Chapter 9 · Straight Lines

9.2 Various Forms of the Equation of a Line

A line is fixed by a point and a slope, by two points, by its slope and $y$-intercept, or by its two intercepts — and each way of fixing it gives a form of its equation: $y-y_1=m(x-x_1)$, $y=mx+c$, $\dfrac xa+\dfrac yb=1$.

Water freezes at $0^\circ\text{C}$, which is $32^\circ\text{F}$, and boils at $100^\circ\text{C}$, which is $212^\circ\text{F}$. Both scales are evenly spaced, so if you plot every temperature as a point $(C,F)$ the points lie on one straight line — and two points are enough to fix a line. From those two facts alone you can find the rule that converts any temperature, the reading of a fever of $37^\circ\text{C}$ in Fahrenheit, and the one temperature that reads the same on both scales. All that is needed is the equation of the line through two given points.

In Section 9.1 we described a line by its slope. Now we describe it completely, by an equation in $x$ and $y$ that its points satisfy and no other points do. There are several standard forms, one for each common way of fixing a line: by a point and a slope, by two points, by the slope and the point where it crosses the $y$-axis, or by the two points where it crosses the axes. You will learn each form, why it is true, and how to choose the quickest one for the information you are given.

Equations of Lines; Horizontal and Vertical Lines

📘 Definition 9.3 — Equation of a line

An equation of a line $L$ is an equation in $x$ and $y$ such that a point $(x,y)$ lies on $L$ if and only if its coordinates satisfy the equation. So every point of $L$ satisfies it, and no point off $L$ does.

Both halves matter. The equation $y=2x$ is satisfied by $(1,2)$, so $(1,2)$ is on that line; it is not satisfied by $(2,1)$, so $(2,1)$ is not. To test whether a point lies on a line, substitute its coordinates.

📐 Theorem 9.5 — Lines parallel to the axes

The horizontal line through $(x_1,y_1)$ — the line parallel to the $x$-axis — has equation $$y=y_1,$$ and the vertical line through $(x_1,y_1)$ — parallel to the $y$-axis — has equation $$x=x_1.$$ In particular the $x$-axis is $y=0$ and the $y$-axis is $x=0$.

Reason. Every point of the horizontal line is at the same height as $(x_1,y_1)$, so it has $y$-coordinate $y_1$, whatever its $x$-coordinate; and any point with $y$-coordinate $y_1$ is on that line. The vertical case is the same with the roles of $x$ and $y$ swapped. $\blacksquare$

✏️ Example 1 — Lines parallel to the axes

(a) Write the equations of the lines through $(-2,3)$ parallel to the $x$-axis and to the $y$-axis.
(b) Find the equation of the line parallel to the $y$-axis and $4$ units to the right of it, and of the line parallel to the $x$-axis and $5$ units below it.

💡 Show solution▶

(a) Parallel to the $x$-axis (horizontal): $y=3$. Parallel to the $y$-axis (vertical): $x=-2$.

(b) $4$ units to the right of the $y$-axis: every point has $x=4$, so the line is $x=4$. Five units below the $x$-axis: $y=-5$.

Check $(-2,3)$ satisfies both $y=3$ and $x=-2$ ✓ — it is where the two lines of (a) cross.

The Point-Slope and Two-Point Forms

Through a given point there is exactly one line with a given inclination, hence exactly one non-vertical line with a given slope. Its equation comes straight from the slope formula.

📐 Theorem 9.6 — Point-slope form

The line through $P(x_1,y_1)$ with slope $m$ has equation $$y-y_1=m(x-x_1).$$

Proof. Call the line $L$, and let $Q(x,y)$ be any point other than $P$. If $Q$ is on $L$, then $Q$ is not directly above or below $P$ (that would make $L$ vertical), so $x\ne x_1$, and by Theorem 9.1 the slope of $PQ$ is $\dfrac{y-y_1}{x-x_1}=m$; multiplying by $x-x_1$ gives $y-y_1=m(x-x_1)$. Conversely, if $Q$ satisfies $y-y_1=m(x-x_1)$ then $x\ne x_1$ (otherwise $y=y_1$ too and $Q=P$), so $PQ$ has slope $m$; the line $PQ$ passes through $P$ with slope $m$, so it is $L$, and $Q$ lies on $L$. Finally $P$ itself satisfies the equation ($0=m\cdot0$). $\blacksquare$

📐 Theorem 9.7 — Two-point form

The line through $P(x_1,y_1)$ and $Q(x_2,y_2)$, where $x_1\ne x_2$, has equation $$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}\,(x-x_1).$$ (If $x_1=x_2$, the line is the vertical line $x=x_1$.)

Proof. By Theorem 9.1 the line has slope $m=\dfrac{y_2-y_1}{x_2-x_1}$; now use the point-slope form at $P$. $\blacksquare$

✏️ Example 2 — A point and a slope; two points

(a) Find the equation of the line through $(-2,3)$ with slope $-4$.
(b) Find the equation of the line through $(1,-1)$ and $(3,5)$. Does $(-1,-7)$ lie on it?

💡 Show solution▶

(a) $y-3=-4(x-(-2))=-4(x+2)$, so $y-3=-4x-8$, i.e. $y=-4x-5$, or $4x+y+5=0$. Watch the sign: $x-x_1$ with $x_1=-2$ is $x+2$.

(b) Slope $m=\dfrac{5-(-1)}{3-1}=3$. Point-slope at $(1,-1)$: $y+1=3(x-1)$, so $y=3x-4$.

(b) Check The other given point: $3\cdot3-4=5$ ✓. Using $(3,5)$ instead gives $y-5=3(x-3)$, the same line $y=3x-4$.

(b) Test At $x=-1$: $3(-1)-4=-7$, so $(-1,-7)$ lies on the line.

✏️ Example 3 — Celsius and Fahrenheit

Temperature in degrees Fahrenheit, $F$, is a linear function of temperature in degrees Celsius, $C$. Water freezes at $(C,F)=(0,32)$ and boils at $(100,212)$. (a) Find the equation connecting $F$ and $C$. (b) Convert a body temperature of $37^\circ\text{C}$ to Fahrenheit. (c) Which temperature has the same reading on both scales?

💡 Show solution▶

(a) Treat $C$ as $x$ and $F$ as $y$. Slope $\dfrac{212-32}{100-0}=\dfrac{180}{100}=\dfrac95$. Two-point form at $(0,32)$: $F-32=\tfrac95(C-0)$, so $F=\tfrac95C+32$.

(b) $F=\tfrac95\times37+32=66.6+32=98.6$, so $37^\circ\text{C}=98.6^\circ\text{F}$.

Meaning The slope $\tfrac95$ says each Celsius degree is $1.8$ Fahrenheit degrees; geometrically, (c) is the point where this line meets the line $F=C$.

Drag $A$ and $B$ (they snap to whole numbers), or press Play to move $B$. The readouts write the same line in every form of this section. The green dot marks the $y$-intercept and the orange dot the $x$-intercept. Look for the positions where a form breaks down: a vertical line has no slope-intercept form, and a line through the origin, or one parallel to an axis, has no intercept form.

point-slope (at A)—

two-point—

slope-intercept—

intercept form—

all terms on one side—

The Slope-Intercept Form

📘 Definition 9.4 — Intercepts

If a line meets the $x$-axis at $(a,0)$, the number $a$ is its $x$-intercept; if it meets the $y$-axis at $(0,b)$, the number $b$ is its $y$-intercept. Intercepts are signed numbers, not lengths: a line crossing the $y$-axis at $(0,-3)$ has $y$-intercept $-3$.

📐 Theorem 9.8 — Slope-intercept form

(i) The line with slope $m$ and $y$-intercept $c$ has equation $$y=mx+c.$$ (ii) The line with slope $m$ and $x$-intercept $d$ has equation $$y=m(x-d).$$ (iii) Conversely, for any real numbers $m$ and $c$, the equation $y=mx+c$ is the equation of the line with slope $m$ and $y$-intercept $c$.

Proof. (i) The line passes through $(0,c)$ with slope $m$, so by Theorem 9.6 its equation is $y-c=m(x-0)$, i.e. $y=mx+c$. (ii) Similarly through $(d,0)$: $y-0=m(x-d)$. (iii) The line with slope $m$ and $y$-intercept $c$ has equation $y=mx+c$ by (i), so the points satisfying $y=mx+c$ are exactly the points of that line. $\blacksquare$

Part (iii) is how you read a line: make $y$ the subject, and the coefficient of $x$ is the slope and the constant is the $y$-intercept. For example $3x-4y+12=0$ gives $4y=3x+12$, $y=\tfrac34x+3$: slope $\tfrac34$, $y$-intercept $3$. Every non-vertical line can be written as $y=mx+c$; a vertical line $x=x_1$ cannot, because it has no slope.

✏️ Example 4 — Writing and reading slope-intercept form

(a) Find the equation of the line with slope $-\tfrac12$ and $y$-intercept $3$.
(b) Find the equation of the line with inclination $60^\circ$ that meets the $x$-axis at $(2,0)$.
(c) Find the slope and both intercepts of the line $3x-4y+12=0$, and use them to sketch it.

💡 Show solution▶

(a) $y=-\tfrac12x+3$; clearing the fraction, $2y=-x+6$, i.e. $x+2y-6=0$.

(b) $m=\tan60^\circ=\sqrt3$ and $x$-intercept $2$: $y=\sqrt3(x-2)$, i.e. $\sqrt3x-y-2\sqrt3=0$.

Sketch Plot $(-4,0)$ and $(0,3)$ and join them. Check: the slope from $(-4,0)$ to $(0,3)$ is $\tfrac34$ ✓.

The Intercept Form

📐 Theorem 9.9 — Intercept form

The line with $x$-intercept $a$ and $y$-intercept $b$, where $a\ne0$ and $b\ne0$, has equation $$\dfrac xa+\dfrac yb=1.$$

Proof. The line passes through $(a,0)$ and $(0,b)$, and $a\ne0$, so by the two-point form $$y-0=\dfrac{b-0}{0-a}(x-a),\quad\text{i.e.}\quad ay=-bx+ab.$$ So $bx+ay=ab$, and dividing by $ab\ne0$ gives $\dfrac xa+\dfrac yb=1$. $\blacksquare$

The intercept form is the quickest way to sketch a line — the intercepts are written in the equation — and the natural form whenever a problem talks about where a line cuts the axes. It needs both intercepts to be non-zero: it cannot describe a line through the origin, or a line parallel to an axis.

✏️ Example 5 — Using intercepts

(a) Find the equation of the line with $x$-intercept $-3$ and $y$-intercept $2$.
(b) Find the equation of the line through $(2,3)$ whose two intercepts are equal and non-zero.
(c) The segment of a line between the axes is bisected at $(2,3)$. Find the line, and the area of the triangle it makes with the axes.

💡 Show solution▶

(a) $\dfrac{x}{-3}+\dfrac y2=1$. Multiply by $6$: $-2x+3y=6$, i.e. $2x-3y+6=0$. Check: $(-3,0)$: $-6+6=0$ ✓; $(0,2)$: $-6+6=0$ ✓.

(b) With both intercepts equal to $a$: $\dfrac xa+\dfrac ya=1$, i.e. $x+y=a$. Through $(2,3)$: $a=5$. The line is $x+y=5$. (If zero intercepts were allowed, the line $y=\tfrac32x$ through the origin would also count.)

(c) The segment runs from $(a,0)$ to $(0,b)$, with mid-point $\left(\tfrac a2,\tfrac b2\right)=(2,3)$. So $a=4$, $b=6$, and the line is $\dfrac x4+\dfrac y6=1$, i.e. $3x+2y=12$.

🧭 Problem-solving strategy — Which form should I use?

A point and a slope (or an inclination, or "parallel/perpendicular to a given line" — which gives the slope): point-slope form.
Two points: find the slope, then point-slope form (this is the two-point form). Check with the second point.
The slope and the $y$-intercept, or a question asking for the slope of a given line: slope-intercept form.
Where the line cuts the axes (intercepts, their sum, a mid-point of the segment between the axes, the triangle with the axes): intercept form.
Parallel to an axis: $x=x_1$ or $y=y_1$ — the slope-based forms cannot give a vertical line.

Whatever form you use, it is good practice to finish by clearing fractions and collecting everything on one side, e.g. $3x+2y-12=0$, so that answers can be compared.

✏️ Example 6 — Lines in a triangle

A triangle has vertices $A(1,4)$, $B(-3,0)$ and $C(5,-2)$. Find the equations of (a) the median from $A$ (b) the altitude from $A$ (c) the perpendicular bisector of $BC$.

💡 Show solution▶

(a) The median joins $A$ to the mid-point $M$ of $BC$: $M=\left(\tfrac{-3+5}{2},\tfrac{0-2}{2}\right)=(1,-1)$. $A$ and $M$ both have $x=1$, so the median is the vertical line $x=1$ — the two-point form would divide by zero.

(b) $m_{BC}=\dfrac{-2-0}{5-(-3)}=-\dfrac14$, so the altitude from $A$ (perpendicular to $BC$) has slope $4$: $y-4=4(x-1)$, i.e. $y=4x$.

Check The altitude and the perpendicular bisector are both perpendicular to $BC$, so they are parallel (both slope $4$), with different intercepts $0$ and $-5$ ✓. They would coincide only if $AB=AC$; here $AB^2=32$, $AC^2=52$.

Choose how to fix the line. Point and slope: the line turns about $P(2,1)$ as $m$ changes, leaving faint traces — every line through $P$ except the vertical one is $y-1=m(x-2)$. Slope and $y$-intercept: changing $c$ slides the line up and down, always parallel to itself. Two intercepts: the line cuts off a triangle from the axes, of area $\tfrac12|ab|$. Press Play to animate the slope, the intercept $c$ or the intercept $a$, depending on the choice.

fix the line bym1c1

equation—

slope—

intercepts ($x$, $y$)—

triangle with axes—

⚠️ Common mistakes with equations of lines

1. Sign slips with negative coordinates. Through $(-3,-2)$: $y-(-2)=m(x-(-3))$, i.e. $y+2=m(x+3)$ — not $y-2=m(x-3)$.

2. Reading the slope before making $y$ the subject. In $2x+3y=6$ the slope is not $2$ or $3$: $y=-\tfrac23x+2$, so $m=-\tfrac23$.

3. Intercept form with the wrong number on the right. $\dfrac x3+\dfrac y2=1$ has intercepts $3$ and $2$; the equation $2x+3y=6$ is the same line, but $2x+3y=1$ is not. Always get "$=1$" before reading off intercepts.

4. Treating an intercept as a length. Intercepts are signed: $\dfrac x{-3}+\dfrac y2=1$ meets the $x$-axis at $(-3,0)$. Lengths (for areas) are $|a|$ and $|b|$.

5. Forgetting the vertical line. "The line through $(2,5)$ and $(2,-1)$" is $x=2$. It has no slope and no $y=mx+c$ form; trying the two-point form divides by zero.

📘 In the CBSE syllabus — school assessment only (not in the year-end board exam)

The general equation of a line and its normal form are taught and tested in school, but not in the year-end board exam; JEE Main and JEE Advanced both list the equation of a line in its various forms, which includes these.

The general equation. Every form in this section rearranges to $$Ax+By+C=0\qquad(A,\ B\text{ not both }0),$$ and every such equation is a line. To read it: if $B\ne0$, then $y=-\tfrac ABx-\tfrac CB$, so the slope is $-\tfrac AB$ and the $y$-intercept is $-\tfrac CB$; if $B=0$ the line is vertical, $x=-\tfrac CA$. If $A$, $B$, $C$ are all non-zero, $\dfrac{x}{-C/A}+\dfrac{y}{-C/B}=1$ is its intercept form.

The normal form. Drop the perpendicular $OM$ from the origin to the line. If $OM$ has length $p\ge0$ and makes the angle $\omega$ with the positive $x$-axis, then $M=(p\cos\omega,\,p\sin\omega)$ and the line is $$x\cos\omega+y\sin\omega=p.$$ Why: when $\sin\omega\ne0$, $OM$ has slope $\tan\omega$, so the line has slope $-\tfrac{\cos\omega}{\sin\omega}$; the point-slope form through $M$ is $y-p\sin\omega=-\tfrac{\cos\omega}{\sin\omega}(x-p\cos\omega)$, and multiplying by $\sin\omega$ and using $\sin^2\omega+\cos^2\omega=1$ gives $x\cos\omega+y\sin\omega=p$. (When $\sin\omega=0$ the line is vertical, $x=\pm p$, which the formula also gives.)

From general to normal form. Write $Ax+By=-C$ and divide by $\sqrt{A^2+B^2}$, with the sign chosen so that the right-hand side is not negative. Then $\cos\omega$ and $\sin\omega$ are the coefficients of $x$ and $y$, and $p=\dfrac{|C|}{\sqrt{A^2+B^2}}$ — the distance of the origin from the line, as Section 9.3 will confirm.

Worked example 1. Reduce $\sqrt3\,x+y-8=0$ to normal form, and find its slope and intercepts.
Solution. $\sqrt3\,x+y=8$ and $\sqrt{3+1}=2$, so $\tfrac{\sqrt3}{2}x+\tfrac12y=4$. Then $\cos\omega=\tfrac{\sqrt3}2$, $\sin\omega=\tfrac12$: $\omega=30^\circ$, $p=4$, i.e. $x\cos30^\circ+y\sin30^\circ=4$. Slope $-\tfrac AB=-\sqrt3$; intercepts $\tfrac{8}{\sqrt3}$ on the $x$-axis and $8$ on the $y$-axis. Check: $M=(4\cos30^\circ,4\sin30^\circ)=(2\sqrt3,2)$ lies on the line, since $\sqrt3\cdot2\sqrt3+2-8=0$ ✓.

Worked example 2. Reduce $x-y+4=0$ to normal form.
Solution. $x-y=-4$ has a negative right side, so divide by $-\sqrt2$: $-\tfrac1{\sqrt2}x+\tfrac1{\sqrt2}y=2\sqrt2$. Now $\cos\omega=-\tfrac1{\sqrt2}$ and $\sin\omega=\tfrac1{\sqrt2}$, so $\omega=135^\circ$ and $p=2\sqrt2$: $x\cos135^\circ+y\sin135^\circ=2\sqrt2$. Check: $\tfrac{|4|}{\sqrt{1+1}}=2\sqrt2$ ✓, and the foot $M=(-2,2)$ satisfies $-2-2+4=0$ ✓.

🚀 Beyond the board syllabus (JEE Main · JEE Advanced)

In JEE Advanced 2026 (lines through the point of intersection of two given lines) and JEE Main 2026 (intersection of lines, concurrence of three lines); not in the CBSE year-end exam. Both use it constantly to avoid finding intersection points.

Lines through the intersection of two lines. Write two intersecting lines as $L_1\equiv a_1x+b_1y+c_1=0$ and $L_2\equiv a_2x+b_2y+c_2=0$. For each number $\lambda$, the equation $$L_1+\lambda L_2=0$$ is a line (it is of the first degree in $x$, $y$ unless the $x$ and $y$ terms cancel), and it passes through the intersection point of $L_1$ and $L_2$, because there $L_1=0$ and $L_2=0$. Choosing $\lambda$ to meet one more condition picks out the required line. Three lines are concurrent (pass through one point) exactly when the third passes through the intersection of the other two.

Worked example. Find the line through the intersection of $x+y-5=0$ and $2x-y-1=0$ that passes through $(1,-1)$.
Solution. Take $(x+y-5)+\lambda(2x-y-1)=0$. At $(1,-1)$: $(1-1-5)+\lambda(2+1-1)=0$, so $-5+2\lambda=0$, $\lambda=\tfrac52$. Multiply through by $2$: $2(x+y-5)+5(2x-y-1)=0$, i.e. $12x-3y-15=0$, or $4x-y-5=0$. Check: the given lines meet at $(2,3)$ (add them: $3x=6$), and $4(2)-3-5=0$ ✓; also $4(1)-(-1)-5=0$ ✓.

Exercises

Write the equation of the line. Give each answer with all terms on one side and no fractions.

1 Parallel to the $x$-axis through $(-3,5)$.
2 Parallel to the $y$-axis through $(-3,5)$.
3 Through the origin with slope $-2$.
4 Through $(4,-2)$ with slope $\tfrac34$.
5 Through $(-1,2)$ with inclination $135^\circ$.
6 Through $(2,-3)$ and $(-4,6)$.
7 Through $(1,1)$ and $(1,-4)$.
8 With slope $2$ and $y$-intercept $-3$.
9 With slope $-\tfrac13$ and $x$-intercept $4$.

💡 Show solutions▶

1 $y=5$, i.e. $y-5=0$.

2 $x=-3$, i.e. $x+3=0$.

3 $y=-2x$, i.e. $2x+y=0$.

4 $y+2=\tfrac34(x-4)$; $4y+8=3x-12$; $3x-4y-20=0$.

5 $m=\tan135^\circ=-1$: $y-2=-(x+1)$, so $x+y-1=0$.

6 $m=\tfrac{6-(-3)}{-4-2}=-\tfrac32$: $y+3=-\tfrac32(x-2)$; $2y+6=-3x+6$; $3x+2y=0$ — the line passes through the origin. Check $(-4,6)$: $-12+12=0$ ✓.

7 Same $x$-coordinate: the vertical line $x=1$, i.e. $x-1=0$.

8 $y=2x-3$, i.e. $2x-y-3=0$.

9 $y=-\tfrac13(x-4)$; $3y=-x+4$; $x+3y-4=0$.

Reading and testing equations.

1 Find the slope and $y$-intercept of $2x+3y=6$.
2 Write $2x+3y=6$ in intercept form and read off its intercepts.
3 Find the intercepts of $4x-5y+20=0$.
4 Find the slope and $y$-intercept of $x=7y+1$.
5 Does the point $(3,-1)$ lie on the line $2x-5y=11$?
6 The point $(k,4)$ lies on the line through $(1,2)$ and $(3,5)$. Find $k$.
7 Show that the lines $y=3x-2$ and $6x-2y+7=0$ are parallel and distinct.

💡 Show solutions▶

1 $3y=-2x+6$, $y=-\tfrac23x+2$: slope $-\tfrac23$, $y$-intercept $2$.

2 Divide by $6$: $\dfrac x3+\dfrac y2=1$; intercepts $3$ and $2$.

3 $4x-5y=-20$; divide by $-20$: $\dfrac{x}{-5}+\dfrac y4=1$. $x$-intercept $-5$, $y$-intercept $4$.

4 $y=\tfrac17x-\tfrac17$: slope $\tfrac17$, $y$-intercept $-\tfrac17$.

5 $2(3)-5(-1)=6+5=11$ ✓: yes.

6 Slope $\tfrac32$; line $y-2=\tfrac32(x-1)$. With $y=4$: $2=\tfrac32(k-1)$, $k-1=\tfrac43$, $k=\tfrac73$.

7 $2y=6x+7$, $y=3x+\tfrac72$: slope $3$, the same as $y=3x-2$, so parallel; the $y$-intercepts $\tfrac72$ and $-2$ differ, so the lines are distinct.

III

Intercepts.

1 Find the line with intercepts $3$ and $-4$ on the $x$- and $y$-axes.
2 Find the line through $(3,4)$ with equal non-zero intercepts.
3 Find the line through $(3,4)$ whose intercepts are equal in size but opposite in sign.
4 Find the line through $(2,1)$ whose $x$-intercept is twice its $y$-intercept (both non-zero).
5 The part of a line between the axes has mid-point $(3,-2)$. Find the line.
6 Find the area of the triangle that the line $3x+4y=12$ forms with the coordinate axes.
7 A line through $(2,2)$ cuts the positive $x$- and $y$-axes and forms with them a triangle of area $9$ square units. Find all such lines.

💡 Show solutions▶

1 $\dfrac x3+\dfrac{y}{-4}=1$; multiply by $12$: $4x-3y=12$.

2 $x+y=a$ with $3+4=a$: $x+y=7$.

3 Intercepts $a$ and $-a$: $\dfrac xa-\dfrac ya=1$, i.e. $x-y=a$. Through $(3,4)$: $a=-1$, so $x-y+1=0$ (intercepts $-1$ and $1$ ✓).

4 Intercepts $2b$ and $b$: $\dfrac x{2b}+\dfrac yb=1$, i.e. $x+2y=2b$. Through $(2,1)$: $2b=4$. The line is $x+2y=4$ (intercepts $4$ and $2$ ✓).

5 Intercepts $a=6$, $b=-4$: $\dfrac x6-\dfrac y4=1$, i.e. $2x-3y=12$.

6 Intercepts $4$ and $3$: area $\tfrac12\times4\times3=6$ square units.

7 $\dfrac2a+\dfrac2b=1$ and $\tfrac12ab=9$. The first gives $2(a+b)=ab=18$, so $a+b=9$, $ab=18$: $a,b$ are the roots of $t^2-9t+18=0$, i.e. $3$ and $6$. Two lines: $\dfrac x3+\dfrac y6=1$ ($2x+y=6$) and $\dfrac x6+\dfrac y3=1$ ($x+2y=6$). Both pass through $(2,2)$ ✓.

Geometry and linear models.

1 Find the perpendicular bisector of the segment joining $(1,3)$ and $(5,-1)$.
2 Find the line through $(2,-1)$ perpendicular to $y=2x+5$.
3 Find the line through $(-3,2)$ parallel to $4x-3y+1=0$.
4 In the triangle $A(-2,0)$, $B(4,2)$, $C(1,5)$, find the altitude from $C$.
5 In the same triangle, find the median from $C$.
6 A cab company's fare is a linear function of the distance travelled. A $3\ \text{km}$ ride costs $\text{₹}\,70$ and a $7\ \text{km}$ ride costs $\text{₹}\,142$. Find the fare $F$ for a ride of $d$ km, the fare for $10\ \text{km}$, and the meaning of the constant term.
7 A spring is $25\ \text{cm}$ long when a $1\ \text{kg}$ mass hangs from it and $31\ \text{cm}$ long with $4\ \text{kg}$. Assuming the length is linear in the mass, find the natural (unloaded) length and the mass that stretches it to $40\ \text{cm}$.
8 A machine bought for $\text{₹}\,6$ lakh is valued at $\text{₹}\,4.2$ lakh after $3$ years, its value falling linearly with time. Find the value $V$ (in lakh) after $t$ years, and when the value reaches zero.

💡 Show solutions▶

1 Mid-point $(3,1)$; segment slope $\tfrac{-4}{4}=-1$, so perpendicular slope $1$: $y-1=x-3$, i.e. $x-y-2=0$.

2 Slope $-\tfrac12$: $y+1=-\tfrac12(x-2)$; $2y+2=-x+2$; $x+2y=0$.

3 $3y=4x+1$ has slope $\tfrac43$: $y-2=\tfrac43(x+3)$; $3y-6=4x+12$; $4x-3y+18=0$.

4 $m_{AB}=\tfrac26=\tfrac13$, so the altitude has slope $-3$: $y-5=-3(x-1)$, i.e. $3x+y-8=0$.

5 Mid-point of $AB$ is $(1,1)$; with $C(1,5)$ the median is the vertical line $x=1$.

6 Slope $\tfrac{142-70}{7-3}=18$ rupees per km: $F-70=18(d-3)$, $F=18d+16$. For $10\ \text{km}$: $\text{₹}\,196$. The constant $\text{₹}\,16$ is a fixed charge paid even for a zero-length ride.

7 Slope $\tfrac{31-25}{4-1}=2\ \text{cm/kg}$: $L=2m+23$. Natural length $23\ \text{cm}$; $40=2m+23$ gives $m=8.5\ \text{kg}$.

8 Slope $\tfrac{4.2-6}{3}=-0.6$ lakh per year: $V=6-0.6t$. $V=0$ when $t=10$ years.

📘 School-assessment practice (general equation and normal form — see the 📘 box; not in the year-end board exam).

1 Reduce $3x-4y+10=0$ to slope-intercept form and to intercept form.
2 Reduce $x+\sqrt3\,y-6=0$ to normal form; find $p$ and $\omega$.
3 Reduce $x-y=4$ to normal form; find $p$ and $\omega$ (with $0^\circ\le\omega\lt360^\circ$).
4 Find the line whose perpendicular from the origin has length $5$ and makes the angle $135^\circ$ with the positive $x$-axis.
5 Use the normal form to find the distance of the origin from $3x+4y+15=0$, and the foot of the perpendicular.

💡 Show solutions▶

1 $4y=3x+10$: $y=\tfrac34x+\tfrac52$ (slope $\tfrac34$, $y$-intercept $\tfrac52$). $3x-4y=-10$; divide by $-10$: $\dfrac{x}{-10/3}+\dfrac{y}{5/2}=1$ (intercepts $-\tfrac{10}3$ and $\tfrac52$).

2 $x+\sqrt3\,y=6$, $\sqrt{1+3}=2$: $\tfrac12x+\tfrac{\sqrt3}2y=3$. So $\cos\omega=\tfrac12$, $\sin\omega=\tfrac{\sqrt3}2$: $\omega=60^\circ$, $p=3$.

3 Divide by $\sqrt2$: $\tfrac1{\sqrt2}x-\tfrac1{\sqrt2}y=2\sqrt2$. $\cos\omega=\tfrac1{\sqrt2}$, $\sin\omega=-\tfrac1{\sqrt2}$: $\omega=315^\circ$, $p=2\sqrt2$.

4 $x\cos135^\circ+y\sin135^\circ=5$: $-\tfrac{x}{\sqrt2}+\tfrac{y}{\sqrt2}=5$, i.e. $x-y+5\sqrt2=0$.

5 $3x+4y=-15$; divide by $-5$: $-\tfrac35x-\tfrac45y=3$. So $p=3$, with $\cos\omega=-\tfrac35$, $\sin\omega=-\tfrac45$; the foot is $M=(3\cos\omega,3\sin\omega)=\left(-\tfrac95,-\tfrac{12}5\right)$. Check: $3\left(-\tfrac95\right)+4\left(-\tfrac{12}5\right)+15=-\tfrac{27}5-\tfrac{48}5+15=0$ ✓.

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

How this is examined: Board papers ask for the equation of a line from a point and a slope, from two points or from intercepts (1–2 marks), and for medians, altitudes and perpendicular bisectors of a triangle (3–5 marks); case studies set up a linear model from two data points. JEE adds intercept conditions that lead to quadratics and asks which of several lines satisfy all the conditions.

E1. Board · 1 mark The $x$- and $y$-intercepts of the line $3x-2y+6=0$ are (A) $-2$ and $3$ (B) $2$ and $-3$ (C) $3$ and $-2$ (D) $-3$ and $2$

💡 Show solution▶

Idea $y=0$ gives $3x+6=0$, $x=-2$; $x=0$ gives $-2y+6=0$, $y=3$. (Or: $3x-2y=-6$, so $\tfrac{x}{-2}+\tfrac y3=1$.)

Trap (B) has both signs wrong — it comes from moving $6$ across without changing its sign; (D) and (C) read the coefficients instead of solving.

Answer (A).

E2. Board · 2 marks Find the equation of the line through $(-2,5)$ parallel to the line joining $(1,1)$ and $(4,-5)$.

💡 Show solution▶

Slope $\dfrac{-5-1}{4-1}=-2$; a parallel line has the same slope. (1 mark)

Answer $y-5=-2(x+2)$, i.e. $2x+y-1=0$. (1 mark)

E3. Board · 3 marks The vertices of a triangle are $A(2,5)$, $B(-4,-1)$ and $C(6,1)$. Find the equations of the median and the altitude through $A$.

💡 Show solution▶

Median Mid-point of $BC$: $M(1,0)$. Slope of $AM$: $\tfrac{5-0}{2-1}=5$. Median: $y=5(x-1)$, i.e. $5x-y-5=0$. (1½ marks)

Altitude $m_{BC}=\tfrac{1-(-1)}{6-(-4)}=\tfrac15$, so the altitude has slope $-5$. (½ mark)

Answer Altitude: $y-5=-5(x-2)$, i.e. $5x+y-15=0$. Check: $A$ satisfies both ($10-5-5=0$, $10+5-15=0$) ✓. (1 mark)

E4. Board · Case study A temperature sensor in a weather station gives an output voltage $V$ (in volts) that is linear in the temperature $T$ (in $^\circ\text{C}$). It reads $0.50\ \text{V}$ at $20^\circ\text{C}$ and $0.80\ \text{V}$ at $80^\circ\text{C}$. (i) Find $V$ in terms of $T$. (1 mark) (ii) What voltage does it give at $50^\circ\text{C}$? (1 mark) (iii) What temperature corresponds to $1.00\ \text{V}$, and what does the $V$-intercept of the line mean? (2 marks)

💡 Show solution▶

(i) Slope $\dfrac{0.80-0.50}{80-20}=0.005\ \text{V per }^\circ\text{C}$. $V-0.50=0.005(T-20)$, so $V=0.005T+0.40$. (1 mark)

(ii) $V=0.005\times50+0.40=0.65\ \text{V}$. (1 mark)

(iii) Temperature $1.00=0.005T+0.40$ gives $T=\tfrac{0.60}{0.005}=120^\circ\text{C}$. (1 mark)

Answer (iii) The $V$-intercept $0.40\ \text{V}$ is the output at $0^\circ\text{C}$ (assuming the straight-line behaviour continues that far down). (1 mark)

E5. JEE Main A line through $(2,3)$ meets the axes at $A$ and $B$, and $(2,3)$ is the mid-point of $AB$. The equation of the line is (A) $3x+2y=12$ (B) $2x+3y=13$ (C) $3x-2y=0$ (D) $x+y=5$

💡 Show solution▶

Method With $A(a,0)$ and $B(0,b)$, the mid-point is $\left(\tfrac a2,\tfrac b2\right)=(2,3)$, so $a=4$, $b=6$ and the line is $\tfrac x4+\tfrac y6=1$, i.e. $3x+2y=12$.

Trap All four lines pass through $(2,3)$! (B) is the line through $(2,3)$ perpendicular to the line joining it to the origin; its intercepts $\tfrac{13}2$, $\tfrac{13}3$ have mid-point $\left(\tfrac{13}4,\tfrac{13}6\right)$. (C) passes through the origin, so $A=B=O$. (D) has equal intercepts $5$, mid-point $(2.5,2.5)$.

⏱ Fast route If $(h,k)$ bisects the intercepts, the line is $\tfrac xh+\tfrac yk=2$: here $\tfrac x2+\tfrac y3=2$, i.e. $3x+2y=12$.

Answer (A).

E6. JEE Advanced One or more options correct. A line passes through $(1,2)$ and the sum of its intercepts on the axes is $6$. The line can be (A) $2x+y=4$ (B) $x+y=3$ (C) $x+2y=5$ (D) $4x+y=6$

💡 Show solution▶

Set up Intercepts $a$ and $6-a$ (both non-zero): $\dfrac1a+\dfrac2{6-a}=1$. Multiply by $a(6-a)$: $6-a+2a=6a-a^2$, so $a^2-5a+6=0$, $a=2$ or $a=3$.

Lines $a=2$, $b=4$: $\tfrac x2+\tfrac y4=1$, i.e. $2x+y=4$. $a=3$, $b=3$: $x+y=3$.

Other options (C) and (D) also pass through $(1,2)$, but their intercepts are $5,\tfrac52$ and $\tfrac32,6$ — sums $7.5$, not $6$.

Answer (A) and (B).

🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. Does your phone charge at a steady rate — is battery percentage a linear function of time — and can a two-point line predict when it will reach $80\%$?

Why it matters. Engineers, economists and scientists constantly fit straight lines to data and use them to predict: fuel use against distance, sales against price, sensor voltage against temperature. The two-point form is the simplest such model. Knowing where a straight-line model works and where it fails is just as important as the model itself — phone batteries are a good example, because charging deliberately slows down near full.

What to do.

Start charging a phone at around $20\%$. Every $5$ minutes for about $45$ minutes, record the time $t$ (minutes) and the battery percentage $p$.
Plot the points $(t,p)$ on squared paper or in a spreadsheet.
Choose two of your early readings and find the line through them with the two-point form, $p-p_1=\dfrac{p_2-p_1}{t_2-t_1}(t-t_1)$. Interpret the slope (percentage points per minute).
Use your line to predict when the phone reaches $80\%$, then keep charging and record when it actually does.
Compare: where do the data points leave your line? Repeat with two different starting points and see how much the prediction changes.

Tools: A phone and charger, a clock or timer, squared paper or a spreadsheet

Hand in: Your table and graph with the fitted line drawn, the equation and the meaning of its slope, your prediction and the actual time, and a paragraph on when the linear model is good and when it is not.

Going further. Use a spreadsheet's trend-line (least squares) tool and compare its slope with your two-point slope. Why is a line through all the data better than a line through two chosen points?

Instructor note: with a typical fast charger the readings between about $20\%$ and $70\%$ lie close to a line with slope of roughly $1$–$2$ percentage points per minute; above about $80\%$ the rate drops sharply, so predictions of the time to reach $100\%$ from an early line are much too early. Different pairs of points give slightly different lines — a natural lead-in to fitting by least squares.

🧪 Self-check quiz

0/5 correct

Q1. The line through $(1,-3)$ with slope $2$ is

Q2. The slope and $y$-intercept of $3x+y=7$ are

Q3. The line with $x$-intercept $2$ and $y$-intercept $-5$ is

Q4. The line through $(-4,7)$ parallel to the $y$-axis is

Q5. Which line passes through both $(2,-1)$ and $(-1,5)$?

← 9.1 Slope of a Line and the Angle between Two Lines 9.3 Distance of a Point from a Line →