Equations of a line — slope-intercept, point-slope, two-point, standard (ax+by+c=0): Exercises

Before we start, let me re-anchor the four tools you will keep reaching for, in plain words:

Reminder on the words: the $x$ -axis is the horizontal number line; the $y$ -axis is the vertical one. A point is written $(x, y)$ = (how far right, how far up). Slope $m = \frac{rise}{run} = \frac{Δ y}{Δ x}$ where the triangle symbol $Δ$ (Greek "delta") just means "the change in". A positive slope climbs to the right; a negative one falls to the right; zero is flat; undefined is a vertical wall.

Level 1 — Recognition

Goal: read a form and pull out its numbers. No manipulation yet.

Exercise 1.1

State the slope $m$ and $y$ -intercept $c$ of the line $y = - 2 x + 7$ .

Recall Solution 1.1

The form $y = m x + c$ is a template. Line up the given line against it symbol-by-symbol: $y = m - 2 x + c 7 .$ So $m = - 2$ (falls 2 units of height for every 1 step right) and $c = 7$ (crosses the $y$ -axis at the point $(0, 7)$ ). Answer: $m = - 2, c = 7$ .

Exercise 1.2

For the line $5 x + 2 y - 8 = 0$ , state $a$ , $b$ , $c$ , then its slope and $y$ -intercept.

Recall Solution 1.2

Match against $a x + b y + c = 0$ : here $a = 5, b = 2, c = - 8$ .

Slope $m = - \frac{a}{b} = - \frac{5}{2}$ . (Why the minus? Solving $a x + b y + c = 0$ for $y$ gives $y = - \frac{a}{b} x - \frac{c}{b}$ — the minus is forced by moving the $a x$ term across.)
$y$ -intercept $= - \frac{c}{b} = - \frac{- 8}{2} = 4$ , i.e. the point $(0, 4)$ . Answer: $a = 5, b = 2, c = - 8, m = - \frac{5}{2}, y -int = 4$ .

Exercise 1.3

Which of these are lines you cannot write as $y = m x + c$ , and why? (a) $x = 3$ , (b) $y = 4$ , (c) $2 x - y = 0$ .

Recall Solution 1.3

(a) $x = 3$ is a vertical wall: every point has $x = 3$ , $y$ can be anything. Its run is $0$ , so slope $= \frac{Δ y}{0}$ is undefined — division by zero is illegal. Cannot be $y = m x + c$ .
(b) $y = 4$ is horizontal: $m = 0$ , $c = 4$ . This is $y = m x + c$ with $m = 0$ . Can be written.
(c) $2 x - y = 0 \Rightarrow y = 2 x$ , so $m = 2, c = 0$ . Can be written. Answer: only (a) cannot.

Level 2 — Application

Goal: apply one form to build an equation.

Exercise 2.1

Find the equation (in slope–intercept form) of the line through $(2, 3)$ with slope $4$ .

Recall Solution 2.1

Use point–slope $y - y_{1} = m (x - x_{1})$ with $(x_{1}, y_{1}) = (2, 3)$ , $m = 4$ : $y - 3 = 4 (x - 2) .$ Distribute the $4$ (multiply it into both terms in the bracket): $y - 3 = 4 x - 8$ . Add $3$ to both sides to isolate $y$ : $y = 4 x - 5$ . Answer: $y = 4 x - 5$ (so $c = - 5$ ).

Exercise 2.2

Find the line through the two points $(1, 2)$ and $(3, 8)$ , and verify your answer against the second point.

Recall Solution 2.2

Step 1 — get the slope (two points → we must compute $m$ first): $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{8 - 2}{3 - 1} = \frac{6}{2} = 3.$ Step 2 — plug into point–slope using $(1, 2)$ : $y - 2 = 3 (x - 1) \Rightarrow y - 2 = 3 x - 3 \Rightarrow y = 3 x - 1$ . Step 3 — verify with the other point $(3, 8)$ : $3 (3) - 1 = 9 - 1 = 8$ ✓. Answer: $y = 3 x - 1$ .

Exercise 2.3

Write the line through $(5, - 2)$ that is (a) vertical, (b) horizontal.

Recall Solution 2.3

(a) A vertical line freezes $x$ at the point's $x$ -value: $x = 5$ , i.e. $x - 5 = 0$ . Slope undefined.
(b) A horizontal line freezes $y$ at the point's $y$ -value: $y = - 2$ . Slope $0$ . Answer: (a) $x = 5$ ; (b) $y = - 2$ .

Level 3 — Analysis

Goal: combine two ideas — read one form, use it to find a feature or relationship.

Exercise 3.1

For $3 x + 4 y - 12 = 0$ , find the slope, the $y$ -intercept, and the $x$ -intercept. Sketch what the line looks like (see figure).

Figure — Equations of a line — slope-intercept, point-slope, two-point, standard (ax+by+c=0)

Recall Solution 3.1

Match $a x + b y + c = 0$ : $a = 3, b = 4, c = - 12$ .

Slope $m = - \frac{a}{b} = - \frac{3}{4}$ (falls 3 for every 4 right — a gentle downhill).
$y$ -intercept ( $x = 0$ ): $- \frac{c}{b} = - \frac{- 12}{4} = 3 \Rightarrow (0, 3)$ .
$x$ -intercept ( $y = 0$ ): set $y = 0$ in $3 x - 12 = 0 \Rightarrow x = 4 \Rightarrow (4, 0)$ . Look at the figure: the line drops from $(0, 3)$ on the $y$ -axis down to $(4, 0)$ on the $x$ -axis — a negative slope, exactly $- \frac{3}{4}$ . Answer: $m = - \frac{3}{4}$ , $y$ -int $= 3$ , $x$ -int $= 4$ .

Exercise 3.2

Line $L_{1}$ is $y = 2 x + 1$ . Find the equation of the line through $(0, - 3)$ that is (a) parallel to $L_{1}$ , (b) perpendicular to $L_{1}$ .

Recall Solution 3.2

Two parallel lines have equal slopes; two perpendicular lines have slopes multiplying to $- 1$ .

$L_{1}$ has slope $m_{1} = 2$ .
(a) Parallel: same slope $m = 2$ , through $(0, - 3)$ . Since the point is on the $y$ -axis, $c = - 3$ : $y = 2 x - 3$ .
(b) Perpendicular: need $m_{2}$ with $2 \cdot m_{2} = - 1 \Rightarrow m_{2} = - \frac{1}{2}$ . Through $(0, - 3)$ : $y = - \frac{1}{2} x - 3$ . Answer: (a) $y = 2 x - 3$ ; (b) $y = - \frac{1}{2} x - 3$ .

Exercise 3.3

Find the acute angle between $L_{1} : y = x$ and $L_{2} : y = 0$ (the $x$ -axis).

Recall Solution 3.3

Slopes: $m_{1} = 1$ , $m_{2} = 0$ . The angle formula is $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} = \frac{1 - 0}{1 + 0} = 1.$ The angle whose tangent is $1$ is $4 5^{\circ}$ . (Why $tan$ ? Slope is literally $tan$ of the angle each line makes with the $x$ -axis; the formula packages the difference of those two angles.) Answer: $θ = 4 5^{\circ}$ .

Level 4 — Synthesis

Goal: build a full solution from several parts you must arrange yourself.

Exercise 4.1

A line passes through $(- 2, 5)$ and is parallel to $4 x - 2 y + 7 = 0$ . Give its equation in standard form.

Recall Solution 4.1

Step 1 — slope of the given line. From $a x + b y + c = 0$ with $a = 4, b = - 2$ : $m = - \frac{a}{b} = - \frac{4}{- 2} = 2$ . Step 2 — parallel means same slope: our line also has $m = 2$ . Step 3 — point–slope with $(- 2, 5)$ : $y - 5 = 2 (x - (- 2)) = 2 (x + 2) = 2 x + 4$ , so $y = 2 x + 9$ . Step 4 — to standard form: move all to one side: $2 x - y + 9 = 0$ . Answer: $2 x - y + 9 = 0$ .

Exercise 4.2

Find the line in intercept form $\frac{x}{p} + \frac{y}{q} = 1$ that has $x$ -intercept $3$ and $y$ -intercept $- 6$ . Then convert it to standard form.

Recall Solution 4.2

In intercept form, $p$ is the $x$ -intercept and $q$ the $y$ -intercept. So $p = 3, q = - 6$ : $\frac{x}{3} + \frac{y}{- 6} = 1.$ Convert: multiply through by the common denominator $6$ (to clear fractions): $2 x - y = 6$ , i.e. $2 x - y - 6 = 0$ . Check: $x$ -intercept, set $y = 0 \Rightarrow 2 x = 6 \Rightarrow x = 3$ ✓; $y$ -intercept, set $x = 0 \Rightarrow - y = 6 \Rightarrow y = - 6$ ✓. Answer: intercept form $\frac{x}{3} - \frac{y}{6} = 1$ ; standard $2 x - y - 6 = 0$ .

Exercise 4.3

The line $a x + 3 y - 6 = 0$ passes through $(3, - 1)$ . Find $a$ , then the slope of the line.

Recall Solution 4.3

Step 1 — "passes through" means the point satisfies the equation. Substitute $x = 3, y = - 1$ : $a (3) + 3 (- 1) - 6 = 0 \Rightarrow 3 a - 3 - 6 = 0 \Rightarrow 3 a = 9 \Rightarrow a = 3.$ Step 2 — slope with $a = 3, b = 3$ : $m = - \frac{a}{b} = - \frac{3}{3} = - 1$ . Answer: $a = 3$ , slope $= - 1$ .

Level 5 — Mastery

Goal: a multi-step puzzle mixing several tools, including a degenerate case.

Exercise 5.1

Triangle $A B C$ has vertices $A (0, 0)$ , $B (4, 0)$ , $C (4, 3)$ . (a) Find the equation of each side in standard form. (b) Find the equation of the altitude from $A$ (the line through $A$ perpendicular to $B C$ ). See the figure.

Recall Solution 5.1

Side $A B$ : through $A (0, 0)$ and $B (4, 0)$ . Both have $y = 0$ — a horizontal line. Slope $0$ . Equation: $y = 0$ . Side $B C$ : through $B (4, 0)$ and $C (4, 3)$ . Both have $x = 4$ — a vertical wall, slope undefined; here slope–intercept is impossible (this is the degenerate case). Equation: $x = 4$ , i.e. $x - 4 = 0$ . Side $C A$ : through $C (4, 3)$ and $A (0, 0)$ . Slope $m = \frac{0 - 3}{0 - 4} = \frac{- 3}{- 4} = \frac{3}{4}$ . Through the origin ( $c = 0$ ): $y = \frac{3}{4} x$ , standard $3 x - 4 y = 0$ .

(b) Altitude from $A$ perpendicular to $B C$ : $B C$ is vertical, so the line perpendicular to it is horizontal. A horizontal line through $A (0, 0)$ is $y = 0$ . (Notice: this coincides with side $A B$ — because in this right-angled triangle the right angle is at $B$ , and $A B ⊥ B C$ already.)

Look at the figure: $A B$ runs flat along the $x$ -axis, $B C$ shoots straight up, and the hypotenuse $C A$ climbs at slope $\frac{3}{4}$ . Answers: $A B : y = 0$ ; $B C : x - 4 = 0$ ; $C A : 3 x - 4 y = 0$ ; altitude from $A : y = 0$ .

Exercise 5.2

Find the value of $k$ for which the lines $2 x + k y - 4 = 0$ and $3 x - y + 2 = 0$ are perpendicular.

Recall Solution 5.2

Slopes: first line $m_{1} = - \frac{2}{k}$ ; second line ( $a = 3, b = - 1$ ) $m_{2} = - \frac{3}{- 1} = 3$ . Perpendicular condition: $m_{1} m_{2} = - 1$ : $(- \frac{2}{k}) (3) = - 1 \Rightarrow - \frac{6}{k} = - 1 \Rightarrow \frac{6}{k} = 1 \Rightarrow k = 6.$ Check: $m_{1} = - \frac{2}{6} = - \frac{1}{3}$ , and $- \frac{1}{3} \times 3 = - 1$ ✓. Answer: $k = 6$ .

One-line recap

Recall What each level trained

L1 read numbers off a form ::: L2 build one equation from a point + slope (or two points) L3 combine forms with slope/angle/perpendicularity ::: L4 assemble multi-part solutions and convert forms L5 juggle several tools at once and survive the vertical-line degenerate case ::: the through-line is always a point + a slope

Connections

Parent topic — all four forms
Slope of a line — the number every exercise leans on.
Angle between two lines — used in Ex 3.3.
Parallel and perpendicular lines — used in Ex 3.2, 4.1, 5.2.
Intercept form of a line — used in Ex 4.2.
Distance of a point from a line — next natural step from standard form.
Straight line as a first-degree equation — why degree-1 always draws a line.