Mixing
2 points coordinates such that two lines to be medians in a triangle
$begingroup$
I have the point and the equations of two lines:
and
.Also, I know that
and
.
I have to find the coordinates of B and C such that d1 and d2 to be medians in ABC triangle.
I found the intersection of d1 and d2 point G(1,2)
geometry analytic-geometry
asked Jan 11 at 17:29
Vali ROVali RO
736
$endgroup$
add a comment |
$begingroup$
I have the point and the equations of two lines:
and
.Also, I know that
and
.
I have to find the coordinates of B and C such that d1 and d2 to be medians in ABC triangle.
I found the intersection of d1 and d2 point G(1,2)
geometry analytic-geometry
asked Jan 11 at 17:29
Vali ROVali RO
736
$endgroup$
add a comment |
$begingroup$
I have the point and the equations of two lines:
and
.Also, I know that
and
.
I have to find the coordinates of B and C such that d1 and d2 to be medians in ABC triangle.
I found the intersection of d1 and d2 point G(1,2)
geometry analytic-geometry
asked Jan 11 at 17:29
Vali ROVali RO
736
$endgroup$
I have the point and the equations of two lines:
and
.Also, I know that
and
.
I have to find the coordinates of B and C such that d1 and d2 to be medians in ABC triangle.
I found the intersection of d1 and d2 point G(1,2)
geometry analytic-geometry
geometry analytic-geometry
asked Jan 11 at 17:29
Vali ROVali RO
736
asked Jan 11 at 17:29
Vali ROVali RO
736
asked Jan 11 at 17:29
Vali ROVali RO
736
asked Jan 11 at 17:29
Vali ROVali RO
736
asked Jan 11 at 17:29
Vali ROVali RO
736
736
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
@ValiRo
You know that your point G(1,2) is the center of gravity of the triangle. Denoting $B(x_B, y_B)$ and $C(x_C,y_C)$ you must have (using the formula for the center of gravity)
$(0+x_B+x_C)/3=1$ and $(-1+y_B+y_C)/3=2$.
If you add these two equations to the equations of $d1$ and $d2$ (obviously stisfied by $B(x_B, y_B)$ and $C(x_C,y_C)$ respectively) you get a 4 by 4 linear system that you can easily solve. My solution is $B(0,1)$ and $C(3,6)$.
Cheers.
answered Jan 11 at 17:48
GReyesGReyes
1,0895
$endgroup$
$begingroup$
Thanks a lot! I totally forgot about the formula for the center of gravity.Thanks again!
$endgroup$
– Vali RO
Jan 11 at 17:51
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
@ValiRo
You know that your point G(1,2) is the center of gravity of the triangle. Denoting $B(x_B, y_B)$ and $C(x_C,y_C)$ you must have (using the formula for the center of gravity)
$(0+x_B+x_C)/3=1$ and $(-1+y_B+y_C)/3=2$.
If you add these two equations to the equations of $d1$ and $d2$ (obviously stisfied by $B(x_B, y_B)$ and $C(x_C,y_C)$ respectively) you get a 4 by 4 linear system that you can easily solve. My solution is $B(0,1)$ and $C(3,6)$.
Cheers.
answered Jan 11 at 17:48
GReyesGReyes
1,0895
$endgroup$
$begingroup$
Thanks a lot! I totally forgot about the formula for the center of gravity.Thanks again!
$endgroup$
– Vali RO
Jan 11 at 17:51
add a comment |
$begingroup$
@ValiRo
You know that your point G(1,2) is the center of gravity of the triangle. Denoting $B(x_B, y_B)$ and $C(x_C,y_C)$ you must have (using the formula for the center of gravity)
$(0+x_B+x_C)/3=1$ and $(-1+y_B+y_C)/3=2$.
If you add these two equations to the equations of $d1$ and $d2$ (obviously stisfied by $B(x_B, y_B)$ and $C(x_C,y_C)$ respectively) you get a 4 by 4 linear system that you can easily solve. My solution is $B(0,1)$ and $C(3,6)$.
Cheers.
answered Jan 11 at 17:48
GReyesGReyes
1,0895
$endgroup$
$begingroup$
Thanks a lot! I totally forgot about the formula for the center of gravity.Thanks again!
$endgroup$
– Vali RO
Jan 11 at 17:51
add a comment |
$begingroup$
@ValiRo
You know that your point G(1,2) is the center of gravity of the triangle. Denoting $B(x_B, y_B)$ and $C(x_C,y_C)$ you must have (using the formula for the center of gravity)
$(0+x_B+x_C)/3=1$ and $(-1+y_B+y_C)/3=2$.
If you add these two equations to the equations of $d1$ and $d2$ (obviously stisfied by $B(x_B, y_B)$ and $C(x_C,y_C)$ respectively) you get a 4 by 4 linear system that you can easily solve. My solution is $B(0,1)$ and $C(3,6)$.
Cheers.
answered Jan 11 at 17:48
GReyesGReyes
1,0895
$endgroup$
@ValiRo
You know that your point G(1,2) is the center of gravity of the triangle. Denoting $B(x_B, y_B)$ and $C(x_C,y_C)$ you must have (using the formula for the center of gravity)
$(0+x_B+x_C)/3=1$ and $(-1+y_B+y_C)/3=2$.
If you add these two equations to the equations of $d1$ and $d2$ (obviously stisfied by $B(x_B, y_B)$ and $C(x_C,y_C)$ respectively) you get a 4 by 4 linear system that you can easily solve. My solution is $B(0,1)$ and $C(3,6)$.
Cheers.
answered Jan 11 at 17:48
GReyesGReyes
1,0895
answered Jan 11 at 17:48
GReyesGReyes
1,0895
answered Jan 11 at 17:48
GReyesGReyes
1,0895
answered Jan 11 at 17:48
GReyesGReyes
1,0895
1,0895
$begingroup$
Thanks a lot! I totally forgot about the formula for the center of gravity.Thanks again!
$endgroup$
– Vali RO
Jan 11 at 17:51
add a comment |
$begingroup$
Thanks a lot! I totally forgot about the formula for the center of gravity.Thanks again!
$endgroup$
– Vali RO
Jan 11 at 17:51
$begingroup$
Thanks a lot! I totally forgot about the formula for the center of gravity.Thanks again!
$endgroup$
– Vali RO
Jan 11 at 17:51
$begingroup$
Thanks a lot! I totally forgot about the formula for the center of gravity.Thanks again!
$endgroup$
– Vali RO
Jan 11 at 17:51
add a comment |
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