Analytic Geometry: straight line

Bisectors of $4x-3y+1=0$ and $12x-5y+7=0$.

A point $(x,y)$ on a bisector is equidistant from both lines:

$\frac{4x-3y+1}{\sqrt{16+9}}=\pm\frac{12x-5y+7}{\sqrt{144+25}}\;\Rightarrow\;\frac{4x-3y+1}{5}=\pm\frac{12x-5y+7}{13}.$

Positive sign: $13(4x-3y+1)=5(12x-5y+7)$ $52x-39y+13=60x-25y+35\;\Rightarrow\;8x+14y+22=0\;\Rightarrow\;4x+7y+11=0.$

Negative sign: $13(4x-3y+1)=-5(12x-5y+7)$ $52x-39y+13=-60x+25y-35\;\Rightarrow\;112x-64y+48=0\;\Rightarrow\;7x-4y+3=0.$

The bisectors are $4x+7y+11=0$ and $7x-4y+3=0$.

They are perpendicular. Slopes: $m_1=-\dfrac47$ and $m_2=\dfrac74$, and $m_1m_2=-\dfrac47\cdot\dfrac74=-1$. So the bisectors are at right angles. $\blacksquare$

Bisector containing the origin. Both given lines have positive constant terms ($1$ and $7$), so the origin gives positive values of $4x-3y+1$ and $12x-5y+7$. The bisector of the angle that contains the origin is the one with the same sign on both sides, i.e. the positive-sign case:

$4x+7y+11=0.$

Check: the lines meet at $(-1,-1)$, which lies on both bisectors. The point $(6,-5)$ on $4x+7y+11=0$ makes both expressions positive ($40$ and $104$), the same side as the origin, so this bisector lies inside the angle containing the origin.

OR

Substituting $y=mx$ in $ax^2+2hxy+by^2=0$ gives $bm^2+2hm+a=0$, so the lines are $y=m_1x$ and $y=m_2x$ with $m_1+m_2=-\dfrac{2h}b$, $m_1m_2=\dfrac ab$.

The bisectors are the points equidistant from both lines:

$\frac{y-m_1x}{\sqrt{1+m_1^2}}=\pm\frac{y-m_2x}{\sqrt{1+m_2^2}}.$

Squaring and expanding, $(y-m_1x)^2(1+m_2^2)-(y-m_2x)^2(1+m_1^2)=0$ $(m_2^2-m_1^2)(y^2-x^2)-2xy(m_1-m_2)(1-m_1m_2)=0.$

Divide by $(m_1-m_2)\ne0$: $(m_1+m_2)(x^2-y^2)=2xy(1-m_1m_2).$

Substituting $m_1+m_2=-\dfrac{2h}b$ and $m_1m_2=\dfrac ab$: $-\frac{2h}{b}(x^2-y^2)=2xy\cdot\frac{b-a}{b}\;\Rightarrow\;h(x^2-y^2)=xy(a-b),$

so $\frac{x^2-y^2}{a-b}=\frac{xy}{h}.\ \blacksquare$

Write $4x-8y+3=0$ as $y=\dfrac12x+\dfrac38$, so its slope is $\dfrac12$.

A parallel line has the same slope, $m=\dfrac12$. With $y$-intercept $c=2$, using $y=mx+c$:

$y=\frac12x+2\;\Rightarrow\;2y=x+4.$

The required line is $x-2y+4=0$.

Midpoint of $(2,-4)$ and $(2,4)$: $\left(\dfrac{2+2}2,\dfrac{-4+4}2\right)=(2,0)$.

The line $3x-2y=4$ has slope $\dfrac32$, so the required line also has slope $\dfrac32$. Through $(2,0)$:

$y-0=\frac32(x-2)\;\Rightarrow\;2y=3x-6.$

The required line is $3x-2y-6=0$.

(i) The quadratic part factorises as $2x^2+3xy+y^2=(2x+y)(x+y)$. Assume

$2x^2+3xy+y^2+5x+2y-3=(2x+y+a)(x+y+b).$

Expanding the right side gives $2x^2+3xy+y^2+(a+2b)x+(a+b)y+ab$. Comparing coefficients:

$a+2b=5,\qquad a+b=2,\qquad ab=-3.$

Subtracting, $b=3$, so $a=-1$, and $ab=-3$ is satisfied. Hence the equation is $(2x+y-1)(x+y+3)=0$, so the pair of lines is

$2x+y-1=0 \quad\text{and}\quad x+y+3=0.$

(ii) No. Putting $x=0,\ y=0$ in the equation gives $-3\ne0$, so the origin does not satisfy it. (Also, neither $2x+y-1=0$ nor $x+y+3=0$ passes through the origin.)

(iii) Subtracting $x+y+3=0$ from $2x+y-1=0$ gives $x-4=0$, so $x=4$. Then $4+y+3=0$ gives $y=-7$. The point of intersection is $(4,-7)$.

(iv) Mutually perpendicular unit vectors satisfy

$\vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=0,\qquad |\vec a|=|\vec b|=|\vec c|=1.$

(For a right-handed set, also $\vec a\times\vec b=\vec c$, $\vec b\times\vec c=\vec a$, $\vec c\times\vec a=\vec b$, so $\vec a\cdot(\vec b\times\vec c)=1$.)