Equations, Slope & Family of Lines

Slope $m=\tan\theta$, where $\theta$ is the angle the line makes with the positive x-axis, measured anticlockwise, $0^\circ\le\theta<180^\circ$. So a line leaning to the left has $\theta$ obtuse and $m$ negative. Computed from any two points on the line, $m=\dfrac{y_2-y_1}{x_2-x_1}$ — which pair you pick makes no difference, and that is what makes slope a property of the line rather than of the points. Forms:

• Slope-intercept: $y=mx+c$ — use when you know the slope and the y-intercept $c$.
• Point-slope: $y-y_1=m(x-x_1)$ — use when you know one point and the slope.
• Two-point: $\dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}$ — use when you know two points.
• General: $ax+by+c=0$, slope $-a/b$. Careful: this $c$ is the constant term, not the y-intercept it meant one line above.
• Normal (perpendicular) form: $x\cos\alpha+y\sin\alpha=p$, where $p$ is the distance from the origin to the line and $\alpha$ is the angle that perpendicular makes with the x-axis — a different angle from $\theta$.

An intercept is a number, not a point: the x-intercept is the value of $x$ at which the line crosses the x-axis. That is why a question can ask for their sum.

• Intercept form: $\dfrac{x}{a}+\dfrac{y}{b}=1$, where $a$ is the x-intercept and $b$ the y-intercept — readable off the equation only when the right-hand side is exactly $1$.
• From a general line $px+qy+r=0$: x-intercept $=-r/p$, y-intercept $=-r/q$. The letters are $p,q,r$ here on purpose — in the line above, $a$ and $b$ are intercepts, while in a general equation $a$ and $b$ are coefficients. They are not the same thing, and confusing them is this concept's classic error.
• From a midpoint: if $(h,k)$ is the midpoint of the segment the line cuts between the axes, then $a=2h$ and $b=2k$.
• Distance from the origin to $\tfrac xa+\tfrac yb=1$: $p=\dfrac{|ab|}{\sqrt{a^2+b^2}}$, i.e. $\dfrac1{p^2}=\dfrac1{a^2}+\dfrac1{b^2}$ — the altitude-to-the-hypotenuse relation of the right triangle the line cuts off. A stem that gives the intercepts and asks for $p$ (or gives $p$ and one intercept) is this line.

• Family (pencil): every line through the intersection of $L_1=0$ and $L_2=0$ can be written $L_1+\lambda L_2=0$, so you never have to find the intersection point. Fix $\lambda$ from one extra condition. (The single line it misses is $L_2=0$ itself.)
• Through a point, parallel or perpendicular to a given line: keep the same slope, or the negative-reciprocal one. Both tests are proved in the next block, Angle Between Lines — for now, parallel means equal slopes and perpendicular means the slopes multiply to $-1$.
• Concurrent means all three lines pass through one common point. If three lines are concurrent, then $\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=0$. The converse is not automatic: the determinant also vanishes when two of the lines are parallel, so a zero determinant is evidence, not proof. Expanding a $3\times3$ determinant is in the Matrices & Determinants notes, linked below.
• A fixed point forced by a coefficient condition: if $A,B,C$ are in AP then $C-B=B-A$, so $C=2B-A$. Put that into $Ax+2By+C=0$: $Ax+2By+2B-A=0$, which regroups as $A(x-1)+2B(y+1)=0$. That holds for every $A$ and $B$ only when $x-1=0$ and $y+1=0$ — so every such line passes through $(1,-1)$.
• How many intersection points can $n$ lines make? At most $\binom n2$ (every pair meets once, no two parallel, no three concurrent); $n$ lines and a circle add at most $2n$ more (each line cuts the circle twice); $n$ circles meet in at most $2\binom n2=n(n-1)$ points.

If $P'$ is the image of $P$ in the line $L$, two facts pin it down: the midpoint of $PP'$ lies on $L$, and $PP'$ is perpendicular to $L$. Write $P'=(h,k)$, turn each fact into an equation, and solve the pair.

• Two things borrowed from later blocks, stated here so this method is usable now: the midpoint of $(x_1,y_1)$ and $(x_2,y_2)$ is $\left(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\right)$, and perpendicular means the two slopes multiply to $-1$.
• The foot of the perpendicular $F$ from $P$ to $L$ is the midpoint of $PP'$, so $P'=2F-P$.
• Shortcuts worth memorising: reflecting in $y=x$ swaps coordinates, $(a,b)\to(b,a)$; in the x-axis, $(a,b)\to(a,-b)$; in the y-axis, $(a,b)\to(-a,b)$.
• Backwards: given $P$ and its image $P'$, the mirror is the perpendicular bisector of $PP'$.