Coordinate Geometry Question (1 Viewer)

axwe7 · Dec 29, 2015

The lines x+2y+9=0 and 2x-y+3=0 intersect at B.
(a) Write down the general equation of a line through B.
(b) Hence find the equation of the line l through B and the origin O.

Cheers.

Paradoxica · Dec 29, 2015

axwe7 said:
The lines x+2y+9=0 and 2x-y+3=0 intersect at B.
(a) Write down the general equation of a line through B.
(b) Hence find the equation of the line l through B and the origin O.

Cheers.

$Have you found B yet?$

axwe7 · Dec 29, 2015

Paradoxica said:
$Have you found B yet?$

Nope.

I quoted the question. Am I supposed to find B simultaneously?

Paradoxica · Dec 29, 2015

axwe7 said:
Nope.

I quoted the question. Am I supposed to find B simultaneously?

$\noindent Yes. Then use the point gradient formula for a line to determine the family of lines that lie on B. Substitute the origin into the general line to complete the question.$

InteGrand · Dec 29, 2015

axwe7 said:
The lines x+2y+9=0 and 2x-y+3=0 intersect at B.
(a) Write down the general equation of a line through B.
(b) Hence find the equation of the line l through B and the origin O.

Cheers.

$I think they want you to do it like this:$

$a) The answer wanted is $x+2y + 9 + k(2x-y+3) = 0$, where $k\in \mathbb{R}$.$

$b) We want $(0,0)$ to lie on the line, so this happens iff $0+0+9 +k(0+0+3)=0 \Longleftrightarrow 3k+9=0$, that is, $k=-3$. Substitute this in to our general form in a) and simplify to get the equation of the line.$

leehuan · Dec 29, 2015

Fortunately in the exam they don't usually demand you to use this 'k' method.

But whilst we're here, for the maths geniuses:

Q: What does 'k' represent

Green Yoda · Dec 29, 2015

InteGrand said:
$I think they want you to do it like this:$

$a) The answer wanted is $x+2y + 9 + k(2x-y+3) = 0$, where $k\in \mathbb{R}$.$

$b) We want $(0,0)$ to lie on the line, so this happens iff $0+0+9 +k(0+0+3)=0 \Longleftrightarrow 3k+9=0$, that is, $k=-3$. Substitute this in to our general form in a) and simplify to get the equation of the line.$

This is a 3u method right?

InteGrand · Dec 29, 2015

Rathin said:
This is a 3u method right?

Can't remember, check the syllabus online. It's either 2U or 3U for sure. Iirc it's in the Pender (Cambridge) Year 11 3U textbook.

axwe7 · Dec 31, 2015

leehuan said:
Fortunately in the exam they don't usually demand you to use this 'k' method.

But whilst we're here, for the maths geniuses:

Q: What does 'k' represent

'k' is a constant.

axwe7 · Dec 31, 2015

axwe7 said:
'k' is a constant.

Yep, got the answer, thanks everyone!

leehuan · Dec 31, 2015

Rathin said:
This is a 3u method right?

2U.

axwe7 said:
'k' is a constant.

Nope.

I remember my physics teacher (he taught MX1 last year as well) DISTINCTIVELY saying back in prelim that 'k' ACTUALLY REPRESENTS SOMETHING slightly beyond a mere constant.

InteGrand · Dec 31, 2015

leehuan said:
2U.

Nope.

I remember my physics teacher (he taught MX1 last year as well) DISTINCTIVELY saying back in prelim that 'k' ACTUALLY REPRESENTS SOMETHING slightly beyond a mere constant.

$Explaining this properly requires some linear algebra. Loosely speaking, $k$ represents the ratio of the ``combination'' of the two given lines' equations that add to give the equation of the desired line.$

$Let our two lines be $\ell_1: a_1 x+b_1 y + c_1 = 0$ and $\ell_2: a_2 x + b_2 y + c_2 = 0$. Incidentally, The actual proper set of all lines through the point of intersection actually requires two $k$'s, i.e. $k_1 ( a_1 x+b_1 y + c_1) + k_2 (a_2 x + b_2 y + c_2)=0$, since otherwise we can't get the line $ a_2 x+b_2 y + c_2=0$ if we just do $( a_1 x+b_1 y + c_1) + k (a_2 x + b_2 y + c_2)=0$. But this is a trivial line since it's one given to us already. So assuming we want different lines, we can actually divide through by $k_1$ in the equation with $k_1$ and $k_2$ to get $k\equiv \frac{k_2}{k_1}$, the ratio of the addition of each line. More explanation is below, based on linear algebra.$

$Recall we let our two lines be $\ell_1: a_1 x+b_1 y + c_1 = 0$ and $\ell_2: a_2 x + b_2 y + c_2 = 0$. These lines have a unique point of intersection iff (if and only if) $a_1 b_2 -b_1 a_2 \neq 0$ (which amounts to saying they are non-parallel; in linear-algebra-speak, the determinant of the system is non-zero). So we will assume this is the case, since we always deal with lines having a single point of intersection for these Q's. Let their point of intersection be $P(x^*,y^*)$ (this is a constant point for a given pair of lines).$

$We are interested in representing the set of all coefficients $(A,B,C)$ for an equation $Ax+By+C=0$ such that $(x^*,y^*)$ satisfies this. That is, we want the set of all vectors $\bold{x}:=(A,B,C)$ such that $Ax^*+By^*+C=0 \Longleftrightarrow \bold{x} \cdot \bold{b} =0$, where $ \bold{b}:= (x^*,y^*,1) $ is a constant non-zero vector in $\mathbb{R}^3 $.$

$This set is just the null-space of the linear map $T:\mathbb{R}^3\to \mathbb{R}$ given by $T(\bold{x}) = \bold{b}\cdot \bold{x}$. The rank of this linear map is 1, so its nullity is 2. This means its null-space is a two-dimensional subspace of $\mathbb{R}^3$, and so is the span of two linearly independent vectors; and the vectors $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are two such vectors, because they are linearly independent due to the assumption that they represent two non-parallel lines, and both are in the null space as they both give 0 when dotted with $(x^*,y^*,1)$ (in other words, both pass through $(x^*,y^*)$ in the Cartesian Plane).$

$This means the set of all lines through the point of intersection is simply the span of $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$. This implies that any desired line's coefficients $(A,B,C)$ can be written in the form $(A,B,C) = k_1 (a_1,b_1,c_1) + k_2 (a_2,b_2,c_2)$. Now if we don't want the line $a_2x + b_2y + c_2 = 0$, we can simply write $(A,B,C)$ in the form $(a_1,b_1,c_1) + k (a_2,b_2,c_2)$, as $k_1$ would be non-zero and we could just rewrite the linear combination with $k_1 = 1$ (this would mean that we can't get every single $(A,B,C)$, but we can still get all the line equations; i.e. we may not be able to get $3x + 3y + 3c = 0$ but we'd still get some form of this line, e.g. $x+y+c = 0$, so we can still get all lines in some form).$

$So $k$ essentially represents the ratio of which we add the two vectors of the coefficients of the given lines to get the desired line.$

InteGrand · Dec 31, 2015

leehuan · Nov 6, 2016

That awesome feeling when this makes sense now..

Coordinate Geometry Question (1 Viewer)

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