Chapter 9 · Straight Lines

9.3 Distance of a Point from a Line

The shortest way from a point to a line is along the perpendicular. Its length is $d=\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$, and two parallel lines $Ax+By+C_1=0$, $Ax+By+C_2=0$ are $\dfrac{|C_1-C_2|}{\sqrt{A^2+B^2}}$ apart.

A village stands $10\ \text{km}$ from a district town, and a new straight highway is planned past it. The village wants an access road — as short as possible — down to the highway. Where should it join, and how long will it be? Or think of a ship steaming on a straight course past a lighthouse: how close will it come, and when? Both questions ask for the distance of a point from a line, and both have the same answer: walk off at right angles. Any other route to the line is the hypotenuse of a right triangle whose other side is the perpendicular, so it is longer.

In this section we turn that geometric fact into a formula you can use without drawing: the distance of a point $(x_1,y_1)$ from a line in terms of the coefficients of its equation. We also find the point where the perpendicular meets the line (the foot of the perpendicular), and the distance between two parallel lines — the width of a straight road, a canal or a strip of land.

The Shortest Distance from a Point to a Line

📘 Definition 9.5 — Distance of a point from a line

Let $P$ be a point and $L$ a line. Draw the line through $P$ perpendicular to $L$, meeting $L$ at $N$. The point $N$ is the foot of the perpendicular from $P$ to $L$, and the length $PN$ is the (perpendicular) distance of $P$ from $L$. If $P$ lies on $L$, then $N=P$ and the distance is $0$.

Why the perpendicular is shortest. If $Q$ is any other point of $L$, the triangle $PNQ$ has a right angle at $N$, so by Pythagoras $PQ^2=PN^2+NQ^2\gt PN^2$. Every other path from $P$ to the line is longer.

For the formula we write the line with all its terms on one side, $$Ax+By+C=0\qquad(A,B\text{ not both }0).$$ Every equation in Section 9.2 can be rearranged like this: $y=mx+c$ becomes $mx-y+c=0$, and $x=x_1$ becomes $x-x_1=0$. When $B\ne0$ you can go back with $y=-\tfrac ABx-\tfrac CB$, so the slope is $-\tfrac AB$; when $B=0$ the line is vertical.

📐 Theorem 9.10 — Distance of a point from a line

The distance of the point $P(x_1,y_1)$ from the line $Ax+By+C=0$, where $A$ and $B$ are not both $0$, is $$d=\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.$$

Proof. Case $B=0$. The line is the vertical line $x=-\tfrac CA$, and the perpendicular from $P$ is horizontal, so $d=\left|x_1+\tfrac CA\right|=\dfrac{|Ax_1+C|}{|A|}$, which is the formula with $B=0$.

Case $B\ne0$. Write the line as $y=mx+c$ with $m=-\tfrac AB$ and $c=-\tfrac CB$. The vertical line through $P$ meets $L$ at $Q(x_1,\,mx_1+c)$, so the vertical distance is $PQ=|y_1-mx_1-c|$. If $m=0$ the line is horizontal and $PN=PQ$. Otherwise the triangle $PNQ$ has a right angle at $N$. Let $\varphi$ be the acute angle between $L$ and the $x$-axis, so $\tan\varphi=|m|$. The side $PQ$ is vertical and $PN$ is perpendicular to $L$; turning both the $x$-axis and $L$ through $90^\circ$ turns them into $PQ$ and $PN$, so $\angle QPN=\varphi$. Hence $$PN=PQ\cos\varphi=\dfrac{PQ}{\sqrt{1+\tan^2\varphi}}=\dfrac{|y_1-mx_1-c|}{\sqrt{1+m^2}},$$ which also covers $m=0$. Now substitute $m=-\tfrac AB$, $c=-\tfrac CB$: $$d=\dfrac{\left|y_1+\tfrac ABx_1+\tfrac CB\right|}{\sqrt{1+\tfrac{A^2}{B^2}}}=\dfrac{|Ax_1+By_1+C|/|B|}{\sqrt{A^2+B^2}/|B|}=\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.\ \blacksquare$$

The middle line of the proof is worth remembering on its own: for $y=mx+c$, $$d=\dfrac{|y_1-mx_1-c|}{\sqrt{1+m^2}}=\dfrac{\text{vertical distance}}{\sqrt{1+m^2}}.$$ The steeper the line, the more the vertical distance overstates the true distance.

🧭 Problem-solving strategy — Using the distance formula

Rewrite the line as $Ax+By+C=0$ — everything on one side, "$=0$" on the other.
Substitute the point into the left-hand side: $Ax_1+By_1+C$. Take the modulus.
Divide by $\sqrt{A^2+B^2}$ — the square root of the sum of the squares of the coefficients of $x$ and $y$ only (not $C$).
If the distance is given and something is unknown, $|\ \cdot\ |=k$ means "$=k$ or $=-k$": expect two answers.

✏️ Example 1 — Distances of points from lines

Find the distance of (a) $(2,-1)$ from $3x+4y-12=0$ (b) the origin from $5x-12y+26=0$ (c) $(-3,4)$ from the line $y=2x+5$.

💡 Show solution▶

(a) $d=\dfrac{|3(2)+4(-1)-12|}{\sqrt{9+16}}=\dfrac{|-10|}{5}=2$ units.

(b) $d=\dfrac{|0-0+26|}{\sqrt{25+144}}=\dfrac{26}{13}=2$ units. (For the origin the formula reduces to $\dfrac{|C|}{\sqrt{A^2+B^2}}$.)

(c) Check With the slope form: the vertical distance from $(-3,4)$ to the line is $|4-(2(-3)+5)|=5$, and $\dfrac{5}{\sqrt{1+2^2}}=\sqrt5$ ✓.

✏️ Example 2 — An altitude and an area

For the triangle with vertices $A(4,5)$, $B(1,1)$, $C(7,-1)$, find the length of the altitude from $A$ and hence the area of the triangle.

💡 Show solution▶

Line BC Slope $\dfrac{-1-1}{7-1}=-\dfrac13$: $y-1=-\tfrac13(x-1)$, so $3y-3=-x+1$, i.e. $x+3y-4=0$.

Altitude $h=\dfrac{|4+3(5)-4|}{\sqrt{1+9}}=\dfrac{15}{\sqrt{10}}$.

Area $BC=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}$, so the area is $\tfrac12\times2\sqrt{10}\times\dfrac{15}{\sqrt{10}}=15$ square units.

Check The Class 10 area formula gives $\tfrac12\left|4(1-(-1))+1(-1-5)+7(5-1)\right|=\tfrac12|8-6+28|=15$ ✓.

✏️ Example 3 — An unknown from a distance

Find the values of $k$ for which the point $(k,1)$ is at a distance $2$ from the line $4x+3y-1=0$.

💡 Show solution▶

Set up $\dfrac{|4k+3-1|}{5}=2$, so $|4k+2|=10$.

Solve $4k+2=10$ gives $k=2$; $4k+2=-10$ gives $k=-3$.

Check $(2,1)$: $\tfrac{|8+3-1|}{5}=2$ ✓; $(-3,1)$: $\tfrac{|-12+3-1|}{5}=2$ ✓. The two points lie on opposite sides of the line — the expression $4x+3y-1$ is $+10$ at one and $-10$ at the other.

The Foot of the Perpendicular

To find the point $N$ itself, use Sections 9.1 and 9.2: write the line through $P$ perpendicular to $L$ (its slope is the negative reciprocal of the slope of $L$), and solve the two equations together. Then check that $PN$ agrees with the distance formula.

✏️ Example 4 — Foot of the perpendicular and image

Find the foot of the perpendicular from $P(5,4)$ to the line $x+2y-3=0$, and the image (mirror reflection) of $P$ in the line.

💡 Show solution▶

Perpendicular The line has slope $-\tfrac12$, so the perpendicular through $P$ has slope $2$: $y-4=2(x-5)$, i.e. $y=2x-6$.

Intersect Substitute into $x+2y-3=0$: $x+2(2x-6)-3=0$, $5x=15$, $x=3$, $y=0$. So $N=(3,0)$.

Check $PN=\sqrt{2^2+4^2}=\sqrt{20}=2\sqrt5$, and the formula gives $\dfrac{|5+8-3|}{\sqrt5}=\dfrac{10}{\sqrt5}=2\sqrt5$ ✓.

Image The image $P'$ is as far beyond $N$ as $P$ is before it, so $N$ is the mid-point of $PP'$: $P'=(2\cdot3-5,\ 2\cdot0-4)=(1,-4)$. Check: $1+2(-4)-3=-10$, and $\tfrac{10}{\sqrt5}=2\sqrt5$ again ✓.

✏️ Example 5 — An access road to a highway

On a map with the district town at the origin and distances in km, a village $V$ is at $(6,8)$ and a new straight highway is to run along $3x+4y=12$. How long is the shortest access road from the village to the highway, and where does it meet the highway?

💡 Show solution▶

Length The highway is $3x+4y-12=0$: $d=\dfrac{|18+32-12|}{5}=\dfrac{38}{5}=7.6\ \text{km}$.

Direction The highway has slope $-\tfrac34$, so the access road has slope $\tfrac43$: $y-8=\tfrac43(x-6)$, i.e. $4x-3y=0$.

Junction Solve $3x+4y=12$, $4x-3y=0$: from the second $x=\tfrac34y$, so $\tfrac94y+4y=12$, $y=\tfrac{48}{25}=1.92$, $x=1.44$. The road meets the highway at $(1.44,\ 1.92)$.

Check From $(6,8)$ to $(1.44,1.92)$: $\sqrt{4.56^2+6.08^2}=\sqrt{57.76}=7.6\ \text{km}$ ✓.

Choose a line and drag the point $P$ (or press Play to send it on a loop). The green segment is the perpendicular $PN$; the orange dashed segment is the vertical distance $PQ$ to the line, and the purple angle at $P$ is $\varphi$. The readouts build the formula step by step — the value of $Ax_1+By_1+C$, then $\sqrt{A^2+B^2}$ — and compare it with $PN$ measured on the picture and with $PQ\cos\varphi$. Watch the sign of $Ax_1+By_1+C$ when $P$ crosses the line.

line

$P(x_1,y_1)$—

$Ax_1+By_1+C$—

$\sqrt{A^2+B^2}$—

formula $d$—

measured $PN$—

$PQ\cos\varphi$—

Distance between Parallel Lines

Two parallel lines are the same distance apart everywhere — that is what makes them parallel — so the distance between them is the distance of any point of one from the other. Choosing the point cleverly gives a formula.

📐 Theorem 9.11 — Distance between parallel lines

The distance between the parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ (same $A$ and $B$, not both $0$) is $$d=\dfrac{|C_1-C_2|}{\sqrt{A^2+B^2}}.$$ In slope-intercept form, the distance between $y=mx+c_1$ and $y=mx+c_2$ is $\dfrac{|c_1-c_2|}{\sqrt{1+m^2}}$.

Proof. Let $(x_0,y_0)$ be any point of the first line, so $Ax_0+By_0=-C_1$. By Theorem 9.10 its distance from the second line is $$\dfrac{|Ax_0+By_0+C_2|}{\sqrt{A^2+B^2}}=\dfrac{|-C_1+C_2|}{\sqrt{A^2+B^2}}=\dfrac{|C_1-C_2|}{\sqrt{A^2+B^2}}.$$ The second statement is the case $A=m$, $B=-1$, $C_1=c_1$, $C_2=c_2$ (the lines $mx-y+c_1=0$, $mx-y+c_2=0$). $\blacksquare$

✏️ Example 6 — Parallel lines

(a) Find the distance between the lines $3x-4y+7=0$ and $6x-8y-1=0$.
(b) Find the lines parallel to $3x-4y+7=0$ at a distance $2$ from it.

💡 Show solution▶

(a) Match The slopes are both $\tfrac34$, so the lines are parallel — but the coefficients differ. Divide the second by $2$: $3x-4y-\tfrac12=0$. Now $A=3$, $B=-4$ in both.

(a) Distance $d=\dfrac{\left|7-\left(-\tfrac12\right)\right|}{5}=\dfrac{7.5}{5}=\dfrac32$ units. (Using $|7-(-1)|$ without matching the coefficients would wrongly give $\tfrac85$.)

(b) A parallel line is $3x-4y+C=0$. We need $\dfrac{|C-7|}{5}=2$, so $C-7=\pm10$, $C=17$ or $C=-3$: the lines $3x-4y+17=0$ and $3x-4y-3=0$, one on each side.

Check $(1,0)$ is on $3x-4y-3=0$, and its distance from $3x-4y+7=0$ is $\tfrac{|3+7|}{5}=2$ ✓.

Two parallel lines $y=mx+c_1$ (red) and $y=mx+c_2$ (blue) bound a strip. A point $R$ slides along the red line; the green perpendicular from $R$ to the blue line keeps exactly the same length wherever $R$ is — that common length is the distance between the lines. The orange vertical gap is always $|c_1-c_2|$; dividing it by $\sqrt{1+m^2}$ gives the true width. Make the lines steeper and watch the vertical gap overstate the width more and more.

slope $m$0.5$c_1$2$c_2$−1.5

vertical gap $|c_1-c_2|$—

$\sqrt{1+m^2}$—

formula $d$—

measured from $R$—

⚠️ Common mistakes with distances

1. Using the formula before rearranging. For $y=2x+5$, the expression to substitute into is $2x-y+5$, not $2x+5$. Get "$=0$" first.

2. Putting $C$ under the square root. The denominator is $\sqrt{A^2+B^2}$ — the coefficients of $x$ and $y$ only.

3. Forgetting the modulus. Distances are never negative. Without $|\ |$, a point on the "other side" gives a negative answer.

4. Unmatched coefficients for parallel lines. $|C_1-C_2|/\sqrt{A^2+B^2}$ needs the same $A$ and $B$ in both equations. Scale one equation first, as in Example 6.

5. Confusing the vertical distance with the perpendicular distance. $|y_1-(mx_1+c)|$ is the vertical gap; the distance is that divided by $\sqrt{1+m^2}$. They agree only for horizontal lines.

🚀 Beyond the board syllabus (JEE Advanced)

In JEE Advanced 2026 (angle bisectors and the incentre); not in JEE Main 2026 or the CBSE year-end exam. JEE Advanced uses the distance formula to find angle bisectors and the incentre of a triangle.

Angle bisectors. A point lies on a bisector of an angle between two lines exactly when it is equidistant from them. So the two bisectors of the angles between $A_1x+B_1y+C_1=0$ and $A_2x+B_2y+C_2=0$ are $$\dfrac{A_1x+B_1y+C_1}{\sqrt{A_1^2+B_1^2}}=\pm\dfrac{A_2x+B_2y+C_2}{\sqrt{A_2^2+B_2^2}},$$ one for each sign; the two bisectors are always perpendicular to each other.

Worked example. Find the bisectors of the angles between $3x+4y-7=0$ and $4x-3y+1=0$.
Solution. Both denominators are $5$, so $3x+4y-7=\pm(4x-3y+1)$. With $+$: $-x+7y-8=0$, i.e. $x-7y+8=0$. With $-$: $7x+y-6=0$. Check: the given lines meet where $3x+4y=7$ and $4x-3y=-1$, i.e. at $(0.68,\ 1.24)$; both bisectors pass through it ($0.68-8.68+8=0$ and $4.76+1.24-6=0$ ✓), and their slopes $\tfrac17$ and $-7$ multiply to $-1$ ✓.

Exercises

Find the distance of the point from the line.

1 $(0,0)$ from $3x+4y-10=0$.
2 $(1,2)$ from $4x-3y+7=0$.
3 $(-2,3)$ from $5x+12y+1=0$.
4 $(3,-1)$ from $x=-2$.
5 $(4,1)$ from $y=3x-1$.
6 $(2,5)$ from the line through $(0,1)$ and $(3,-3)$.
7 $(1,1)$ from $12x-5y+6=0$.
8 The origin from the line $\dfrac xa+\dfrac yb=1$ ($a,b\ne0$).

💡 Show solutions▶

1 $\dfrac{|-10|}{5}=2$.

2 $\dfrac{|4-6+7|}{5}=1$.

3 $\dfrac{|-10+36+1|}{13}=\dfrac{27}{13}$.

4 The line is vertical: $|3-(-2)|=5$. (Formula: $x+2=0$, $\tfrac{|3+2|}{1}=5$ ✓.)

5 $3x-y-1=0$: $\dfrac{|12-1-1|}{\sqrt{10}}=\dfrac{10}{\sqrt{10}}=\sqrt{10}\approx3.16$.

6 Slope $-\tfrac43$, $y$-intercept $1$: $y=-\tfrac43x+1$, i.e. $4x+3y-3=0$. Distance $\dfrac{|8+15-3|}{5}=4$.

7 $\dfrac{|12-5+6|}{13}=1$.

8 Multiply by $ab$: $bx+ay-ab=0$. Distance $\dfrac{|ab|}{\sqrt{a^2+b^2}}$.

Unknowns, feet of perpendiculars and triangles.

1 Find $k$ if the distance of $(2,k)$ from $3x-4y+4=0$ is $2$.
2 Find $c$ if the line $5x+12y+c=0$ is at distance $3$ from the origin.
3 Find the foot of the perpendicular from the origin to $3x+4y=25$.
4 Find the foot of the perpendicular from $(1,8)$ to $x-y+1=0$, and check the distance with the formula.
5 Find the image of $(1,8)$ in the line $x-y+1=0$.
6 Find the points on the $y$-axis at distance $3$ from the line $4x-3y+5=0$.
7 For the triangle $A(0,4)$, $B(-2,0)$, $C(4,1)$, find the altitude from $A$ and the area.

💡 Show solutions▶

1 $\dfrac{|6-4k+4|}{5}=2$, so $10-4k=\pm10$: $k=0$ or $k=5$.

2 $\dfrac{|c|}{13}=3$: $c=39$ or $c=-39$.

3 The perpendicular through $O$ has slope $\tfrac43$: $y=\tfrac43x$. Then $3x+\tfrac{16}3x=25$, $x=3$, $y=4$: the foot is $(3,4)$ (and the distance is $5=\tfrac{25}{5}$ ✓).

4 The line has slope $1$; the perpendicular is $y-8=-(x-1)$, $y=-x+9$. Then $x-(-x+9)+1=0$, $x=4$, $y=5$: foot $(4,5)$. $PN=\sqrt{9+9}=3\sqrt2$; formula $\tfrac{|1-8+1|}{\sqrt2}=\tfrac{6}{\sqrt2}=3\sqrt2$ ✓.

5 $(4,5)$ is the mid-point of $(1,8)$ and its image: $(2\cdot4-1,\ 2\cdot5-8)=(7,2)$.

6 $(0,y)$: $\dfrac{|-3y+5|}{5}=3$, so $5-3y=\pm15$: $y=-\tfrac{10}3$ or $y=\tfrac{20}3$. Points $\left(0,-\tfrac{10}3\right)$ and $\left(0,\tfrac{20}3\right)$.

7 $BC$: slope $\tfrac16$, $y=\tfrac16(x+2)$, i.e. $x-6y+2=0$. Altitude $\dfrac{|0-24+2|}{\sqrt{37}}=\dfrac{22}{\sqrt{37}}$. $BC=\sqrt{36+1}=\sqrt{37}$, so the area is $\tfrac12\sqrt{37}\cdot\tfrac{22}{\sqrt{37}}=11$ square units.

III

Parallel lines.

1 Find the distance between $3x+4y-7=0$ and $3x+4y+8=0$.
2 Find the distance between $5x-12y+2=0$ and $10x-24y-22=0$.
3 Find the distance between $y=2x+3$ and $y=2x-7$.
4 Find the distance between $x=4$ and $x=-1$.
5 Find the lines parallel to $8x-15y+1=0$ at a distance $2$ from it.
6 Find the line midway between $3x-4y+2=0$ and $3x-4y-8=0$.
7 The two edges of a straight road lie along $y=0.75x+2$ and $y=0.75x-5.5$ (in metres). How wide is the road?

Check: the distance from a point to a line $Ax+By+C=0$

$A$$B$$C$$x_1$$y_1$

$\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$—

foot of perpendicular $N$—

$PN$ from the coordinates—

💡 Show solutions▶

1 $\dfrac{|-7-8|}{5}=3$.

2 Halve the second: $5x-12y-11=0$. Distance $\dfrac{|2-(-11)|}{13}=1$.

3 $\dfrac{|3-(-7)|}{\sqrt{1+4}}=\dfrac{10}{\sqrt5}=2\sqrt5\approx4.47$.

4 Both vertical: $|4-(-1)|=5$.

5 $8x-15y+C=0$ with $\dfrac{|C-1|}{17}=2$: $C=35$ or $C=-33$. Lines $8x-15y+35=0$ and $8x-15y-33=0$.

6 Midway means $C=\tfrac{2+(-8)}{2}=-3$: $3x-4y-3=0$. Its distance from each line is $\tfrac{5}{5}=1$, half of the gap $\tfrac{10}{5}=2$ ✓.

7 $\dfrac{|2-(-5.5)|}{\sqrt{1+0.5625}}=\dfrac{7.5}{1.25}=6\ \text{m}$.

Applications and problems.

1 A ship sails along the straight line $x+y=10$, with a lighthouse at the origin (distances in km). How close does the ship come to the lighthouse, and where is it then?
2 Show that the point $(3,2)$ is equidistant from the lines $3x-4y+9=0$ and $4x+3y-8=0$.
3 Two sides of a square lie along $2x-y+3=0$ and $2x-y-7=0$. Find the length of a side and the area of the square.
4 A rectangular park is bounded by the lines $x-2y+4=0$, $x-2y-6=0$, $2x+y-2=0$ and $2x+y+8=0$, with one unit representing $10\ \text{m}$. Find the area of the park in square metres.
5 Find the points on the line $y=x$ at distance $2$ from the line $3x+4y-2=0$.
6 Line $l_1$ passes through $(0,3)$ and line $l_2$ through $(4,0)$, both with slope $2$. Find the distance between them.

💡 Show solutions▶

1 $\dfrac{|0+0-10|}{\sqrt2}=5\sqrt2\approx7.07\ \text{km}$. The perpendicular from $O$ is $y=x$, which meets $x+y=10$ at $(5,5)$.

2 $\dfrac{|9-8+9|}{5}=2$ and $\dfrac{|12+6-8|}{5}=2$: equal.

3 The two lines are parallel; side $=\dfrac{|3-(-7)|}{\sqrt5}=\dfrac{10}{\sqrt5}=2\sqrt5$; area $=20$ square units.

4 The pairs are parallel and the slopes $\tfrac12$, $-2$ multiply to $-1$, so the park is a rectangle. Widths $\dfrac{|4-(-6)|}{\sqrt5}=2\sqrt5$ and $\dfrac{|-2-8|}{\sqrt5}=2\sqrt5$ units — in fact a square. Area $20$ square units $=20\times100=2000\ \text{m}^2$.

5 $(t,t)$: $\dfrac{|7t-2|}{5}=2$, so $7t-2=\pm10$, $t=\tfrac{12}7$ or $t=-\tfrac87$: the points $\left(\tfrac{12}7,\tfrac{12}7\right)$ and $\left(-\tfrac87,-\tfrac87\right)$.

6 $l_1$: $y=2x+3$, i.e. $2x-y+3=0$; $l_2$: $y=2(x-4)$, i.e. $2x-y-8=0$. Distance $\dfrac{|3-(-8)|}{\sqrt5}=\dfrac{11}{\sqrt5}\approx4.92$.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask for the distance of a point from a line or between two parallel lines (1–2 marks, often with a line that must first be rearranged or rescaled), and for the foot of the perpendicular or the image of a point (3 marks). JEE combines the distance formula with unknowns, gives two answers through the modulus, and uses equal distances for incentres and bisectors.

E1. Board · 1 mark The distance between the parallel lines $4x+3y=11$ and $8x+6y=15$ is (A) $\tfrac7{10}$ (B) $\tfrac45$ (C) $\tfrac75$ (D) $\tfrac{26}5$

💡 Show solution▶

Idea Halve the second line: $4x+3y-7.5=0$; the first is $4x+3y-11=0$. $d=\dfrac{|-11+7.5|}{5}=\dfrac{3.5}{5}=\dfrac7{10}$.

Trap (B) uses $\tfrac{|11-15|}{5}$ without matching the coefficients; (D) also adds the constants.

Answer (A).

E2. Board · 2 marks Find the distance of the point $(2,-3)$ from the line $4(x-1)=3(y+2)$.

💡 Show solution▶

Rearrange $4x-4=3y+6$, so $4x-3y-10=0$. (1 mark)

Answer $d=\dfrac{|8+9-10|}{\sqrt{16+9}}=\dfrac75$ units. (1 mark)

E3. Board · 3 marks Find the coordinates of the foot of the perpendicular from $P(2,7)$ to the line $x+2y-6=0$, and the image of $P$ in this line.

💡 Show solution▶

Perpendicular The line has slope $-\tfrac12$; the perpendicular through $P$ has slope $2$: $y-7=2(x-2)$, i.e. $y=2x+3$. (1 mark)

Foot $x+2(2x+3)-6=0$ gives $5x=0$: $N=(0,3)$. Check: $PN=\sqrt{4+16}=2\sqrt5=\tfrac{|2+14-6|}{\sqrt5}$ ✓. (1 mark)

Answer Foot $(0,3)$; image $(2\cdot0-2,\ 2\cdot3-7)=(-2,-1)$. (1 mark)

E4. Board · Case study A lighthouse stands at the origin $O$ of a chart marked in kilometres. A cargo ship holds a straight course along the line $3x+4y=60$. (i) How close does the ship come to the lighthouse? (1 mark) (ii) At which point of its course is it closest? (1 mark) (iii) The lighthouse beam can be seen up to $13\ \text{km}$ away. For how many kilometres of its course can the ship see the beam? (2 marks)

💡 Show solution▶

(i) $d=\dfrac{|0+0-60|}{5}=12\ \text{km}$. (1 mark)

(ii) The perpendicular from $O$ has slope $\tfrac43$: $y=\tfrac43x$. Then $3x+\tfrac{16}3x=60$, $x=7.2$, $y=9.6$: the point $(7.2,\ 9.6)$. (1 mark)

(iii) Idea The ship sees the beam while its distance from $O$ is at most $13$. The visible stretch is centred on $N(7.2,9.6)$; at either end, the right triangle $O$–$N$–(end) has hypotenuse $13$ and one leg $12$. (1 mark)

Answer The other leg is $\sqrt{13^2-12^2}=5$, so the ship sees the beam for $2\times5=10\ \text{km}$ of its course. (1 mark)

E5. JEE Main The lines parallel to $3x+4y+5=0$ at a distance of $2$ units from the point $(1,1)$ are $3x+4y+C=0$. The product of the possible values of $C$ is (A) $-51$ (B) $51$ (C) $-14$ (D) $30$

💡 Show solution▶

Method Distance of $(1,1)$ from $3x+4y+C=0$: $\dfrac{|3+4+C|}{5}=2$, so $7+C=\pm10$: $C=3$ or $C=-17$.

Trap (C) is the sum of the values. The given line $3x+4y+5=0$ only fixes the direction; its constant $5$ plays no part.

Answer (A) $3\times(-17)=-51$.

E6. JEE Advanced Numerical answer. The circle inscribed in the triangle formed by the lines $x=0$, $y=0$ and $3x+4y=12$ has centre $(r,r)$, where $r$ is its radius. Find $r$.

💡 Show solution▶

Idea The centre $(r,r)$ is already at distance $r$ from both axes; it must also be at distance $r$ from the third side: $\dfrac{|3r+4r-12|}{5}=r$, i.e. $|7r-12|=5r$.

Two cases $12-7r=5r$ gives $r=1$; $7r-12=5r$ gives $r=6$. The point $(6,6)$ satisfies $3x+4y\gt12$, so it lies outside the triangle (it is the centre of a circle touching the three lines from outside).

⏱ Fast route For a right triangle with legs $3$, $4$ and hypotenuse $5$, $r=\tfrac{3+4-5}{2}=1$.

Answer $r=\mathbf{1}$.

🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. How far is your school (or home) from the nearest straight stretch of a highway, railway line or canal — and does the distance formula agree with a measurement on a map?

Why it matters. "Distance to the nearest line" is one of the most used calculations in planning: noise and pollution buffers along highways, safety zones beside railway tracks, the length of a pipeline to join a main line, flood zones beside rivers and the closest approach of a ship to a coast or of an aircraft to a mountain. Mapping software does exactly this calculation.

What to do.

Open an online map (or a printed town map with a scale) and find a straight stretch of a road, railway line or canal near your school.
Set up axes: put the origin at a convenient landmark, with $x$ east and $y$ north, and use the map scale (or the map's distance tool) to read the coordinates, in metres, of two points on the straight stretch and of your school.
Find the equation of the line through the two points in the form $Ax+By+C=0$, and compute the distance of your school from it with Theorem 9.10. Find the foot of the perpendicular as well.
Measure the shortest distance directly with the map's distance tool and compare.
Repeat for a second landmark (a hospital, a park) and comment on the size and causes of any differences.

Tools: An online map with a distance tool (or a scaled paper map and ruler), a calculator or spreadsheet

Hand in: A sketch of your axes and points, the equation of the line, the computed distance and foot, the measured distance, and a short discussion of the errors.

Going further. Real roads are not perfectly straight. Split a curving road into three straight segments, find the distance of your school from each segment's line, and explain why the true shortest distance might be to an end point of a segment rather than to a foot of a perpendicular.

Instructor note: with coordinates read to the nearest $10\ \text{m}$, computed and measured distances usually agree to within a few per cent for distances of several hundred metres; larger gaps come from misread scales or from a road that bends between the two chosen points. The "going further" question introduces the idea that the foot of the perpendicular may fall outside a segment — then the nearest point of the segment is an end point.

🧪 Self-check quiz

0/5 correct

Q1. The distance of the origin from the line $3x-4y+15=0$ is

Q2. The distance between the lines $2x+y-4=0$ and $4x+2y+2=0$ is

Q3. The distance of the point $(3,4)$ from the $x$-axis is

Q4. To find the distance of $(2,-1)$ from the line $y=3x+2$ with the formula, you first write the line as

Q5. The foot of the perpendicular from $(4,0)$ to the line $y=x$ is

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