Mixing
Equation Of A Straight Line Passing Through A Point and Having an Intercept
$begingroup$
The question is as follows:
Find the equation of the straight line passing through $(-2,-7)$ and having intercept of length 3 units between straight lines $4x + 3y = 12$ and $4x + 3y = 3$.
I really didn't understand what they meant by "intercept of length 3 units". Hence, I couldn't proceed with the question.
I have a basic knowledge about the different forms of equations of a straight line, but none of them seemed to fit here.
Help will be appreciated. Thanks.
geometry euclidean-geometry coordinate-systems
edited May 17 '17 at 3:45
ja72
7,54212044
asked Dec 10 '14 at 15:00
EuclidAteMyBreakfastEuclidAteMyBreakfast
37111
$endgroup$
add a comment |
$begingroup$
The question is as follows:
Find the equation of the straight line passing through $(-2,-7)$ and having intercept of length 3 units between straight lines $4x + 3y = 12$ and $4x + 3y = 3$.
I really didn't understand what they meant by "intercept of length 3 units". Hence, I couldn't proceed with the question.
I have a basic knowledge about the different forms of equations of a straight line, but none of them seemed to fit here.
Help will be appreciated. Thanks.
geometry euclidean-geometry coordinate-systems
edited May 17 '17 at 3:45
ja72
7,54212044
asked Dec 10 '14 at 15:00
EuclidAteMyBreakfastEuclidAteMyBreakfast
37111
$endgroup$
add a comment |
$begingroup$
The question is as follows:
Find the equation of the straight line passing through $(-2,-7)$ and having intercept of length 3 units between straight lines $4x + 3y = 12$ and $4x + 3y = 3$.
I really didn't understand what they meant by "intercept of length 3 units". Hence, I couldn't proceed with the question.
I have a basic knowledge about the different forms of equations of a straight line, but none of them seemed to fit here.
Help will be appreciated. Thanks.
geometry euclidean-geometry coordinate-systems
edited May 17 '17 at 3:45
ja72
7,54212044
asked Dec 10 '14 at 15:00
EuclidAteMyBreakfastEuclidAteMyBreakfast
37111
$endgroup$
The question is as follows:
Find the equation of the straight line passing through $(-2,-7)$ and having intercept of length 3 units between straight lines $4x + 3y = 12$ and $4x + 3y = 3$.
I really didn't understand what they meant by "intercept of length 3 units". Hence, I couldn't proceed with the question.
I have a basic knowledge about the different forms of equations of a straight line, but none of them seemed to fit here.
Help will be appreciated. Thanks.
geometry euclidean-geometry coordinate-systems
geometry euclidean-geometry coordinate-systems
edited May 17 '17 at 3:45
ja72
7,54212044
asked Dec 10 '14 at 15:00
EuclidAteMyBreakfastEuclidAteMyBreakfast
37111
edited May 17 '17 at 3:45
ja72
7,54212044
asked Dec 10 '14 at 15:00
EuclidAteMyBreakfastEuclidAteMyBreakfast
37111
edited May 17 '17 at 3:45
ja72
7,54212044
edited May 17 '17 at 3:45
ja72
7,54212044
edited May 17 '17 at 3:45
ja72
7,54212044
7,54212044
asked Dec 10 '14 at 15:00
EuclidAteMyBreakfastEuclidAteMyBreakfast
37111
asked Dec 10 '14 at 15:00
EuclidAteMyBreakfastEuclidAteMyBreakfast
37111
asked Dec 10 '14 at 15:00
EuclidAteMyBreakfastEuclidAteMyBreakfast
37111
37111
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
I agree it is difficult to interpret the question.
If you write the lines in the more useful $y=mx+c$ form you find
$$L_1:=y=-frac43 x+3$$
and
$$L_2:=y=-frac{4}{3}x+1.$$
This means that the lines have the same slopes and so are parallel. Also the first cuts at $y=3$ and the second at $y=1$.
I think the "intercept of length three" means that the line we are looking for, $L$, cuts $L_1$ and $L_2$ at points $P_1=Lcap L_1$ and $P_2=Lcap L_2$ such that
$$|P_1P_2|=3.$$
There are a number of ways of finishing this off. Synthetic geometry is probably easiest.
Remember: Coordinate Geometry is supposed to make Synthetic Geometry
easier! You are perfectly entitled to use Synthetic Geometry to make
Coordinate Geometry easier also!
Consider the point $R$ on $L_1$ that is found vertically above $P_2$ and the triangle $Delta (RP_1P_2)$.
We know the angle at $R$ (draw a picture and use the slope of $L_1$).
Using the Sine Rule we can thus find the angle at $P_1$ as $|P_2R|=2$ (why?).
Hence you know all the angles in particular the angle at $P_2$, $alpha$.
The slope of $L$ is thus
$$m=tanleft(frac{pi}{2}-alpharight).$$
answered Dec 10 '14 at 15:49
JP McCarthyJP McCarthy
5,73412441
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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votes
$begingroup$
I agree it is difficult to interpret the question.
If you write the lines in the more useful $y=mx+c$ form you find
$$L_1:=y=-frac43 x+3$$
and
$$L_2:=y=-frac{4}{3}x+1.$$
This means that the lines have the same slopes and so are parallel. Also the first cuts at $y=3$ and the second at $y=1$.
I think the "intercept of length three" means that the line we are looking for, $L$, cuts $L_1$ and $L_2$ at points $P_1=Lcap L_1$ and $P_2=Lcap L_2$ such that
$$|P_1P_2|=3.$$
There are a number of ways of finishing this off. Synthetic geometry is probably easiest.
Remember: Coordinate Geometry is supposed to make Synthetic Geometry
easier! You are perfectly entitled to use Synthetic Geometry to make
Coordinate Geometry easier also!
Consider the point $R$ on $L_1$ that is found vertically above $P_2$ and the triangle $Delta (RP_1P_2)$.
We know the angle at $R$ (draw a picture and use the slope of $L_1$).
Using the Sine Rule we can thus find the angle at $P_1$ as $|P_2R|=2$ (why?).
Hence you know all the angles in particular the angle at $P_2$, $alpha$.
The slope of $L$ is thus
$$m=tanleft(frac{pi}{2}-alpharight).$$
answered Dec 10 '14 at 15:49
JP McCarthyJP McCarthy
5,73412441
$endgroup$
add a comment |
$begingroup$
I agree it is difficult to interpret the question.
If you write the lines in the more useful $y=mx+c$ form you find
$$L_1:=y=-frac43 x+3$$
and
$$L_2:=y=-frac{4}{3}x+1.$$
This means that the lines have the same slopes and so are parallel. Also the first cuts at $y=3$ and the second at $y=1$.
I think the "intercept of length three" means that the line we are looking for, $L$, cuts $L_1$ and $L_2$ at points $P_1=Lcap L_1$ and $P_2=Lcap L_2$ such that
$$|P_1P_2|=3.$$
There are a number of ways of finishing this off. Synthetic geometry is probably easiest.
Remember: Coordinate Geometry is supposed to make Synthetic Geometry
easier! You are perfectly entitled to use Synthetic Geometry to make
Coordinate Geometry easier also!
Consider the point $R$ on $L_1$ that is found vertically above $P_2$ and the triangle $Delta (RP_1P_2)$.
We know the angle at $R$ (draw a picture and use the slope of $L_1$).
Using the Sine Rule we can thus find the angle at $P_1$ as $|P_2R|=2$ (why?).
Hence you know all the angles in particular the angle at $P_2$, $alpha$.
The slope of $L$ is thus
$$m=tanleft(frac{pi}{2}-alpharight).$$
answered Dec 10 '14 at 15:49
JP McCarthyJP McCarthy
5,73412441
$endgroup$
add a comment |
$begingroup$
I agree it is difficult to interpret the question.
If you write the lines in the more useful $y=mx+c$ form you find
$$L_1:=y=-frac43 x+3$$
and
$$L_2:=y=-frac{4}{3}x+1.$$
This means that the lines have the same slopes and so are parallel. Also the first cuts at $y=3$ and the second at $y=1$.
I think the "intercept of length three" means that the line we are looking for, $L$, cuts $L_1$ and $L_2$ at points $P_1=Lcap L_1$ and $P_2=Lcap L_2$ such that
$$|P_1P_2|=3.$$
There are a number of ways of finishing this off. Synthetic geometry is probably easiest.
Remember: Coordinate Geometry is supposed to make Synthetic Geometry
easier! You are perfectly entitled to use Synthetic Geometry to make
Coordinate Geometry easier also!
Consider the point $R$ on $L_1$ that is found vertically above $P_2$ and the triangle $Delta (RP_1P_2)$.
We know the angle at $R$ (draw a picture and use the slope of $L_1$).
Using the Sine Rule we can thus find the angle at $P_1$ as $|P_2R|=2$ (why?).
Hence you know all the angles in particular the angle at $P_2$, $alpha$.
The slope of $L$ is thus
$$m=tanleft(frac{pi}{2}-alpharight).$$
answered Dec 10 '14 at 15:49
JP McCarthyJP McCarthy
5,73412441
$endgroup$
I agree it is difficult to interpret the question.
If you write the lines in the more useful $y=mx+c$ form you find
$$L_1:=y=-frac43 x+3$$
and
$$L_2:=y=-frac{4}{3}x+1.$$
This means that the lines have the same slopes and so are parallel. Also the first cuts at $y=3$ and the second at $y=1$.
I think the "intercept of length three" means that the line we are looking for, $L$, cuts $L_1$ and $L_2$ at points $P_1=Lcap L_1$ and $P_2=Lcap L_2$ such that
$$|P_1P_2|=3.$$
There are a number of ways of finishing this off. Synthetic geometry is probably easiest.
Remember: Coordinate Geometry is supposed to make Synthetic Geometry
easier! You are perfectly entitled to use Synthetic Geometry to make
Coordinate Geometry easier also!
Consider the point $R$ on $L_1$ that is found vertically above $P_2$ and the triangle $Delta (RP_1P_2)$.
We know the angle at $R$ (draw a picture and use the slope of $L_1$).
Using the Sine Rule we can thus find the angle at $P_1$ as $|P_2R|=2$ (why?).
Hence you know all the angles in particular the angle at $P_2$, $alpha$.
The slope of $L$ is thus
$$m=tanleft(frac{pi}{2}-alpharight).$$
answered Dec 10 '14 at 15:49
JP McCarthyJP McCarthy
5,73412441
edited Dec 10 '14 at 16:03
answered Dec 10 '14 at 15:49
JP McCarthyJP McCarthy
5,73412441
answered Dec 10 '14 at 15:49
JP McCarthyJP McCarthy
5,73412441
answered Dec 10 '14 at 15:49
JP McCarthyJP McCarthy
5,73412441
5,73412441
add a comment |
add a comment |
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