Changing the side of a triangle without changing area?
$begingroup$
$triangle ABC$ has vertices $A=(8,2)$, $B=(0,6)$ and $C=(-3,2)$. Point $C$ can be moved along a certain line with points $A$ and $B$ remaining stationary so that the area of $ABC$ will not change? What is the slope of that line?
The answer is $-frac12$, but I don't understand why or even how to come to this conclusion. Working backwards, I see that $-frac12$ is also the slope of line $AB$, but I don't know why this information is related or how to solve another problem of this type but not the exact same.
analytic-geometry triangle area slope
asked Apr 17 '16 at 5:16
JuliaJulia
163
$endgroup$
add a comment |
$begingroup$
$triangle ABC$ has vertices $A=(8,2)$, $B=(0,6)$ and $C=(-3,2)$. Point $C$ can be moved along a certain line with points $A$ and $B$ remaining stationary so that the area of $ABC$ will not change? What is the slope of that line?
The answer is $-frac12$, but I don't understand why or even how to come to this conclusion. Working backwards, I see that $-frac12$ is also the slope of line $AB$, but I don't know why this information is related or how to solve another problem of this type but not the exact same.
analytic-geometry triangle area slope
asked Apr 17 '16 at 5:16
JuliaJulia
163
$endgroup$
1
$begingroup$
Draw a line through C parallel to AB. So the height of the triangle remains constant.
$endgroup$
– N.S.JOHN
Apr 17 '16 at 5:22
$begingroup$
The line that height is constant is parallel.
$endgroup$
– Takahiro Waki
Apr 17 '16 at 7:09
add a comment |
$begingroup$
$triangle ABC$ has vertices $A=(8,2)$, $B=(0,6)$ and $C=(-3,2)$. Point $C$ can be moved along a certain line with points $A$ and $B$ remaining stationary so that the area of $ABC$ will not change? What is the slope of that line?
The answer is $-frac12$, but I don't understand why or even how to come to this conclusion. Working backwards, I see that $-frac12$ is also the slope of line $AB$, but I don't know why this information is related or how to solve another problem of this type but not the exact same.
analytic-geometry triangle area slope
asked Apr 17 '16 at 5:16
JuliaJulia
163
$endgroup$
$triangle ABC$ has vertices $A=(8,2)$, $B=(0,6)$ and $C=(-3,2)$. Point $C$ can be moved along a certain line with points $A$ and $B$ remaining stationary so that the area of $ABC$ will not change? What is the slope of that line?
The answer is $-frac12$, but I don't understand why or even how to come to this conclusion. Working backwards, I see that $-frac12$ is also the slope of line $AB$, but I don't know why this information is related or how to solve another problem of this type but not the exact same.
analytic-geometry triangle area slope
analytic-geometry triangle area slope
asked Apr 17 '16 at 5:16
JuliaJulia
163
asked Apr 17 '16 at 5:16
JuliaJulia
163
edited Apr 17 '16 at 5:44
user249332
asked Apr 17 '16 at 5:16
JuliaJulia
163
asked Apr 17 '16 at 5:16
JuliaJulia
163
asked Apr 17 '16 at 5:16
JuliaJulia
163
163
1
$begingroup$
Draw a line through C parallel to AB. So the height of the triangle remains constant.
$endgroup$
– N.S.JOHN
Apr 17 '16 at 5:22
$begingroup$
The line that height is constant is parallel.
$endgroup$
– Takahiro Waki
Apr 17 '16 at 7:09
add a comment |
1
$begingroup$
Draw a line through C parallel to AB. So the height of the triangle remains constant.
$endgroup$
– N.S.JOHN
Apr 17 '16 at 5:22
$begingroup$
The line that height is constant is parallel.
$endgroup$
– Takahiro Waki
Apr 17 '16 at 7:09
1
1
$begingroup$
Draw a line through C parallel to AB. So the height of the triangle remains constant.
$endgroup$
– N.S.JOHN
Apr 17 '16 at 5:22
$begingroup$
Draw a line through C parallel to AB. So the height of the triangle remains constant.
$endgroup$
– N.S.JOHN
Apr 17 '16 at 5:22
$begingroup$
The line that height is constant is parallel.
$endgroup$
– Takahiro Waki
Apr 17 '16 at 7:09
$begingroup$
The line that height is constant is parallel.
$endgroup$
– Takahiro Waki
Apr 17 '16 at 7:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're almost there. You can move the vertex of a triangle along a line parallel to its base (the opposite side) without changing the area. This is called a shear transform. In this case, just treat point $C$ as the vertex and $AB$ as the base. The base has slope $-frac 12$, which is also the slope of all lines parallel to this.
The reason is simple: the perpendicular height $h$ (dropped from the vertex to the base extended as needed) does not change. The base $b$ remains the same, so the area $A = frac 12 bcdot h$ also doesn't change.
answered Apr 17 '16 at 5:19
DeepakDeepak
16.8k11436
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add a comment |
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$begingroup$
You're almost there. You can move the vertex of a triangle along a line parallel to its base (the opposite side) without changing the area. This is called a shear transform. In this case, just treat point $C$ as the vertex and $AB$ as the base. The base has slope $-frac 12$, which is also the slope of all lines parallel to this.
The reason is simple: the perpendicular height $h$ (dropped from the vertex to the base extended as needed) does not change. The base $b$ remains the same, so the area $A = frac 12 bcdot h$ also doesn't change.
answered Apr 17 '16 at 5:19
DeepakDeepak
16.8k11436
$endgroup$
add a comment |
$begingroup$
You're almost there. You can move the vertex of a triangle along a line parallel to its base (the opposite side) without changing the area. This is called a shear transform. In this case, just treat point $C$ as the vertex and $AB$ as the base. The base has slope $-frac 12$, which is also the slope of all lines parallel to this.
The reason is simple: the perpendicular height $h$ (dropped from the vertex to the base extended as needed) does not change. The base $b$ remains the same, so the area $A = frac 12 bcdot h$ also doesn't change.
answered Apr 17 '16 at 5:19
DeepakDeepak
16.8k11436
$endgroup$
add a comment |
$begingroup$
You're almost there. You can move the vertex of a triangle along a line parallel to its base (the opposite side) without changing the area. This is called a shear transform. In this case, just treat point $C$ as the vertex and $AB$ as the base. The base has slope $-frac 12$, which is also the slope of all lines parallel to this.
The reason is simple: the perpendicular height $h$ (dropped from the vertex to the base extended as needed) does not change. The base $b$ remains the same, so the area $A = frac 12 bcdot h$ also doesn't change.
answered Apr 17 '16 at 5:19
DeepakDeepak
16.8k11436
$endgroup$
You're almost there. You can move the vertex of a triangle along a line parallel to its base (the opposite side) without changing the area. This is called a shear transform. In this case, just treat point $C$ as the vertex and $AB$ as the base. The base has slope $-frac 12$, which is also the slope of all lines parallel to this.
The reason is simple: the perpendicular height $h$ (dropped from the vertex to the base extended as needed) does not change. The base $b$ remains the same, so the area $A = frac 12 bcdot h$ also doesn't change.
answered Apr 17 '16 at 5:19
DeepakDeepak
16.8k11436
answered Apr 17 '16 at 5:19
DeepakDeepak
16.8k11436
answered Apr 17 '16 at 5:19
DeepakDeepak
16.8k11436
answered Apr 17 '16 at 5:19
DeepakDeepak
16.8k11436
16.8k11436
add a comment |
add a comment |
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$begingroup$
Draw a line through C parallel to AB. So the height of the triangle remains constant.
$endgroup$
– N.S.JOHN
Apr 17 '16 at 5:22
$begingroup$
The line that height is constant is parallel.
$endgroup$
– Takahiro Waki
Apr 17 '16 at 7:09