Equations of a Line and Distances, Worked Solutions

Ques­tion 1: Four lines in four dif­fer­ent forms#

The prob­lem#

Find the equa­tion: (a) through $(0, -2)$ with slope $\displaystyle \tfrac{3}{4}$; (b) through $(-3, 5)$ and $(1, -3)$; (c) inter­cepts $-2$ and $5$; (d) through $(4, -3)$ par­al­lel to the $y$-axis.

Under­stand­ing the prob­lem#

Each part gives you a dif­fer­ent kind of infor­ma­tion about a line. Your job is to choose the right stan­dard form for each and write the equa­tion, prefer­ably in the gen­eral form $Ax + By + C = 0$.

The idea#

(a) point and slope → point–slope form; (b) two points → find the slope, then point–slope; (c) inter­cepts → inter­cept form; (d) par­al­lel to the $y$-axis → ver­ti­cal line $x = \text{constant}$.

Step-by-step solu­tion#

Part (a)

Step 1. Use $y - y_0 = m(x - x_0)$ with $(x_0, y_0) = (0, -2)$ and $\displaystyle m = \tfrac{3}{4}$.

$$\displaystyle y + 2 = \frac{3}{4}(x - 0).$$

Step 2. Mul­ti­ply by $4$ to clear the frac­tion, then bring every­thing to one side.

$$4y + 8 = 3x \quad\Longrightarrow\quad 3x - 4y - 8 = 0.$$

Part (b)

Step 1. Find the slope through $(-3, 5)$ and $(1, -3)$.

$$\displaystyle m = \frac{-3 - 5}{1 - (-3)} = \frac{-8}{4} = -2.$$

Step 2. Point–slope form through $(-3, 5)$.

$$y - 5 = -2(x + 3) = -2x - 6.$$

Step 3. Rearrange.

$$2x + y + 1 = 0.$$

Part (c)

Step 1. The $x$-inter­cept is $a = -2$ and the $y$-inter­cept is $b = 5$.

$$\displaystyle \frac{x}{-2} + \frac{y}{5} = 1.$$

Step 2. Mul­ti­ply by $-10$ (the LCM of the denom­i­na­tors, with a sign that makes the $x$-coef­fi­cient pos­i­tive).

$$5x - 2y = -10 \quad\Longrightarrow\quad 5x - 2y + 10 = 0.$$

Part (d)

Step 1. A line par­al­lel to the $y$-axis is ver­ti­cal: every point on it has the same $x$-coor­di­nate. Through $(4, -3)$ that $x$-coor­di­nate is $4$.

$$x = 4.$$

Check­ing the answer#

(a) $(0, -2)$: $0 + 8 - 8 = 0$ ✓, slope $\displaystyle -\tfrac{3}{-4} = \tfrac{3}{4}$ ✓. (b) $(-3, 5)$: $-6 + 5 + 1 = 0$ ✓; $(1, -3)$: $2 - 3 + 1 = 0$ ✓. (c) $(-2, 0)$: $-10 + 10 = 0$ ✓; $(0, 5)$: $-10 + 10 = 0$ ✓.

Answer#

(a) $3x - 4y - 8 = 0$; (b) $2x + y + 1 = 0$; (c) $5x - 2y + 10 = 0$; (d) $x = 4$.

Ques­tion 2: Con­vert­ing a gen­eral equa­tion#

The prob­lem#

Write $5x - 2y + 10 = 0$ in slope–inter­cept and inter­cept forms.

Under­stand­ing the prob­lem#

Slope–inter­cept form is $y = mx + c$. Inter­cept form is $\displaystyle \tfrac{x}{a} + \tfrac{y}{b} = 1$. You must rearrange the same line into each.

The idea#

For slope–inter­cept form, make $y$ the sub­ject. For inter­cept form, move the con­stant to the right and divide so that the right side is $1$.

Step-by-step solu­tion#

Slope–inter­cept form

Step 1. Keep $-2y$ on the left and move the rest to the right.

$$-2y = -5x - 10.$$

Step 2. Divide by $-2$.

$$\displaystyle y = \frac{5}{2}x + 5.$$

So the slope is $\displaystyle \tfrac{5}{2}$ and the $y$-inter­cept is $5$.

Inter­cept form

Step 1. Move the con­stant to the right.

$$5x - 2y = -10.$$

Step 2. Divide both sides by $-10$.

$$\displaystyle \frac{5x}{-10} - \frac{2y}{-10} = 1 \quad\Longrightarrow\quad \frac{x}{-2} + \frac{y}{5} = 1.$$

So the inter­cepts are $-2$ on the $x$-axis and $5$ on the $y$-axis.

Check­ing the answer#

From the gen­eral form: slope $\displaystyle -\tfrac{A}{B} = -\tfrac{5}{-2} = \tfrac{5}{2}$ ✓, $x$-inter­cept $\displaystyle -\tfrac{C}{A} = -\tfrac{10}{5} = -2$ ✓, $y$-inter­cept $\displaystyle -\tfrac{C}{B} = -\tfrac{10}{-2} = 5$ ✓.

Answer#

$\displaystyle y = \tfrac{5}{2}x + 5$; $\displaystyle \frac{x}{-2} + \frac{y}{5} = 1$.

Ques­tion 3: A per­pen­dic­u­lar line through a point#

The prob­lem#

Find the line through $(1, -2)$ per­pen­dic­u­lar to $2x - 3y + 7 = 0$.

Under­stand­ing the prob­lem#

You know one point on the required line. You also know it meets the given line at a right angle, which fixes its slope.

The idea#

Find the slope of the given line, take the neg­a­tive rec­i­p­ro­cal (prod­uct $-1$), then use point–slope form.

Step-by-step solu­tion#

Step 1. Slope of $2x - 3y + 7 = 0$ is

$$\displaystyle m_1 = -\frac{A}{B} = -\frac{2}{-3} = \frac{2}{3}.$$

Step 2. Per­pen­dic­u­lar slope: $m_1 m_2 = -1$.

$$\displaystyle m_2 = -\frac{3}{2}.$$

Step 3. Point–slope form through $(1, -2)$.

$$\displaystyle y + 2 = -\frac{3}{2}(x - 1).$$

Step 4. Mul­ti­ply by $2$ and rearrange.

$$2y + 4 = -3x + 3 \quad\Longrightarrow\quad 3x + 2y + 1 = 0.$$

Check­ing the answer#

$(1, -2)$: $3 - 4 + 1 = 0$ ✓. Slopes: $\displaystyle \tfrac{2}{3} \times \left(-\tfrac{3}{2}\right) = -1$ ✓.

Answer#

$3x + 2y + 1 = 0$

Ques­tion 4: A par­al­lel line through a point#

The prob­lem#

Find the line through $(2, 3)$ par­al­lel to $x + 4y = 9$.

Under­stand­ing the prob­lem#

Par­al­lel lines have the same slope, so the new line has the same $x$ and $y$ coef­fi­cients as the given one; only the con­stant changes.

The idea#

Write the required line as $x + 4y = k$ and choose $k$ so that $(2, 3)$ lies on it.

Step-by-step solu­tion#

Step 1. Any line par­al­lel to $x + 4y = 9$ has the form

$$x + 4y = k.$$

Step 2. Sub­sti­tute $(2, 3)$.

$$2 + 4(3) = k \quad\Longrightarrow\quad k = 14.$$

Step 3. Write the line.

$$x + 4y = 14, \quad\text{i.e.}\quad x + 4y - 14 = 0.$$

Check­ing the answer#

$(2, 3)$: $2 + 12 - 14 = 0$ ✓. Both lines have slope $\displaystyle -\tfrac{1}{4}$ ✓.

Answer#

$x + 4y - 14 = 0$

Ques­tion 5: Dis­tance of a point from a line#

The prob­lem#

Find the dis­tance of $(3, -5)$ from $4x - 3y + 8 = 0$.

Under­stand­ing the prob­lem#

"Dis­tance from a point to a line" always means the per­pen­dic­u­lar (short­est) dis­tance.

The idea#

Use $\displaystyle d = \frac{\lvert Ax_1 + By_1 + C \rvert}{\sqrt{A^2 + B^2}}$ with $A = 4$, $B = -3$, $C = 8$, $(x_1, y_1) = (3, -5)$.

Step-by-step solu­tion#

Step 1. Sub­sti­tute the point into the left side of the line's equa­tion.

$$4(3) - 3(-5) + 8 = 12 + 15 + 8 = 35.$$

Step 2. Com­pute the denom­i­na­tor.

$$\sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5.$$

Step 3. Divide.

$$\displaystyle d = \frac{\lvert 35 \rvert}{5} = 7.$$

Check­ing the answer#

The answer is pos­i­tive (a dis­tance) and the num­bers come out whole, which is typ­i­cal when $A^2 + B^2$ is a per­fect square.

Answer#

$7$ units

Com­mon mis­take to avoid#

$-3 \times (-5) = +15$, not $-15$. Keep track of the signs before tak­ing the mod­u­lus.

Ques­tion 6: Dis­tance between par­al­lel lines#

The prob­lem#

Find the dis­tance between $x + 2y = 3$ and $2x + 4y = 11$.

Under­stand­ing the prob­lem#

These lines are par­al­lel (both have slope $\displaystyle -\tfrac{1}{2}$). The dis­tance between them is the same every­where.

The idea#

Use $\displaystyle d = \frac{\lvert C_1 - C_2 \rvert}{\sqrt{A^2 + B^2}}$. This for­mula needs both equa­tions to have the same $A$ and $B$, so first rescale one of them.

Step-by-step solu­tion#

Step 1. Mul­ti­ply the first equa­tion by $2$ so both have $2x + 4y$.

$$2x + 4y - 6 = 0, \qquad 2x + 4y - 11 = 0.$$

So $A = 2$, $B = 4$, $C_1 = -6$, $C_2 = -11$.

Step 2. Apply the for­mula.

$$\displaystyle d = \frac{\lvert -6 - (-11) \rvert}{\sqrt{2^2 + 4^2}} = \frac{5}{\sqrt{20}}.$$

Step 3. Sim­plify: $\sqrt{20} = 2\sqrt{5}$, and $\displaystyle \tfrac{5}{2\sqrt{5}} = \tfrac{\sqrt{5}}{2}$.

$$\displaystyle d = \frac{\sqrt{5}}{2}.$$

Check­ing the answer#

Take the point $(3, 0)$ on the first line. Its dis­tance from $2x + 4y - 11 = 0$ is $\displaystyle \tfrac{\lvert 6 - 11 \rvert}{\sqrt{20}} = \tfrac{5}{\sqrt{20}} = \tfrac{\sqrt{5}}{2}$ ✓.

Answer#

$\displaystyle \frac{\sqrt{5}}{2}$ units

Com­mon mis­take to avoid#

Using $C_1 = -3$ and $C_2 = -11$ directly is wrong, because the $x$ and $y$ coef­fi­cients were not yet matched.

Ques­tion 7: Line whose inter­cepted seg­ment is bisected at a point#

The prob­lem#

Find the equa­tion of the line whose seg­ment between the axes is bisected at $(2, 3)$.

Under­stand­ing the prob­lem#

The line cuts the $x$-axis at $A(a, 0)$ and the $y$-axis at $B(0, b)$. The piece $AB$ has its mid­point at $(2, 3)$.

The idea#

Use the mid­point for­mula to find $a$ and $b$, then use the inter­cept form.

Step-by-step solu­tion#

Step 1. Mid­point of $A(a, 0)$ and $B(0, b)$ is $\displaystyle \left(\tfrac{a}{2}, \tfrac{b}{2}\right)$. Set it equal to $(2, 3)$.

$$\displaystyle \frac{a}{2} = 2 \Rightarrow a = 4, \qquad \frac{b}{2} = 3 \Rightarrow b = 6.$$

Step 2. Inter­cept form.

$$\displaystyle \frac{x}{4} + \frac{y}{6} = 1.$$

Step 3. Mul­ti­ply by $12$.

$$3x + 2y = 12.$$

Check­ing the answer#

The line meets the axes at $(4, 0)$ and $(0, 6)$; their mid­point is $(2, 3)$ ✓. Also $(2, 3)$ is on the line: $6 + 6 = 12$ ✓.

Answer#

$3x + 2y = 12$

Ques­tion 8: Find­ing $k$ for per­pen­dic­u­lar lines#

The prob­lem#

Find $k$ if the lines $kx + 2y = 5$ and $3x - 6y = 1$ are per­pen­dic­u­lar.

Under­stand­ing the prob­lem#

Two lines are per­pen­dic­u­lar when the prod­uct of their slopes is $-1$. One slope con­tains the unknown $k$.

The idea#

Find both slopes using $\displaystyle -\tfrac{A}{B}$, set their prod­uct to $-1$, solve for $k$.

Step-by-step solu­tion#

Step 1. Slope of $kx + 2y = 5$:

$$\displaystyle m_1 = -\frac{k}{2}.$$

Step 2. Slope of $3x - 6y = 1$:

$$\displaystyle m_2 = -\frac{3}{-6} = \frac{1}{2}.$$

Step 3. Per­pen­dic­u­lar con­di­tion.

$$\displaystyle \left(-\frac{k}{2}\right)\left(\frac{1}{2}\right) = -1 \quad\Longrightarrow\quad -\frac{k}{4} = -1 \quad\Longrightarrow\quad k = 4.$$

Check­ing the answer#

With $k = 4$, the first line has slope $-2$; $\displaystyle (-2) \times \tfrac{1}{2} = -1$ ✓.

Answer#

$k = 4$

Ques­tion 9: Equa­tion of an alti­tude of a tri­an­gle#

The prob­lem#

Find the equa­tion of the alti­tude from $A$ in tri­an­gle $A(2, 4)$, $B(-1, 1)$, $C(5, -2)$.

Under­stand­ing the prob­lem#

The alti­tude from $A$ is the line through $A$ that is per­pen­dic­u­lar to the oppo­site side $BC$.

The idea#

Find the slope of $BC$, take the per­pen­dic­u­lar slope, and use point–slope form through $A$.

Step-by-step solu­tion#

Step 1. Slope of $BC$.

$$\displaystyle m_{BC} = \frac{-2 - 1}{5 - (-1)} = \frac{-3}{6} = -\frac{1}{2}.$$

Step 2. The alti­tude is per­pen­dic­u­lar to $BC$, so its slope is the neg­a­tive rec­i­p­ro­cal.

$$m = 2.$$

Step 3. Point–slope form through $A(2, 4)$.

$$y - 4 = 2(x - 2) = 2x - 4 \quad\Longrightarrow\quad y = 2x.$$

Step 4. Gen­eral form.

$$2x - y = 0.$$

Check­ing the answer#

$A(2, 4)$: $4 - 4 = 0$ ✓. Slopes $\displaystyle 2 \times \left(-\tfrac{1}{2}\right) = -1$ ✓.

Answer#

$2x - y = 0$

Ques­tion 10: Fahren­heit and Cel­sius as a lin­ear func­tion#

The prob­lem#

The Fahren­heit tem­per­a­ture is a lin­ear func­tion of the Cel­sius tem­per­a­ture, with $F = 32$ at $C = 0$ and $F = 212$ at $C = 100$. Write the equa­tion and find $C$ when $F = 104$.

Under­stand­ing the prob­lem#

Think of $C$ as the hor­i­zon­tal vari­able and $F$ as the ver­ti­cal one. You have two points, $(0, 32)$ and $(100, 212)$, on a straight line. Find the line, then use it back­wards.

The idea#

Two-point form: find the slope, then use the $F$-inter­cept $32$.

Step-by-step solu­tion#

Step 1. Slope.

$$\displaystyle m = \frac{212 - 32}{100 - 0} = \frac{180}{100} = \frac{9}{5}.$$

Step 2. The inter­cept (value of $F$ when $C = 0$) is $32$.

$$\displaystyle F = \frac{9}{5}C + 32.$$

Step 3. Put $F = 104$ and solve for $C$.

$$\displaystyle \begin{aligned} 104 &= \frac{9}{5}C + 32\\ 72 &= \frac{9}{5}C\\ C &= 72 \times \frac{5}{9} = 40. \end{aligned}$$

Check­ing the answer#

$\displaystyle \tfrac{9}{5}(100) + 32 = 180 + 32 = 212$ ✓, and $\displaystyle \tfrac{9}{5}(40) + 32 = 72 + 32 = 104$ ✓.

Answer#

$\displaystyle F = \tfrac{9}{5}C + 32$; when $F = 104$, $C = 40$.