Straight Lines: Mixed Practice, Worked Solutions

Ques­tion 1: Line from a point and an incli­na­tion#

The prob­lem#

Find the equa­tion of the line through $(-2, 3)$ with incli­na­tion $135^\circ$.

Under­stand­ing the prob­lem#

The incli­na­tion is the angle the line makes with the pos­i­tive $x$-axis, mea­sured anti­clock­wise. Its tan­gent is the slope. You also know one point on the line.

The idea#

Find the slope $m = \tan 135^\circ$, then use the point–slope form $y - y_0 = m(x - x_0)$.

Step-by-step solu­tion#

Step 1. Find the slope.

$$m = \tan 135^\circ = \tan(180^\circ - 45^\circ) = -\tan 45^\circ = -1$$

Step 2. Use point–slope form with $(x_0, y_0) = (-2, 3)$.

$$y - 3 = -1\,(x + 2)$$

Step 3. Rearrange into gen­eral form.

$$y - 3 = -x - 2 \;\Rightarrow\; x + y - 1 = 0$$

Check­ing the answer#

$(-2) + 3 - 1 = 0$, so the point lies on the line ✓, and the slope of $x + y - 1 = 0$ is $\displaystyle -\tfrac{1}{1} = -1$. ✓

Answer#

$x + y - 1 = 0$

Ques­tion 2: A value that makes three points collinear#

The prob­lem#

Find $p$ if $(p, 1)$, $(3, 5)$ and $(7, 13)$ are collinear.

Under­stand­ing the prob­lem#

Collinear means all three points lie on one straight line. Two of the points are fully known, so they fix the line; the third must also lie on it.

The idea#

Points are collinear exactly when the slope between any two pairs is the same. Find the slope from the two known points and match it.

Step-by-step solu­tion#

Step 1. Slope through $(3, 5)$ and $(7, 13)$.

$$\displaystyle m = \frac{13 - 5}{7 - 3} = \frac{8}{4} = 2$$

Step 2. The slope through $(p, 1)$ and $(3, 5)$ must also be $2$.

$$\displaystyle \frac{5 - 1}{3 - p} = 2$$

Step 3. Solve.

$$4 = 2(3 - p) \;\Rightarrow\; 2 = 3 - p \;\Rightarrow\; p = 1$$

Check­ing the answer#

The line through $(3, 5)$ with slope 2 is $y = 2x - 1$. At $x = 1$, $y = 1$, so $(1, 1)$ is on it. ✓

Answer#

$p = 1$

Ques­tion 3: The medi­ans of a tri­an­gle#

The prob­lem#

Find the equa­tions of the medi­ans of the tri­an­gle $A(0, 0)$, $B(6, 0)$, $C(2, 8)$ and ver­ify that they meet at $\displaystyle \left(\tfrac{8}{3}, \tfrac{8}{3}\right)$.

Under­stand­ing the prob­lem#

A median joins a ver­tex to the mid­point of the oppo­site side. There are three medi­ans. You must find each one's equa­tion and then show the point $\displaystyle \left(\tfrac83, \tfrac83\right)$ lies on all three.

The idea#

Find the three mid­points with the mid­point for­mula, then use the two-point form for each median. Finally sub­sti­tute the given point.

Step-by-step solu­tion#

Step 1. Mid­points of the sides.

$$\displaystyle \text{mid } BC = \left(\tfrac{6 + 2}{2}, \tfrac{0 + 8}{2}\right) = (4, 4), \quad \text{mid } CA = (1, 4), \quad \text{mid } AB = (3, 0)$$

Step 2. Median from $A(0, 0)$ to $(4, 4)$: slope $\displaystyle \tfrac{4}{4} = 1$, through the ori­gin.

$$y = x$$

Step 3. Median from $B(6, 0)$ to $(1, 4)$: slope $\displaystyle \tfrac{4 - 0}{1 - 6} = -\tfrac45$.

$$\displaystyle y - 0 = -\tfrac45(x - 6) \;\Rightarrow\; 5y = -4x + 24 \;\Rightarrow\; 4x + 5y = 24$$

Step 4. Median from $C(2, 8)$ to $(3, 0)$: slope $\displaystyle \tfrac{0 - 8}{3 - 2} = -8$.

$$y - 0 = -8(x - 3) \;\Rightarrow\; 8x + y = 24$$

Step 5. Sub­sti­tute $\displaystyle \left(\tfrac83, \tfrac83\right)$ into each.

$$\displaystyle \begin{aligned} y = x &: \tfrac83 = \tfrac83 \\ 4x + 5y &= \tfrac{32}{3} + \tfrac{40}{3} = \tfrac{72}{3} = 24 \\ 8x + y &= \tfrac{64}{3} + \tfrac{8}{3} = \tfrac{72}{3} = 24 \end{aligned}$$

All three are sat­is­fied, so the medi­ans meet at $\displaystyle \left(\tfrac83, \tfrac83\right)$.

Check­ing the answer#

The cen­troid is the aver­age of the ver­tices: $\displaystyle \left(\tfrac{0 + 6 + 2}{3}, \tfrac{0 + 0 + 8}{3}\right) = \left(\tfrac83, \tfrac83\right)$. ✓

Answer#

Medi­ans $y = x$, $4x + 5y = 24$, $8x + y = 24$; all pass through $\displaystyle \left(\tfrac83, \tfrac83\right)$.

Ques­tion 4: Point on the x-axis equidis­tant from two points#

The prob­lem#

Find the point on the $x$-axis equidis­tant from $(1, 4)$ and $(5, 2)$.

Under­stand­ing the prob­lem#

Any point on the $x$-axis has the form $(x, 0)$. "Equidis­tant" means its dis­tance to $(1, 4)$ equals its dis­tance to $(5, 2)$.

The idea#

Equal dis­tances means equal squared dis­tances, which avoids square roots. Use the dis­tance for­mula.

Step-by-step solu­tion#

Step 1. Set the squared dis­tances equal.

$$(x - 1)^2 + (0 - 4)^2 = (x - 5)^2 + (0 - 2)^2$$

Step 2. Expand.

$$x^2 - 2x + 1 + 16 = x^2 - 10x + 25 + 4$$

Step 3. Can­cel $x^2$ and col­lect terms.

$$\displaystyle -2x + 17 = -10x + 29 \;\Rightarrow\; 8x = 12 \;\Rightarrow\; x = \frac32$$

Check­ing the answer#

Dis­tance$^2$ to $(1, 4)$: $\displaystyle \left(\tfrac12\right)^2 + 16 = \tfrac{65}{4}$. To $(5, 2)$: $\displaystyle \left(\tfrac72\right)^2 + 4 = \tfrac{49}{4} + \tfrac{16}{4} = \tfrac{65}{4}$. ✓

Answer#

$\displaystyle \left(\tfrac32, 0\right)$

Ques­tion 5: A line through an inter­sec­tion, per­pen­dic­u­lar to a given line#

The prob­lem#

Find the equa­tion of the line through the inter­sec­tion of $2x + y = 5$ and $x - 3y = -1$ and per­pen­dic­u­lar to $x + y = 0$.

Under­stand­ing the prob­lem#

The required line passes through the point where the first two lines cross, and it must be at right angles to $x + y = 0$.

The idea#

Solve the two equa­tions for the point; per­pen­dic­u­lar lines have slopes whose prod­uct is $-1$; then use point–slope form.

Step-by-step solu­tion#

Step 1. From $x - 3y = -1$, $x = 3y - 1$. Sub­sti­tute into $2x + y = 5$.

$$2(3y - 1) + y = 5 \;\Rightarrow\; 7y = 7 \;\Rightarrow\; y = 1, \quad x = 2$$

Step 2. The slope of $x + y = 0$ is $-1$, so the per­pen­dic­u­lar slope is $1$ (because $-1 \times 1 = -1$).

Step 3. Line through $(2, 1)$ with slope $1$.

$$y - 1 = x - 2 \;\Rightarrow\; x - y - 1 = 0$$

Check­ing the answer#

$(2, 1)$: $2 - 1 - 1 = 0$ ✓; both orig­i­nal lines pass through $(2, 1)$: $4 + 1 = 5$ ✓ and $2 - 3 = -1$ ✓.

Answer#

$x - y - 1 = 0$

Ques­tion 6: Angle between two lines#

The prob­lem#

Find the angle between $\sqrt{3}x - y = 2$ and $x - \sqrt{3}y = 5$.

Under­stand­ing the prob­lem#

"The angle between two lines" means the acute angle. You need the slope of each line first.

The idea#

Use $\displaystyle \tan\theta = \left\lvert \frac{m_2 - m_1}{1 + m_1 m_2} \right\rvert$.

Step-by-step solu­tion#

Step 1. Slopes. From $y = \sqrt3 x - 2$, $m_1 = \sqrt3$. From $\sqrt3 y = x - 5$, $\displaystyle m_2 = \frac{1}{\sqrt3}$.

Step 2. Sub­sti­tute.

$$\displaystyle \tan\theta = \left\lvert \frac{\tfrac{1}{\sqrt3} - \sqrt3}{1 + \sqrt3 \cdot \tfrac{1}{\sqrt3}} \right\rvert = \left\lvert \frac{\tfrac{1 - 3}{\sqrt3}}{2} \right\rvert = \frac{1}{\sqrt3}$$

Step 3. So $\theta = 30^\circ$.

Check­ing the answer#

The first line has incli­na­tion $60^\circ$ ($\tan 60^\circ = \sqrt3$), the sec­ond $30^\circ$ ($\displaystyle \tan 30^\circ = \tfrac{1}{\sqrt3}$); the dif­fer­ence is $30^\circ$. ✓

Answer#

$30^\circ$

Ques­tion 7: Image of a point in a line#

The prob­lem#

Find the image of $(1, 2)$ in the line $x - 3y + 4 = 0$.

Under­stand­ing the prob­lem#

The image $Q$ of a point $P$ in a line is its mir­ror reflec­tion: the line is the per­pen­dic­u­lar bisec­tor of $PQ$. So the foot of the per­pen­dic­u­lar from $P$ is the mid­point of $P$ and its image.

The idea#

Find the foot $F$ of the per­pen­dic­u­lar from $P(1, 2)$, then use $F =$ mid­point of $PQ$ to find $Q$.

Step-by-step solu­tion#

Step 1. The given line has slope $\displaystyle \tfrac13$, so the per­pen­dic­u­lar through $(1, 2)$ has slope $-3$.

$$y - 2 = -3(x - 1) \;\Rightarrow\; 3x + y - 5 = 0$$

Step 2. Solve with $x - 3y + 4 = 0$. From the per­pen­dic­u­lar, $y = 5 - 3x$.

$$\displaystyle x - 3(5 - 3x) + 4 = 0 \;\Rightarrow\; 10x - 11 = 0 \;\Rightarrow\; x = \frac{11}{10}, \quad y = 5 - \frac{33}{10} = \frac{17}{10}$$

So the foot is $\displaystyle F\left(\tfrac{11}{10}, \tfrac{17}{10}\right)$.

Step 3. $F$ is the mid­point of $P(1, 2)$ and $Q(h, k)$.

$$\displaystyle \frac{1 + h}{2} = \frac{11}{10} \;\Rightarrow\; h = \frac{6}{5}, \qquad \frac{2 + k}{2} = \frac{17}{10} \;\Rightarrow\; k = \frac{7}{5}$$

Check­ing the answer#

$P$ and $Q$ must be the same dis­tance from the line: $\displaystyle \tfrac{\lvert 1 - 6 + 4 \rvert}{\sqrt{10}} = \tfrac{1}{\sqrt{10}}$ and $\displaystyle \tfrac{\lvert \tfrac65 - \tfrac{21}{5} + 4 \rvert}{\sqrt{10}} = \tfrac{1}{\sqrt{10}}$ ✓ (on oppo­site sides, since the signs $-1$ and $+1$ dif­fer).

Answer#

The image is $\displaystyle \left(\tfrac65, \tfrac75\right)$.

Ques­tion 8: Lines at 45° to a given line#

The prob­lem#

Find the equa­tions of the lines through $(1, 2)$ mak­ing an angle of $45^\circ$ with $3x - y = 7$.

Under­stand­ing the prob­lem#

There are usu­ally two such lines, one on each side. You know a point on each, so you need their slopes $m$.

The idea#

The given line has slope $3$. Use $\displaystyle \tan 45^\circ = \left\lvert \frac{m - 3}{1 + 3m} \right\rvert = 1$ and solve both sign cases.

Step-by-step solu­tion#

Step 1. Set up the equa­tion.

$$\displaystyle \left\lvert \frac{m - 3}{1 + 3m} \right\rvert = 1$$

Step 2. Case $+1$: $m - 3 = 1 + 3m \Rightarrow -2m = 4 \Rightarrow m = -2$.

Step 3. Case $-1$: $\displaystyle m - 3 = -1 - 3m \Rightarrow 4m = 2 \Rightarrow m = \tfrac12$.

Step 4. Lines through $(1, 2)$.

$$\displaystyle \begin{aligned} m = -2&: \; y - 2 = -2(x - 1) \;\Rightarrow\; 2x + y - 4 = 0 \\ m = \tfrac12&: \; y - 2 = \tfrac12(x - 1) \;\Rightarrow\; x - 2y + 3 = 0 \end{aligned}$$

Check­ing the answer#

For $\displaystyle m = \tfrac12$: $\displaystyle \left\lvert \tfrac{0.5 - 3}{1 + 1.5} \right\rvert = \tfrac{2.5}{2.5} = 1$ ✓. The two lines are per­pen­dic­u­lar to each other ($\displaystyle -2 \times \tfrac12 = -1$), as the two $45^\circ$ lines always are.

Answer#

$2x + y - 4 = 0$ and $x - 2y + 3 = 0$

Ques­tion 9: Dis­tance mea­sured along a given direc­tion#

The prob­lem#

Find the dis­tance of the point $(2, 3)$ from the line $2x - 3y + 9 = 0$ mea­sured along the line $x - y + 1 = 0$.

Under­stand­ing the prob­lem#

This is not the per­pen­dic­u­lar dis­tance. You travel from $(2, 3)$ along the line $x - y + 1 = 0$ until you hit $2x - 3y + 9 = 0$, and mea­sure that stretch. Note that $(2, 3)$ lies on $x - y + 1 = 0$ ($2 - 3 + 1 = 0$).

The idea#

Find where the two lines meet, then use the dis­tance for­mula between $(2, 3)$ and that point.

Step-by-step solu­tion#

Step 1. From $x - y + 1 = 0$, $y = x + 1$. Sub­sti­tute into $2x - 3y + 9 = 0$.

$$2x - 3(x + 1) + 9 = 0 \;\Rightarrow\; -x + 6 = 0 \;\Rightarrow\; x = 6, \quad y = 7$$

Step 2. Dis­tance from $(2, 3)$ to $(6, 7)$.

$$\sqrt{(6 - 2)^2 + (7 - 3)^2} = \sqrt{16 + 16} = 4\sqrt2$$

Check­ing the answer#

The per­pen­dic­u­lar dis­tance is $\displaystyle \tfrac{\lvert 4 - 9 + 9 \rvert}{\sqrt{13}} = \tfrac{4}{\sqrt{13}} \approx 1.11$, which is less than $4\sqrt2 \approx 5.66$, as it must be, since the per­pen­dic­u­lar is the short­est route.

Answer#

$4\sqrt2$ units

Ques­tion 10: Area of a tri­an­gle formed by two lines and the y-axis#

The prob­lem#

Show that the area of the tri­an­gle formed by $y = m_1x + c_1$, $y = m_2x + c_2$ and $x = 0$ is $\displaystyle \frac{(c_1 - c_2)^2}{2\lvert m_1 - m_2 \rvert}$.

Under­stand­ing the prob­lem#

This is a proof. The three lines are the two given lines and the $y$-axis ($x = 0$). The tri­an­gle's ver­tices are where these lines meet in pairs. Assume $m_1 \ne m_2$ so the lines do cross.

The idea#

Take the side on the $y$-axis as the base. The height is then the hor­i­zon­tal dis­tance from the third ver­tex to the $y$-axis, which is $\lvert x \rvert$ of that ver­tex.

Step-by-step solu­tion#

Step 1. Where each line meets $x = 0$: put $x = 0$.

$$(0, c_1) \quad\text{and}\quad (0, c_2)$$

Step 2. The base is the gap between these points on the $y$-axis.

$$\text{base} = \lvert c_1 - c_2 \rvert$$

Step 3. Where the two lines meet: set $m_1x + c_1 = m_2x + c_2$.

$$\displaystyle (m_1 - m_2)x = c_2 - c_1 \;\Rightarrow\; x = \frac{c_2 - c_1}{m_1 - m_2}$$

Step 4. The height is the dis­tance of this ver­tex from the $y$-axis.

$$\displaystyle \text{height} = \frac{\lvert c_1 - c_2 \rvert}{\lvert m_1 - m_2 \rvert}$$

Step 5. Area $\displaystyle = \tfrac12 \times$ base $\times$ height.

$$\displaystyle \text{Area} = \frac12 \lvert c_1 - c_2 \rvert \cdot \frac{\lvert c_1 - c_2 \rvert}{\lvert m_1 - m_2 \rvert} = \frac{(c_1 - c_2)^2}{2\lvert m_1 - m_2 \rvert}$$

This is the required result.

Check­ing the answer#

Take $y = x$ and $y = -x + 2$: ver­tices $(0, 0)$, $(0, 2)$, $(1, 1)$, area $\displaystyle \tfrac12 \times 2 \times 1 = 1$. The for­mula gives $\displaystyle \tfrac{(0 - 2)^2}{2 \times 2} = 1$. ✓

Answer#

Area $\displaystyle = \frac{(c_1 - c_2)^2}{2\lvert m_1 - m_2 \rvert}$, as required.

Ques­tion 11: Fixed charge from a lin­ear phone bill#

The prob­lem#

A per­son's monthly phone bill is ₹250 when she uses $300$ min­utes and ₹370 when she uses $500$ min­utes. Assum­ing a lin­ear rela­tion, what is the fixed charge?

Under­stand­ing the prob­lem#

"Lin­ear rela­tion" means bill $=$ (fixed charge) $+$ (rate per minute) $\times$ (min­utes), i.e. $y = mx + c$ with $x$ in min­utes and $y$ in rupees. The fixed charge is $c$, the bill at $0$ min­utes.

The idea#

The two data points $(300, 250)$ and $(500, 370)$ lie on the line. Find the slope, then the $y$-inter­cept.

Step-by-step solu­tion#

Step 1. Slope (rate per minute).

$$\displaystyle m = \frac{370 - 250}{500 - 300} = \frac{120}{200} = 0.6$$

Step 2. Use $(300, 250)$ in $y = mx + c$.

$$250 = 0.6 \times 300 + c = 180 + c \;\Rightarrow\; c = 70$$

Check­ing the answer#

At $500$ min­utes: $0.6 \times 500 + 70 = 300 + 70 = 370$. ✓

Answer#

The fixed charge is ₹70 (and the rate is ₹0.60 per minute).

Ques­tion 12: Line in nor­mal form#

The prob­lem#

Find the equa­tion of the line whose per­pen­dic­u­lar dis­tance from the ori­gin is $4$ and whose nor­mal makes $30^\circ$ with the pos­i­tive $x$-axis.

Under­stand­ing the prob­lem#

The nor­mal is the per­pen­dic­u­lar from the ori­gin to the line. Its length is $p = 4$ and it makes angle $\omega = 30^\circ$ with the $x$-axis.

The idea#

Use the nor­mal form $x\cos\omega + y\sin\omega = p$.

Step-by-step solu­tion#

Step 1. Sub­sti­tute $\omega = 30^\circ$, $p = 4$.

$$x\cos 30^\circ + y\sin 30^\circ = 4$$

Step 2. Use $\displaystyle \cos 30^\circ = \tfrac{\sqrt3}{2}$, $\displaystyle \sin 30^\circ = \tfrac12$.

$$\displaystyle \frac{\sqrt3}{2}x + \frac12 y = 4$$

Step 3. Mul­ti­ply by 2.

$$\sqrt3 x + y = 8$$

Check­ing the answer#

Dis­tance from the ori­gin: $\displaystyle \tfrac{\lvert -8 \rvert}{\sqrt{3 + 1}} = \tfrac82 = 4$. ✓

Answer#

$\sqrt3 x + y = 8$