection of the first and the third line.
analytic-geometry
edited Dec 10 '18 at 11:46
Martin Sleziak
44.9k10122275
asked Dec 10 '18 at 5:38
englishworkvgsenglishworkvgs
32
$endgroup$
closed as off-topic by Isaac Browne, Shailesh, Gibbs, John B, Brahadeesh Dec 10 '18 at 12:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Isaac Browne, Shailesh, Gibbs, John B, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Welcome to MSE! What have you tried so far?
$endgroup$
– platty
Dec 10 '18 at 5:40
$begingroup$
I applied that the multiplication of the slopes of the 3 lines is -1, but I do not know if this is correct.
$endgroup$
– englishworkvgs
Dec 10 '18 at 5:46
1
$begingroup$
Okay, so how does that help you? Could you find the intersection of the two lines that don't depend on $k$, and then figure out what to make $k$ so that it also goes through this point?
$endgroup$
– platty
Dec 10 '18 at 5:47
$begingroup$
I got this result: -1/5, 13/5 But what should I do to get k and have the condition?
$endgroup$
– englishworkvgs
Dec 10 '18 at 6:02
$begingroup$
sum first and third lines. So $5x=5$ and $x=1$. Put $x=1$, in the first line and find $y=-1$. Thus take $x=1$ and $y=-1$ in second line.
$endgroup$
– 1ENİGMA1
Dec 10 '18 at 6:07
|
show 2 more comments
$begingroup$
Find the value of k, so that the following lines intersect at the same point:
$$3x + y - 2 = 0$$
$$kx + 2y - 3 = 0$$
$$2x - y + 3 = 0$$
How can I resolve this?
thanks
I was able to find that $(-frac15,frac{13}5)$ is the intersection of the first and the third line.
analytic-geometry
edited Dec 10 '18 at 11:46
Martin Sleziak
44.9k10122275
asked Dec 10 '18 at 5:38
englishworkvgsenglishworkvgs
32
$endgroup$
Find the value of k, so that the following lines intersect at the same point:
$$3x + y - 2 = 0$$
$$kx + 2y - 3 = 0$$
$$2x - y + 3 = 0$$
How can I resolve this?
thanks
I was able to find that $(-frac15,frac{13}5)$ is the intersection of the first and the third line.
analytic-geometry
analytic-geometry
edited Dec 10 '18 at 11:46
Martin Sleziak
44.9k10122275
asked Dec 10 '18 at 5:38
englishworkvgsenglishworkvgs
32
edited Dec 10 '18 at 11:46
Martin Sleziak
44.9k10122275
asked Dec 10 '18 at 5:38
englishworkvgsenglishworkvgs
32
edited Dec 10 '18 at 11:46
Martin Sleziak
44.9k10122275
edited Dec 10 '18 at 11:46
Martin Sleziak
44.9k10122275
edited Dec 10 '18 at 11:46
Martin Sleziak
44.9k10122275
44.9k10122275
asked Dec 10 '18 at 5:38
englishworkvgsenglishworkvgs
32
asked Dec 10 '18 at 5:38
englishworkvgsenglishworkvgs
32
asked Dec 10 '18 at 5:38
englishworkvgsenglishworkvgs
32
32
closed as off-topic by Isaac Browne, Shailesh, Gibbs, John B, Brahadeesh Dec 10 '18 at 12:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Isaac Browne, Shailesh, Gibbs, John B, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Isaac Browne, Shailesh, Gibbs, John B, Brahadeesh Dec 10 '18 at 12:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Isaac Browne, Shailesh, Gibbs, John B, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Welcome to MSE! What have you tried so far?
$endgroup$
– platty
Dec 10 '18 at 5:40
$begingroup$
I applied that the multiplication of the slopes of the 3 lines is -1, but I do not know if this is correct.
$endgroup$
– englishworkvgs
Dec 10 '18 at 5:46
1
$begingroup$
Okay, so how does that help you? Could you find the intersection of the two lines that don't depend on $k$, and then figure out what to make $k$ so that it also goes through this point?
$endgroup$
– platty
Dec 10 '18 at 5:47
$begingroup$
I got this result: -1/5, 13/5 But what should I do to get k and have the condition?
$endgroup$
– englishworkvgs
Dec 10 '18 at 6:02
$begingroup$
sum first and third lines. So $5x=5$ and $x=1$. Put $x=1$, in the first line and find $y=-1$. Thus take $x=1$ and $y=-1$ in second line.
$endgroup$
– 1ENİGMA1
Dec 10 '18 at 6:07
|
show 2 more comments
$begingroup$
Welcome to MSE! What have you tried so far?
$endgroup$
– platty
Dec 10 '18 at 5:40
$begingroup$
I applied that the multiplication of the slopes of the 3 lines is -1, but I do not know if this is correct.
$endgroup$
– englishworkvgs
Dec 10 '18 at 5:46
1
$begingroup$
Okay, so how does that help you? Could you find the intersection of the two lines that don't depend on $k$, and then figure out what to make $k$ so that it also goes through this point?
$endgroup$
– platty
Dec 10 '18 at 5:47
$begingroup$
I got this result: -1/5, 13/5 But what should I do to get k and have the condition?
$endgroup$
– englishworkvgs
Dec 10 '18 at 6:02
$begingroup$
sum first and third lines. So $5x=5$ and $x=1$. Put $x=1$, in the first line and find $y=-1$. Thus take $x=1$ and $y=-1$ in second line.
$endgroup$
– 1ENİGMA1
Dec 10 '18 at 6:07
$begingroup$
Welcome to MSE! What have you tried so far?
$endgroup$
– platty
Dec 10 '18 at 5:40
$begingroup$
Welcome to MSE! What have you tried so far?
$endgroup$
– platty
Dec 10 '18 at 5:40
$begingroup$
I applied that the multiplication of the slopes of the 3 lines is -1, but I do not know if this is correct.
$endgroup$
– englishworkvgs
Dec 10 '18 at 5:46
$begingroup$
I applied that the multiplication of the slopes of the 3 lines is -1, but I do not know if this is correct.
$endgroup$
– englishworkvgs
Dec 10 '18 at 5:46
1
1
$begingroup$
Okay, so how does that help you? Could you find the intersection of the two lines that don't depend on $k$, and then figure out what to make $k$ so that it also goes through this point?
$endgroup$
– platty
Dec 10 '18 at 5:47
$begingroup$
Okay, so how does that help you? Could you find the intersection of the two lines that don't depend on $k$, and then figure out what to make $k$ so that it also goes through this point?
$endgroup$
– platty
Dec 10 '18 at 5:47
$begingroup$
I got this result: -1/5, 13/5 But what should I do to get k and have the condition?
$endgroup$
– englishworkvgs
Dec 10 '18 at 6:02
$begingroup$
I got this result: -1/5, 13/5 But what should I do to get k and have the condition?
$endgroup$
– englishworkvgs
Dec 10 '18 at 6:02
$begingroup$
sum first and third lines. So $5x=5$ and $x=1$. Put $x=1$, in the first line and find $y=-1$. Thus take $x=1$ and $y=-1$ in second line.
$endgroup$
– 1ENİGMA1
Dec 10 '18 at 6:07
$begingroup$
sum first and third lines. So $5x=5$ and $x=1$. Put $x=1$, in the first line and find $y=-1$. Thus take $x=1$ and $y=-1$ in second line.
$endgroup$
– 1ENİGMA1
Dec 10 '18 at 6:07
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Hint :
Solve the system of the first and third equation to find their intersection point. Then, do the same for the pair of the second and third equation oe the first and second. You will find an intersection point dependent on $k$. But you want thwt point to be the same as the first one you calculated. Thus?
answered Dec 10 '18 at 6:05
RebellosRebellos
15.5k31250
$endgroup$
$begingroup$
Note that after finding the intersection of the first and third lines, plugging this point straight into the second equation gives a single equation with one unknown (geometrically, choosing $k$ so that the intersection of the other two lines lies on this one as well).
$endgroup$
– platty
Dec 10 '18 at 8:06
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint :
Solve the system of the first and third equation to find their intersection point. Then, do the same for the pair of the second and third equation oe the first and second. You will find an intersection point dependent on $k$. But you want thwt point to be the same as the first one you calculated. Thus?
answered Dec 10 '18 at 6:05
RebellosRebellos
15.5k31250
$endgroup$
$begingroup$
Note that after finding the intersection of the first and third lines, plugging this point straight into the second equation gives a single equation with one unknown (geometrically, choosing $k$ so that the intersection of the other two lines lies on this one as well).
$endgroup$
– platty
Dec 10 '18 at 8:06
add a comment |
$begingroup$
Hint :
Solve the system of the first and third equation to find their intersection point. Then, do the same for the pair of the second and third equation oe the first and second. You will find an intersection point dependent on $k$. But you want thwt point to be the same as the first one you calculated. Thus?
answered Dec 10 '18 at 6:05
RebellosRebellos
15.5k31250
$endgroup$
$begingroup$
Note that after finding the intersection of the first and third lines, plugging this point straight into the second equation gives a single equation with one unknown (geometrically, choosing $k$ so that the intersection of the other two lines lies on this one as well).
$endgroup$
– platty
Dec 10 '18 at 8:06
add a comment |
$begingroup$
Hint :
Solve the system of the first and third equation to find their intersection point. Then, do the same for the pair of the second and third equation oe the first and second. You will find an intersection point dependent on $k$. But you want thwt point to be the same as the first one you calculated. Thus?
answered Dec 10 '18 at 6:05
RebellosRebellos
15.5k31250
$endgroup$
Hint :
Solve the system of the first and third equation to find their intersection point. Then, do the same for the pair of the second and third equation oe the first and second. You will find an intersection point dependent on $k$. But you want thwt point to be the same as the first one you calculated. Thus?
answered Dec 10 '18 at 6:05
RebellosRebellos
15.5k31250
answered Dec 10 '18 at 6:05
RebellosRebellos
15.5k31250
answered Dec 10 '18 at 6:05
RebellosRebellos
15.5k31250
answered Dec 10 '18 at 6:05
RebellosRebellos
15.5k31250
15.5k31250
$begingroup$
Note that after finding the intersection of the first and third lines, plugging this point straight into the second equation gives a single equation with one unknown (geometrically, choosing $k$ so that the intersection of the other two lines lies on this one as well).
$endgroup$
– platty
Dec 10 '18 at 8:06
add a comment |
$begingroup$
Note that after finding the intersection of the first and third lines, plugging this point straight into the second equation gives a single equation with one unknown (geometrically, choosing $k$ so that the intersection of the other two lines lies on this one as well).
$endgroup$
– platty
Dec 10 '18 at 8:06
$begingroup$
Note that after finding the intersection of the first and third lines, plugging this point straight into the second equation gives a single equation with one unknown (geometrically, choosing $k$ so that the intersection of the other two lines lies on this one as well).
$endgroup$
– platty
Dec 10 '18 at 8:06
$begingroup$
Note that after finding the intersection of the first and third lines, plugging this point straight into the second equation gives a single equation with one unknown (geometrically, choosing $k$ so that the intersection of the other two lines lies on this one as well).
$endgroup$
– platty
Dec 10 '18 at 8:06
add a comment |
$begingroup$
Welcome to MSE! What have you tried so far?
$endgroup$
– platty
Dec 10 '18 at 5:40
$begingroup$
I applied that the multiplication of the slopes of the 3 lines is -1, but I do not know if this is correct.
$endgroup$
– englishworkvgs
Dec 10 '18 at 5:46
1
$begingroup$
Okay, so how does that help you? Could you find the intersection of the two lines that don't depend on $k$, and then figure out what to make $k$ so that it also goes through this point?
$endgroup$
– platty
Dec 10 '18 at 5:47
$begingroup$
I got this result: -1/5, 13/5 But what should I do to get k and have the condition?
$endgroup$
– englishworkvgs
Dec 10 '18 at 6:02
$begingroup$
sum first and third lines. So $5x=5$ and $x=1$. Put $x=1$, in the first line and find $y=-1$. Thus take $x=1$ and $y=-1$ in second line.
$endgroup$
– 1ENİGMA1
Dec 10 '18 at 6:07