Slope and the Angle Between Lines, Worked Solutions

Ques­tion 1: Slope and incli­na­tion from two points#

The prob­lem#

Find the slope of the line through $(4, -1)$ and $(-2, 5)$, and its incli­na­tion.

Under­stand­ing the prob­lem#

Two points on a line are given. The slope $m$ mea­sures steep­ness; the incli­na­tion $\theta$ is the angle the line makes with the pos­i­tive $x$-axis, with $0 \le \theta < 180^\circ$, and $m = \tan\theta$.

The idea#

Use $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$, then find the angle in $[0^\circ, 180^\circ)$ whose tan­gent is $m$.

Step-by-step solu­tion#

Step 1. Take $(x_1, y_1) = (4, -1)$ and $(x_2, y_2) = (-2, 5)$.

$$\displaystyle m = \frac{5 - (-1)}{-2 - 4} = \frac{6}{-6} = -1.$$

Step 2. Find $\theta$ with $\tan\theta = -1$ and $0^\circ \le \theta < 180^\circ$. Since $\tan 45^\circ = 1$ and tan­gent is neg­a­tive in the sec­ond quad­rant,

$$\theta = 180^\circ - 45^\circ = 135^\circ.$$

Check­ing the answer#

Going from $(4, -1)$ to $(-2, 5)$, $x$ decreases while $y$ increases, so the line falls to the right: a neg­a­tive slope and an obtuse incli­na­tion are both expected.

Answer#

Slope $-1$; incli­na­tion $135^\circ$.

Com­mon mis­take to avoid#

Giv­ing $\theta = -45^\circ$. Incli­na­tion is always taken between $0^\circ$ and $180^\circ$.

Ques­tion 2: Slope from the incli­na­tion#

The prob­lem#

Find the slope of (a) a line with incli­na­tion $150^\circ$; (b) the line bisect­ing the first quad­rant.

Under­stand­ing the prob­lem#

Here the angle is known and we need $m = \tan\theta$. In (b), "bisect­ing the first quad­rant" means the line through the ori­gin cut­ting the $90^\circ$ between the pos­i­tive axes into two equal halves.

The idea#

$m = \tan\theta$. For (b), the incli­na­tion is $\displaystyle \tfrac{90^\circ}{2} = 45^\circ$.

Step-by-step solu­tion#

Part (a)

Step 1. $m = \tan 150^\circ = \tan(180^\circ - 30^\circ) = -\tan 30^\circ$.

Step 2. $\displaystyle \tan 30^\circ = \tfrac{1}{\sqrt{3}}$, so $\displaystyle m = -\tfrac{1}{\sqrt{3}}$.

Part (b)

Step 1. The bisec­tor makes $45^\circ$ with the pos­i­tive $x$-axis.

Step 2. $m = \tan 45^\circ = 1$. (It is the line $y = x$.)

Check­ing the answer#

$150^\circ$ is obtuse, so the slope must be neg­a­tive, and it is.

Answer#

(a) $\displaystyle -\frac{1}{\sqrt{3}}$; (b) $1$.

Ques­tion 3: Prov­ing three points are collinear#

The prob­lem#

Show that the points $(-3, 7)$, $(1, 1)$ and $(5, -5)$ are collinear.

Under­stand­ing the prob­lem#

"Collinear" means all three lie on one straight line. You must prove it.

The idea#

$A$, $B$, $C$ are collinear exactly when slope $AB$ = slope $BC$: the two seg­ments share the point $B$, so equal slopes mean they are parts of the same line.

Step-by-step solu­tion#

Step 1. Name the points $A(-3, 7)$, $B(1, 1)$, $C(5, -5)$.

Step 2. Slope of $AB$:

$$\displaystyle m_{AB} = \frac{1 - 7}{1 - (-3)} = \frac{-6}{4} = -\frac{3}{2}.$$

Step 3. Slope of $BC$:

$$\displaystyle m_{BC} = \frac{-5 - 1}{5 - 1} = \frac{-6}{4} = -\frac{3}{2}.$$

Step 4. The slopes are equal and $B$ is com­mon to both seg­ments, so $AB$ and $BC$ lie on the same line. Hence $A$, $B$, $C$ are collinear. $\blacksquare$

Check­ing the answer#

Area test: $\displaystyle \tfrac12\lvert -3(1 + 5) + 1(-5 - 7) + 5(7 - 1)\rvert = \tfrac12\lvert -18 - 12 + 30\rvert = 0$. Zero area con­firms collinear­ity.

Answer#

Slope $AB$ = slope $BC$ = $\displaystyle -\tfrac{3}{2}$, so the points are collinear.

Ques­tion 4: Find­ing an unknown coor­di­nate from a slope#

The prob­lem#

Find $x$ if the slope of the line through $(x, 3)$ and $(2, -1)$ is $\displaystyle \tfrac{4}{5}$.

Under­stand­ing the prob­lem#

One coor­di­nate is unknown. The slope is given, which gives us an equa­tion in $x$.

The idea#

Write the slope for­mula with the two points, set it equal to $\displaystyle \tfrac45$, and solve.

Step-by-step solu­tion#

Step 1. Slope through $(x, 3)$ and $(2, -1)$:

$$\displaystyle \frac{-1 - 3}{2 - x} = \frac{-4}{2 - x}.$$

Step 2. Set it equal to $\displaystyle \tfrac45$ and cross-mul­ti­ply.

$$\displaystyle \frac{-4}{2 - x} = \frac{4}{5} \;\Longrightarrow\; -20 = 4(2 - x) \;\Longrightarrow\; 2 - x = -5.$$

Step 3. Solve: $x = 7$.

Check­ing the answer#

Slope through $(7, 3)$ and $(2, -1)$: $\displaystyle \tfrac{-1 - 3}{2 - 7} = \tfrac{-4}{-5} = \tfrac45$ ✓.

Answer#

$x = 7$.

Ques­tion 5: A right-angled tri­an­gle by slopes#

The prob­lem#

Show, using slopes, that $(2, 1)$, $(6, 3)$ and $(4, 7)$ are the ver­tices of a right-angled tri­an­gle.

Under­stand­ing the prob­lem#

You must prove that one angle of the tri­an­gle is $90^\circ$, using slopes rather than lengths.

The idea#

Two non-ver­ti­cal lines are per­pen­dic­u­lar exactly when $m_1 m_2 = -1$. Find the slopes of the three sides and look for a pair whose prod­uct is $-1$.

Step-by-step solu­tion#

Step 1. Name the points $A(2, 1)$, $B(6, 3)$, $C(4, 7)$.

Step 2. Slopes of the sides:

$$\displaystyle m_{AB} = \frac{3 - 1}{6 - 2} = \frac{2}{4} = \frac{1}{2}, \qquad m_{BC} = \frac{7 - 3}{4 - 6} = \frac{4}{-2} = -2, \qquad m_{CA} = \frac{1 - 7}{2 - 4} = \frac{-6}{-2} = 3.$$

Step 3. Test the prod­uct of $m_{AB}$ and $m_{BC}$:

$$\displaystyle \frac{1}{2} \times (-2) = -1.$$

Step 4. So $AB \perp BC$, and the angle at $B$ is $90^\circ$. Since $m_{CA} \ne m_{AB}$, the three points are not collinear, so they do form a tri­an­gle, and it is right-angled at $B$. $\blacksquare$

Check­ing the answer#

Pythago­ras: $AB^2 = 16 + 4 = 20$, $BC^2 = 4 + 16 = 20$, $CA^2 = 4 + 36 = 40 = 20 + 20$ ✓. (It is even isosce­les.)

Answer#

$\displaystyle m_{AB} \cdot m_{BC} = \tfrac12 \cdot (-2) = -1$, so the tri­an­gle is right-angled at $B(6, 3)$.

Ques­tion 6: Acute angle between two lines#

The prob­lem#

Find the acute angle between two lines whose slopes are $\displaystyle \tfrac{1}{2}$ and $3$.

Under­stand­ing the prob­lem#

Two inter­sect­ing lines form two pairs of angles; we want the acute one.

The idea#

Use $\displaystyle \tan\theta = \left\lvert \frac{m_2 - m_1}{1 + m_1 m_2} \right\rvert$; the mod­u­lus guar­an­tees the acute angle.

Step-by-step solu­tion#

Step 1. Take $\displaystyle m_1 = \tfrac12$, $m_2 = 3$. Numer­a­tor: $\displaystyle 3 - \tfrac12 = \tfrac52$. Denom­i­na­tor: $\displaystyle 1 + \tfrac12 \cdot 3 = \tfrac52$.

Step 2.

$$\displaystyle \tan\theta = \left\lvert \frac{5/2}{5/2} \right\rvert = 1.$$

Step 3. So $\theta = 45^\circ$.

Check­ing the answer#

The incli­na­tions are $\tan^{-1} 3 \approx 71.57^\circ$ and $\displaystyle \tan^{-1}\tfrac12 \approx 26.57^\circ$; their dif­fer­ence is $45.00^\circ$ ✓.

Answer#

$\theta = 45^\circ$.

Ques­tion 7: Per­pen­dic­u­lar lines through given points#

The prob­lem#

The line through $(-1, 3)$ and $(5, 5)$ is per­pen­dic­u­lar to the line through $(6, 10)$ and $(x, 22)$. Find $x$.

Under­stand­ing the prob­lem#

The first line is fully known. The sec­ond has an unknown $x$. Per­pen­dic­u­lar­ity links their slopes.

The idea#

Find the first slope, deduce the sec­ond from $m_1 m_2 = -1$, then solve for $x$ (as in Exam­ple 3).

Step-by-step solu­tion#

Step 1. First slope:

$$\displaystyle m_1 = \frac{5 - 3}{5 - (-1)} = \frac{2}{6} = \frac{1}{3}.$$

Step 2. Per­pen­dic­u­lar slope: $\displaystyle m_2 = -\frac{1}{m_1} = -3$.

Step 3. Slope of the sec­ond line from its points:

$$\displaystyle m_2 = \frac{22 - 10}{x - 6} = \frac{12}{x - 6}.$$

Step 4. Solve.

$$\displaystyle \frac{12}{x - 6} = -3 \;\Longrightarrow\; 12 = -3x + 18 \;\Longrightarrow\; 3x = 6 \;\Longrightarrow\; x = 2.$$

Check­ing the answer#

Sec­ond slope: $\displaystyle \tfrac{12}{2 - 6} = -3$; prod­uct $\displaystyle \tfrac13 \times (-3) = -1$ ✓.

Answer#

$x = 2$.

Ques­tion 8: Area of a tri­an­gle from its ver­tices#

The prob­lem#

Find the area of the tri­an­gle with ver­tices $(1, 2)$, $(6, 5)$ and $(3, 9)$.

Under­stand­ing the prob­lem#

All three ver­tices are known; we need the area in square units.

The idea#

Use the recap for­mula

$$\displaystyle \text{Area} = \tfrac{1}{2}\left\lvert x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right\rvert.$$

Step-by-step solu­tion#

Step 1. Label $(x_1, y_1) = (1, 2)$, $(x_2, y_2) = (6, 5)$, $(x_3, y_3) = (3, 9)$.

Step 2. Work out each bracket: $y_2 - y_3 = -4$, $y_3 - y_1 = 7$, $y_1 - y_2 = -3$.

Step 3. Sub­sti­tute.

$$\displaystyle \text{Area} = \tfrac12\left\lvert 1(-4) + 6(7) + 3(-3) \right\rvert = \tfrac12\lvert -4 + 42 - 9 \rvert = \tfrac12 \times 29 = 14.5.$$

Check­ing the answer#

Enclose the tri­an­gle in the rec­tan­gle $1 \le x \le 6$, $2 \le y \le 9$ (area $35$) and sub­tract the three cor­ner right tri­an­gles: $\displaystyle \tfrac12(5)(3) + \tfrac12(3)(4) + \tfrac12(2)(7) = 7.5 + 6 + 7 = 20.5$. $35 - 20.5 = 14.5$ ✓.

Answer#

The area is $\displaystyle \tfrac{29}{2} = 14.5$ square units.

Ques­tion 9: A line at 60° to y = √3 x#

The prob­lem#

A line has slope $m$ and makes an angle of $60^\circ$ with the line $y = \sqrt{3}x$. Find $m$.

Under­stand­ing the prob­lem#

The line $y = \sqrt3 x$ has slope $\sqrt3$, so its incli­na­tion is $60^\circ$. We need every slope $m$ of a line mak­ing $60^\circ$ with it. Expect two answers, one on each side.

The idea#

Two routes agree: use incli­na­tions directly, or use the angle for­mula $\displaystyle \tan\theta = \left\lvert\tfrac{m - m_1}{1 + m m_1}\right\rvert$ with $\theta = 60^\circ$.

Step-by-step solu­tion#

Step 1 (incli­na­tions). The given line has incli­na­tion $60^\circ$. A line at $60^\circ$ to it has incli­na­tion $60^\circ + 60^\circ = 120^\circ$ or $60^\circ - 60^\circ = 0^\circ$.

Step 2. Their slopes: $\tan 120^\circ = -\sqrt3$ and $\tan 0^\circ = 0$.

Step 3 (check with the for­mula). With $m_1 = \sqrt3$, $\tan 60^\circ = \sqrt3$:

$$\displaystyle \left\lvert \frac{m - \sqrt3}{1 + \sqrt3 m} \right\rvert = \sqrt3.$$

Case $+$: $m - \sqrt3 = \sqrt3 + 3m \Rightarrow -2m = 2\sqrt3 \Rightarrow m = -\sqrt3$.

Case $-$: $m - \sqrt3 = -\sqrt3 - 3m \Rightarrow 4m = 0 \Rightarrow m = 0$.

Check­ing the answer#

$m = 0$: $\displaystyle \left\lvert\tfrac{-\sqrt3}{1}\right\rvert = \sqrt3$ ✓. $m = -\sqrt3$: $\displaystyle \left\lvert\tfrac{-2\sqrt3}{1 - 3}\right\rvert = \sqrt3$ ✓.

Answer#

$m = 0$ or $m = -\sqrt{3}$ (incli­na­tions $0^\circ$ or $120^\circ$).

Com­mon mis­take to avoid#

Drop­ping the mod­u­lus and keep­ing only one case, that loses one of the two lines.

Ques­tion 10: A par­al­lel­o­gram that is a rhom­bus#

The prob­lem#

Using slopes, show that the points $(1, 1)$, $(5, 2)$, $(6, 6)$ and $(2, 5)$, taken in order, form a par­al­lel­o­gram, in fact a rhom­bus.

Under­stand­ing the prob­lem#

A par­al­lel­o­gram has both pairs of oppo­site sides par­al­lel. A rhom­bus is a par­al­lel­o­gram with all four sides equal. So we must show (i) oppo­site sides are par­al­lel (equal slopes), and (ii) all sides have equal length.

The idea#

Find the four side slopes and com­pare oppo­site pairs; then use the dis­tance for­mula for the lengths.

Step-by-step solu­tion#

Step 1. Name $A(1, 1)$, $B(5, 2)$, $C(6, 6)$, $D(2, 5)$.

Step 2. Slopes of the sides:

$$\displaystyle m_{AB} = \frac{2 - 1}{5 - 1} = \frac14, \qquad m_{BC} = \frac{6 - 2}{6 - 5} = 4,$$

$$\displaystyle m_{CD} = \frac{5 - 6}{2 - 6} = \frac{-1}{-4} = \frac14, \qquad m_{DA} = \frac{1 - 5}{1 - 2} = \frac{-4}{-1} = 4.$$

Step 3. $m_{AB} = m_{CD}$, so $AB \parallel CD$; $m_{BC} = m_{DA}$, so $BC \parallel DA$. Both pairs of oppo­site sides are par­al­lel, so $ABCD$ is a par­al­lel­o­gram.

Step 4. Lengths of the sides:

$$AB = \sqrt{4^2 + 1^2} = \sqrt{17}, \quad BC = \sqrt{1^2 + 4^2} = \sqrt{17}, \quad CD = \sqrt{4^2 + 1^2} = \sqrt{17}, \quad DA = \sqrt{1^2 + 4^2} = \sqrt{17}.$$

Step 5. All four sides are equal, so the par­al­lel­o­gram is a rhom­bus. $\blacksquare$

Check­ing the answer#

Since $\displaystyle \tfrac14 \times 4 = 1 \ne -1$, adja­cent sides are not per­pen­dic­u­lar, so it is a rhom­bus but not a square. The diag­o­nals $AC$ (slope $1$) and $BD$ (slope $\displaystyle \tfrac{5 - 2}{2 - 5} = -1$) are per­pen­dic­u­lar, as they must be in a rhom­bus ✓.

Answer#

Oppo­site sides have slopes $\displaystyle \tfrac14, \tfrac14$ and $4, 4$ (par­al­lel), and all four sides equal $\sqrt{17}$, so $ABCD$ is a rhom­bus.