Exercises — Distance from a point to a line

What we do: just read the coefficients off the standard form $ax+by+c=0$.

• $a=5$ (the number in front of $x$)
• $b=-12$ (front of $y$ — keep the minus sign)
• $c=26$ (the lone constant)

Since $a^2+b^2=25+144=169>0$, this really is a line and the formula applies. The normal length: $\sqrt{a^2+b^2}=\sqrt{5^2+(-12{)}^2}=\sqrt{25+144}=\sqrt{169}=13.$ This $13$ is the length of the arrow $(5,-12)$ that sticks out perpendicular to the line. We will divide by it in every distance calculation.

What we do: plug the point into the left side $ax+by+c$. If we get $0$, the point is on the line. $2(3)-(4)-2=6-4-2=0.$ The result is exactly $0$, so $P$ sits on the line. A point on the line has zero distance: $d=\frac{|0|}{\sqrt{2^2+(-1{)}^2}}=0.$ This is the degenerate case: no perpendicular to drop, because you are already there.

$a=3,b=4,c=-5,(x_1,y_1)=(7,-2)$. $d=\frac{|3(7)+4(-2)+(-5)|}{\sqrt{3^2+4^2}}=\frac{|21-8-5|}{\sqrt{25}}=\frac{|8|}{5}=\frac{8}{5}=\mathrm{1.6.}$ Why each step: substitute into the numerator (the line's "leftover" at $P$), divide by the normal length $5$, take absolute value because distance cannot be negative. Answer: $\frac{8}{5}$ units.

$a=8,b=-15,c=34,(x_1,y_1)=(0,0)$. $d=\frac{|8(0)-15(0)+34|}{\sqrt{8^2+(-15{)}^2}}=\frac{|34|}{\sqrt{64+225}}=\frac{34}{\sqrt{289}}=\frac{34}{17}=2.$ Note: when the point is the origin, the numerator collapses to just $|c|$. So distance from the origin is always $\frac{|c|}{\sqrt{a^2+b^2}}$.

Step 1 — convert to standard form. The formula only accepts $ax+by+c=0$. Move everything to one side: $y=2x+1\Rightarrow 2x-y+1=0.$ So $a=2,b=-1,c=1$.

Step 2 — apply. $d=\frac{|2(-1)+(-1)(3)+1|}{\sqrt{2^2+(-1{)}^2}}=\frac{|-2-3+1|}{\sqrt{5}}=\frac{|-4|}{\sqrt{5}}=\frac{4}{\sqrt{5}}=\frac{4\sqrt{5}}{5}\approx \mathrm{1.789.}$

Step 1 — build the line. Slope through $(1,1)$ and $(4,5)$: $m=\frac{5-1}{4-1}=\frac{4}{3}.$ Point–slope, then clear fractions: $y-1=\frac{4}{3}(x-1)\Rightarrow 3y-3=4x-4\Rightarrow 4x-3y-1=0.$ Step 2 — apply. $a=4,b=-3,c=-1$, point $(2,-3)$: $d=\frac{|4(2)+(-3)(-3)+(-1)|}{\sqrt{16+9}}=\frac{|8+9-1|}{5}=\frac{16}{5}=\mathrm{3.2.}$

Why they are parallel: both have the same $a,b=(6,8)$, so the same normal direction — they never meet. Trick: distance between parallels needs only the two constants (with matched $a,b$): $d=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}=\frac{|-9-5|}{\sqrt{6^2+8^2}}=\frac{14}{\sqrt{100}}=\frac{14}{10}=\mathrm{1.4.}$ Check the logic: pick any point on the first line, say where $x=0$: $8y-9=0\Rightarrow y=\frac{9}{8}$. Its distance to the second line is $\frac{|6(0)+8(9/8)+5|}{10}=\frac{|9+5|}{10}=1.4$. ✓ Same answer.

Step 1 — clear denominators to reach standard form. Multiply by $12$: $4x+3y=12\Rightarrow 4x+3y-12=0.$ $a=4,b=3,c=-12$. Step 2 — apply at $(6,6)$: $d=\frac{|4(6)+3(6)-12|}{\sqrt{16+9}}=\frac{|24+18-12|}{5}=\frac{30}{5}=6.$

Set up the equation from the formula: $2=\frac{|3(1)+4(2)+k|}{\sqrt{3^2+4^2}}=\frac{|11+k|}{5}.$ The denominator $\sqrt{3^2+4^2}=5$ is a fixed positive number, so we may multiply both sides by it without flipping anything: $|11+k|=10.$ The absolute value splits into two cases: $11+k=10\Rightarrow k=-1,11+k=-10\Rightarrow k=-21.$ Why two answers: the line can sit at distance $2$ on either side of $P$. Both $k=-1$ and $k=-21$ are valid — geometrically these are two parallel lines straddling the point.

A point on the $x$-axis is $(t,0)$. Set the two distances equal. $\frac{|3t-4(0)+12|}{\sqrt{9+16}}=\frac{|12t-5(0)-26|}{\sqrt{144+25}}$ $\frac{|3t+12|}{5}=\frac{|12t-26|}{13}.$ Why we may cross-multiply: both denominators, $\sqrt{25}=5$ and $\sqrt{169}=13$, are strictly positive numbers ($a^2+b^2>0$ for each line). Multiplying an equation through by a positive quantity preserves it exactly, so: $13|3t+12|=5|12t-26|,\text{i.e.}|39t+156|=|60t-130|.$ Remove the bars via two cases.

Case A (same sign): $39t+156=60t-130\Rightarrow 286=21t\Rightarrow t=\frac{286}{21}$. Case B (opposite sign): $39t+156=-(60t-130)\Rightarrow 99t=-26\Rightarrow t=-\frac{26}{99}$.

So there are two such points: $\left(\frac{286}{21},0\right)$ and $\left(-\frac{26}{99},0\right)$. Sense check on Case B: at $t=-\frac{26}{99}$, left side $=\frac{|3(-26/99)+12|}{5}=\frac{|-78/99+12|}{5}=\frac{1110/99}{5}=\frac{222}{99}=\frac{74}{33}$, and the right side gives the same $\frac{74}{33}$. ✓

Step 1 — line $AB$. $A(0,0)$ and $B(8,0)$ both have $y=0$, so line $AB$ is $y=0$, i.e. $0\cdot x+1\cdot y+0=0$: $a=0,b=1,c=0$. Here $a^2+b^2=0+1=1>0$, so it is a valid line. Step 2 — distance from $C(3,6)$: $d=\frac{|0(3)+1(6)+0|}{\sqrt{0^2+1^2}}=\frac{6}{1}=6.$ So the altitude from $C$ is $6$. Step 3 — area two ways. Base $AB=8$, height $6$: $\text{Area}=\frac{1}{2}\cdot 8\cdot 6=24.$ Cross-check with the coordinate area formula (see Area of Triangle Using Coordinates): $\text{Area}=\frac{1}{2}|x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)|=\frac{1}{2}|0+8(6)+3(0)|=\frac{1}{2}(48)=24.✓$

Set up: distance to $L_1$ is $2\times$ distance to $L_2$. Both normals have length $\sqrt{4^2+(-3{)}^2}=5$ and $\sqrt{3^2+4^2}=5$ — happily equal (and both positive, so we may cancel them safely). $\frac{|4x-3y+1|}{5}=2\cdot \frac{|3x+4y-7|}{5}.$ The $5$'s cancel: $|4x-3y+1|=2|3x+4y-7|.$ Squaring (or splitting the two sign cases) gives two straight lines (the locus is a pair):

Case A: $4x-3y+1=2(3x+4y-7)\Rightarrow 4x-3y+1=6x+8y-14\Rightarrow -2x-11y+15=0$, i.e. $2x+11y-15=0$. Case B: $4x-3y+1=-2(3x+4y-7)\Rightarrow 4x-3y+1=-6x-8y+14\Rightarrow 10x+5y-13=0$.

Locus: the two lines $2x+11y-15=0$ and $10x+5y-13=0$.

Step 1 — standard form. Multiply through by $pq$: $qx+py-pq=0$, so $a=q,b=p,c=-pq$. (With $p,q\ne 0$ we have $a^2+b^2=q^2+p^2>0$, a valid line.) Step 2 — origin distance (numerator is just $|c|$): $d=\frac{|-pq|}{\sqrt{q^2+p^2}}=\frac{|pq|}{\sqrt{p^2+q^2}}.$ Step 3 — match the target form. Divide top and bottom by $|pq|$: $d=\frac{1}{\sqrt{\frac{p^2+q^2}{p^2{q}^2}}}=\frac{1}{\sqrt{\frac{1}{q^2}+\frac{1}{p^2}}}.✓$ Numeric: $p=3,q=4$: $d=\frac{3\cdot 4}{\sqrt{9+16}}=\frac{12}{5}=2.4$.

The lines below are written as Prompt ::: Answer — read the part before the triple colon, try to answer it, then check against the part after. Distance from $(x_1,y_1)$ to $ax+by+c=0$ ::: $d=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$ Distance between parallels $ax+by+c_1=0$ and $ax+by+c_2=0$ (matched $a,b$) ::: $\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$ Why two answers when solving $|11+k|=10$ ::: the line can lie at that distance on either side of the point When is the formula valid at all ::: only when $a^2+b^2>0$, so that $ax+by+c=0$ is a genuine line