Solved Examples

Solved Examples — the Full Chapter Workout

Thirty-plus worked problems: slope and coordinate drills, every equation form, distances, images and reflections, and the full range of miscellaneous-level classics. Attempt each before reading the solution.

Example 1: Coordinate warm-ups

(i) The base of an equilateral triangle of side $2a$ lies along the $y$-axis with its midpoint at the origin. Find the vertices. (ii) Find the distance between $P(x_1, y_1)$ and $Q(x_2, y_2)$ when $PQ$ is parallel to the $y$-axis; parallel to the $x$-axis.

Solution:

Step 1 — (i) Place the base. Ends at $(0, a)$ and $(0, -a)$.

Step 2 — (i) Locate the apex. It lies on the $x$-axis (perpendicular bisector of the base) at distance $\sqrt{(2a)^2 - a^2} = \sqrt{3}a$ from the origin: $\left(\pm\sqrt{3}a, 0\right)$.

Step 3 — (ii) Axis-parallel distances. Parallel to the $y$-axis ($x_1 = x_2$): $|y_2 - y_1|$; parallel to the $x$-axis: $|x_2 - x_1|$.

Takeaway: Placing figures symmetrically about the origin turns geometry into single-coordinate arithmetic.

Example 2: The simplest equations

Write the equations of (i) the $x$-axis and $y$-axis, (ii) the line through $(-4, 3)$ with slope $\frac{1}{2}$, (iii) the line through the origin with slope $m$.

Solution:

Step 1 — (i). $x$-axis: $y = 0$; $y$-axis: $x = 0$.

Step 2 — (ii). $y - 3 = \frac{1}{2}(x + 4)$; multiply by 2: $x - 2y + 10 = 0$.

Step 3 — (iii). Point-slope at the origin: $y = mx$.

Takeaway: Every one of these is point-slope form with progressively simpler anchors.

Example 3: Two-point form

Find the equation of the line through $(-1, 1)$ and $(2, -4)$.

Solution:

Step 1 — Slope. $m = \frac{-4 - 1}{2 - (-1)} = -\frac{5}{3}$.

Step 2 — Point-slope. $y - 1 = -\frac{5}{3}(x + 1)$; multiply by 3: $3y - 3 = -5x - 5$.

Step 3 — General form. $5x + 3y + 2 = 0$. Check $(2, -4)$: $10 - 12 + 2 = 0$ ✓.

Takeaway: Clear fractions before rearranging — sign slips live in the $\times 3$ step.

Example 4: A ratio condition

A line perpendicular to the segment joining $(1, 0)$ and $(2, 3)$ divides it in the ratio $1 : n$. Find its equation.

Solution:

Step 1 — Section point. Dividing $1 : n$ from $(1,0)$: $\left(\frac{n + 2}{n + 1}, \frac{3}{n + 1}\right)$.

Step 2 — Perpendicular slope. Segment slope 3 → line slope $-\frac{1}{3}$.

Step 3 — Point-slope and clear. $y - \frac{3}{n+1} = -\frac{1}{3}\left(x - \frac{n+2}{n+1}\right)$ multiplies out to $(1 + n)x + 3(1 + n)y = n + 11$.

Takeaway: Keep the parameter $n$ symbolic throughout — it cancels into a single tidy family of lines.

Example 5: An angle of $\frac{2\pi}{3}$

Find the line through $(0, 2)$ making angle $\frac{2\pi}{3}$ with the positive $x$-axis, and the line parallel to it crossing the $y$-axis 2 units below the origin.

Solution:

Step 1 — Slope from the angle. $m = \tan\frac{2\pi}{3} = -\sqrt{3}$.

Step 2 — First line. Through $(0, 2)$: $y = -\sqrt{3}x + 2$, i.e. $\sqrt{3}x + y - 2 = 0$.

Step 3 — Parallel line. Through $(0, -2)$: $\sqrt{3}x + y + 2 = 0$.

Takeaway: Parallel lines differ only in the constant — the slope work is done once.

Example 6: The copper rod model

The length $L$ (cm) of a copper rod is a linear function of its temperature $C$ (°C): $L = 124.942$ at $C = 20$ and $L = 125.134$ at $C = 110$. Express $L$ in terms of $C$.

Solution:

Step 1 — Slope from the data. $\frac{125.134 - 124.942}{110 - 20} = \frac{0.192}{90}$ cm per °C.

Step 2 — Point-slope. $L = 124.942 + \frac{0.192}{90}(C - 20)$.

Takeaway: "Linear function of" = two-point form on the data pairs; leave the slope as an exact fraction.

Example 7: Lines at $60°$ to a given line

Two lines through $(2, 3)$ meet at $60°$; one has slope 2. Find the equation of the other.

Solution:

Step 1 — Angle equation. $\left|\frac{m - 2}{1 + 2m}\right| = \tan 60° = \sqrt{3}$.

Step 2 — Solve both signs. $m - 2 = \pm\sqrt{3}(1 + 2m)$ gives $m = \frac{5\sqrt{3} - 8}{11}$ (from $-$) and $m = -\frac{8 + 5\sqrt{3}}{11}$ (from $+$).

Step 3 — Write the lines. $y - 3 = \frac{5\sqrt{3} - 8}{11}(x - 2)$ and $y - 3 = -\frac{8 + 5\sqrt{3}}{11}(x - 2)$.

Takeaway: Surd slopes are normal here — resist the urge to decimalise; the exact forms are the expected answers.

Example 8: A trigonometric distance identity

If $p$ and $q$ are the distances of the origin from $x\cos\theta - y\sin\theta = k\cos 2\theta$ and $x\sec\theta + y\,\mathrm{cosec}\,\theta = k$, prove $p^2 + 4q^2 = k^2$.

Solution:

Step 1 — Compute $p$. Coefficient vector $(\cos\theta, -\sin\theta)$ has length 1: $p = |k\cos 2\theta|$.

Step 2 — Compute $q$. $\sqrt{\sec^2\theta + \mathrm{cosec}^2\theta} = \frac{1}{|\sin\theta\cos\theta|}$, so $q = |k\sin\theta\cos\theta| = \frac{|k\sin 2\theta|}{2}$.

Step 3 — Combine. $p^2 + 4q^2 = k^2\cos^2 2\theta + k^2\sin^2 2\theta = k^2$. ∎

Takeaway: Distances built from unit or normalisable coefficient vectors collapse via $\sin^2 + \cos^2 = 1$.

Example 9: Concurrency

If $2x + y - 3 = 0$, $5x + ky - 3 = 0$ and $3x - y - 2 = 0$ are concurrent, find $k$.

Solution:

Step 1 — Intersect the two known lines. $2x + y = 3$ and $3x - y = 2$: adding, $5x = 5$, so the point is $(1, 1)$.

Step 2 — Force the third line through it. $5(1) + k(1) - 3 = 0$.

Step 3 — Solve. $k = -2$.

Takeaway: Concurrency = intersect two, substitute into the third — never solve all three simultaneously.

Example 10: Distance measured along a direction

Find the distance of the line $4x - y = 0$ from $P(4, 1)$ measured along the line making $135°$ with the positive $x$-axis.

Solution:

Step 1 — Draw the measuring line. Slope $\tan 135° = -1$ through $P$: $y - 1 = -(x - 4)$, i.e. $x + y - 5 = 0$.

Step 2 — Intersect with the target line. $4x - y = 0$ and $x + y = 5$: $5x = 5$, so $Q(1, 4)$.

Step 3 — Measure the segment. $PQ = \sqrt{(1-4)^2 + (4-1)^2} = \sqrt{18} = 3\sqrt{2}$.

Takeaway: "Distance along a direction" is NOT the perpendicular distance — draw, intersect, measure.

Example 11: Image in a mirror line

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

Solution:

Step 1 — Perpendicularity condition. $PQ \perp$ line (line slope $\frac{1}{3}$): $\frac{k - 2}{h - 1} = -3$, i.e. $3h + k = 5$.

Step 2 — Midpoint condition. $\left(\frac{h+1}{2}, \frac{k+2}{2}\right)$ on the line: $\frac{h+1}{2} - \frac{3(k+2)}{2} + 4 = 0$, i.e. $h - 3k = -3$.

Step 3 — Solve the pair. $h = \frac{6}{5}$, $k = \frac{7}{5}$: the image is $\left(\frac{6}{5}, \frac{7}{5}\right)$.

Takeaway: The mirror line is the perpendicular bisector of point and image — two conditions, two unknowns, always.

Example 12: Area of a triangle from three lines

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

Solution:

Step 1 — Find the vertices. On $x = 0$: $P(0, c_1)$ and $Q(0, c_2)$. The slanted lines meet at $R$ with $x_R = \frac{c_2 - c_1}{m_1 - m_2}$.

Step 2 — Base and height. Base $PQ = |c_1 - c_2|$ along the $y$-axis; height $= |x_R| = \frac{|c_1 - c_2|}{|m_1 - m_2|}$.

Step 3 — Area. $\frac{1}{2}|c_1 - c_2| \cdot \frac{|c_1 - c_2|}{|m_1 - m_2|} = \frac{(c_1 - c_2)^2}{2|m_1 - m_2|}$. ∎

Takeaway: With one side on an axis, base-times-height beats the determinant formula.

Example 13: A bisected segment

A line's segment between $5x - y + 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at $(1, 5)$. Find its equation.

Solution:

Step 1 — Parametrise the endpoints. On the lines: $(\alpha_1, 5\alpha_1 + 4)$ and $\left(\alpha_2, \frac{4 - 3\alpha_2}{4}\right)$.

Step 2 — Midpoint equations. $\alpha_1 + \alpha_2 = 2$ and $\frac{5\alpha_1 + 4}{1} + \frac{4 - 3\alpha_2}{4} = 10$, i.e. $20\alpha_1 - 3\alpha_2 = 20$.

Step 3 — Solve. $\alpha_1 = \frac{26}{23}$; the first endpoint is $\left(\frac{26}{23}, \frac{222}{23}\right)$.

Step 4 — Line through the midpoint and endpoint. Slope $= \frac{222/23 - 5}{26/23 - 1} = \frac{107}{3}$: $y - 5 = \frac{107}{3}(x - 1)$, i.e. $107x - 3y - 92 = 0$.

Takeaway: Put unknown endpoints ON their lines (one parameter each) and let the midpoint supply the equations.

Example 14: An equidistance locus

Show that the path of a point whose distances from $3x - 2y = 5$ and $3x + 2y = 5$ are equal is a straight line.

Solution:

Step 1 — Equate the distances. Both denominators are $\sqrt{13}$: $|3h - 2k - 5| = |3h + 2k - 5|$.

Step 2 — Open the moduli. Same-sign case: $3h - 2k - 5 = 3h + 2k - 5$ gives $k = 0$; opposite-sign case: $3h - 2k - 5 = -(3h + 2k - 5)$ gives $6h = 10$, $h = \frac{5}{3}$.

Step 3 — Conclude. The locus is $y = 0$ or $x = \frac{5}{3}$ — straight lines (the two angle bisectors). ∎

Takeaway: Equidistance from two lines always yields the pair of angle bisectors — expect two answers.

Example 15: Reading a parametrised line

For which $k$ is $(k - 3)x - (4 - k^2)y + k^2 - 7k + 6 = 0$ (a) parallel to the $x$-axis, (b) parallel to the $y$-axis, (c) through the origin?

Solution:

Step 1 — (a) No $x$-term. Parallel to the $x$-axis means the $x$-coefficient vanishes: $k - 3 = 0$, so $k = 3$.

Step 2 — (b) No $y$-term. $4 - k^2 = 0$: $k = \pm 2$.

Step 3 — (c) Zero constant. $k^2 - 7k + 6 = (k-1)(k-6) = 0$: $k = 1$ or $6$.

Takeaway: Each geometric condition switches off exactly one coefficient — read the equation structurally.

Example 16: Intercepts from sum and product

Find the lines whose intercepts on the axes have sum 1 and product $-6$.

Solution:

Step 1 — Build the quadratic. $a + b = 1$, $ab = -6$: $t^2 - t - 6 = 0$.

Step 2 — Solve. $(t - 3)(t + 2) = 0$: $\{a, b\} = \{3, -2\}$.

Step 3 — Write both lines. $\frac{x}{3} + \frac{y}{-2} = 1 \Rightarrow 2x - 3y - 6 = 0$; $\frac{x}{-2} + \frac{y}{3} = 1 \Rightarrow 3x - 2y + 6 = 0$.

Takeaway: Sum-and-product data is a quadratic in disguise — and the two orderings give two lines.

Example 17: Points on the $y$-axis at distance 4

What are the points on the $y$-axis whose distance from $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units?

Solution:

Step 1 — General form. $4x + 3y - 12 = 0$.

Step 2 — Distance equation for $(0, b)$. $\frac{|3b - 12|}{5} = 4$.

Step 3 — Solve both signs. $3b - 12 = \pm 20$: $b = \frac{32}{3}$ or $-\frac{8}{3}$ — points $\left(0, \frac{32}{3}\right)$ and $\left(0, -\frac{8}{3}\right)$.

Takeaway: Same template as the $x$-axis version — only the substituted coordinate changes.

Example 18: A trigonometric chord

Find the perpendicular distance from the origin to the line joining $(\cos\theta, \sin\theta)$ and $(\cos\varphi, \sin\varphi)$.

Solution:

Step 1 — Equation of the chord. Two-point form plus sum-to-product identities give

$$x\cos\frac{\theta + \varphi}{2} + y\sin\frac{\theta + \varphi}{2} = \cos\frac{\theta - \varphi}{2}$$

Step 2 — Distance from the origin. The coefficient vector has length $1$, so $d = \left|\cos\frac{\theta - \varphi}{2}\right|$.

Takeaway: Both points sit on the unit circle — the chord's normalised equation hands over the distance for free.

Example 19: A vertical line through an intersection

Find the line parallel to the $y$-axis through the intersection of $x - 7y + 5 = 0$ and $3x + y = 0$.

Solution:

Step 1 — Intersect. Substitute $y = -3x$: $x + 21x + 5 = 0$, so $x = -\frac{5}{22}$.

Step 2 — Write the vertical line. Parallel to the $y$-axis: $x = -\frac{5}{22}$.

Takeaway: A vertical line needs only the $x$-coordinate — don't bother computing $y$.

Example 20: Perpendicular at the $y$-intercept

Find the line perpendicular to $\frac{x}{4} + \frac{y}{6} = 1$ through the point where it meets the $y$-axis.

Solution:

Step 1 — General form and slope. $3x + 2y - 12 = 0$: slope $-\frac{3}{2}$; $y$-intercept point $(0, 6)$.

Step 2 — Perpendicular slope. $\frac{2}{3}$.

Step 3 — Write the line. $y = \frac{2}{3}x + 6$, i.e. $2x - 3y + 18 = 0$.

Takeaway: "Where it meets the $y$-axis" hands you the anchor point — set $x = 0$ and read it off.

Example 21: A triangle's area and a concurrency value

(i) Find the area of the triangle formed by $y - x = 0$, $x + y = 0$ and $x - k = 0$. (ii) Find $p$ so that $3x + y - 2 = 0$, $px + 2y - 3 = 0$ and $2x - y - 3 = 0$ are concurrent.

Solution:

Step 1 — (i) Vertices. Pairwise intersections: $(0, 0)$, $(k, k)$, $(k, -k)$.

Step 2 — (i) Area. Vertical base of length $2|k|$ at $x = k$; height $|k|$: area $= \frac{1}{2} \cdot 2|k| \cdot |k| = k^2$.

Step 3 — (ii) Concurrency. First and third lines: adding $3x + y = 2$ and $2x - y = 3$ gives $5x = 5$, point $(1, -1)$. Then $p(1) + 2(-1) - 3 = 0$: $p = 5$.

Takeaway: Two independent mini-problems, one method each — vertices then base-height; intersect then substitute.

Example 22: The concurrency identity

If $y = m_1x + c_1$, $y = m_2x + c_2$, $y = m_3x + c_3$ are concurrent, show that $m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0$.

Solution:

Step 1 — Intersect the first two. $x = \frac{c_2 - c_1}{m_1 - m_2}$, $y = \frac{m_1c_2 - m_2c_1}{m_1 - m_2}$.

Step 2 — Substitute into the third. $\frac{m_1c_2 - m_2c_1}{m_1 - m_2} = m_3 \cdot \frac{c_2 - c_1}{m_1 - m_2} + c_3$.

Step 3 — Clear and rearrange. $m_1c_2 - m_2c_1 = m_3(c_2 - c_1) + c_3(m_1 - m_2)$, which regroups to $m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0$. ∎

Takeaway: The identity is cyclic in the indices — a structure check on your final rearrangement.

Example 23: Lines at $45°$

Find the lines through $(3, 2)$ making $45°$ with the line $x - 2y = 3$.

Solution:

Step 1 — Angle equation. Given slope $\frac{1}{2}$: $\left|\frac{m - \frac{1}{2}}{1 + \frac{m}{2}}\right| = 1$.

Step 2 — Solve both signs. $m = 3$ or $m = -\frac{1}{3}$.

Step 3 — Write the lines through $(3, 2)$. $y - 2 = 3(x - 3) \Rightarrow 3x - y - 7 = 0$; $y - 2 = -\frac{1}{3}(x - 3) \Rightarrow x + 3y - 9 = 0$.

Takeaway: The two answers are perpendicular to each other ($3 \times -\frac{1}{3} = -1$) — a built-in sanity check for $45°$ problems.

Example 24: Equal intercepts through an intersection

Find the line through the intersection of $4x + 7y - 3 = 0$ and $2x - 3y + 1 = 0$ with equal intercepts on the axes.

Solution:

Step 1 — Find the intersection. From $4x + 7y = 3$ and $2x - 3y = -1$: eliminating, $13y = 5$, so $y = \frac{5}{13}$, $x = \frac{1}{13}$.

Step 2 — Equal-intercept form. $x + y = a$ through $\left(\frac{1}{13}, \frac{5}{13}\right)$: $a = \frac{6}{13}$.

Step 3 — Write the line. $x + y = \frac{6}{13}$, i.e. $13x + 13y = 6$.

Takeaway: The family trick $L_1 + \lambda L_2 = 0$ also works — impose equal intercepts to find $\lambda$; either route lands on $13x + 13y = 6$.

Example 25: A section ratio

In what ratio does the line $x + y = 4$ divide the segment joining $(-1, 1)$ and $(5, 7)$?

Solution:

Step 1 — Parametrise the section point. Ratio $k : 1$: $\left(\frac{5k - 1}{k + 1}, \frac{7k + 1}{k + 1}\right)$.

Step 2 — Put it on the line. $\frac{5k - 1 + 7k + 1}{k + 1} = 4$, i.e. $\frac{12k}{k+1} = 4$.

Step 3 — Solve. $12k = 4k + 4$: $k = \frac{1}{2}$ — the ratio is $1 : 2$.

Takeaway: Let the ratio be $k : 1$ and push the section point through the line — one linear equation in $k$.

Example 26: Distance along $2x - y = 0$

Find the distance of the line $4x + 7y + 5 = 0$ from the point $(1, 2)$, measured along the line $2x - y = 0$.

Solution:

Step 1 — Check the direction line. $(1, 2)$ satisfies $2x - y = 0$ — the measuring line is $2x - y = 0$ itself.

Step 2 — Intersect with the target. Substitute $y = 2x$: $4x + 14x + 5 = 0$, so $x = -\frac{5}{18}$, $y = -\frac{5}{9}$.

Step 3 — Measure. $d = \sqrt{\left(1 + \frac{5}{18}\right)^2 + \left(2 + \frac{5}{9}\right)^2} = \frac{23}{18}\sqrt{1 + 4} = \frac{23\sqrt{5}}{18}$.

Takeaway: Factor the common $\frac{23}{18}$ out of both coordinate differences before rooting — the surd stays clean.

Example 27: Direction and axis-parallel legs

(i) In what direction should a line through $(-1, 2)$ be drawn so that it meets $x + y = 4$ at distance 3 from $(-1, 2)$? (ii) The hypotenuse of a right triangle has ends $(1, 3)$ and $(-4, 1)$; find the legs, given they are parallel to the axes.

Solution:

Step 1 — (i) Parametrise by direction. A point at distance 3 in direction $\theta$: $(-1 + 3\cos\theta,\ 2 + 3\sin\theta)$.

Step 2 — (i) Put it on the line. $(-1 + 3\cos\theta) + (2 + 3\sin\theta) = 4$ gives $\cos\theta + \sin\theta = 1$.

Step 3 — (i) Solve. $\sqrt{2}\sin\left(\theta + 45°\right) = 1$: $\theta = 0°$ or $90°$ — the line must run parallel to an axis.

Step 4 — (ii) Two possible corner vertices. Right angle at $(1, 1)$: legs $x = 1$ and $y = 1$; or at $(-4, 3)$: legs $x = -4$ and $y = 3$.

Takeaway: The polar parametrisation $(x_0 + r\cos\theta, y_0 + r\sin\theta)$ turns "at distance $r$" conditions into trigonometry.

Example 28: Image in $x + 3y = 7$

Find the image of $(3, 8)$ with respect to the line $x + 3y = 7$ assumed to be a plane mirror.

Solution:

Step 1 — Perpendicularity. Line slope $-\frac{1}{3}$, so $PQ$ has slope 3: $\frac{k - 8}{h - 3} = 3$, i.e. $3h - k = 1$.

Step 2 — Midpoint on the mirror. $\frac{h + 3}{2} + 3 \cdot \frac{k + 8}{2} = 7$, i.e. $h + 3k = -13$.

Step 3 — Solve. From $3h - k = 1$ and $h + 3k = -13$: $h = -1$, $k = -4$ — image $(-1, -4)$.

Step 4 — Check. Midpoint $(1, 2)$: $1 + 6 = 7$ ✓.

Takeaway: Always check the midpoint lands on the mirror — it catches sign slips instantly.

Example 29: Equally inclined lines

If $y = 3x + 1$ and $2y = x + 3$ are equally inclined to $y = mx + 4$, find $m$.

Solution:

Step 1 — Set the two angle-tangents equal and opposite. $\frac{3 - m}{1 + 3m} = -\frac{\frac{1}{2} - m}{1 + \frac{m}{2}}$ (the same-sign case would force $3 = \frac{1}{2}$ — impossible).

Step 2 — Cross-multiply. $(3 - m)(2 + m) = -(1 - 2m)(1 + 3m)$.

Step 3 — Expand and solve. $6 + m - m^2 = -(1 + m - 6m^2)$ gives $7m^2 - 2m - 7 = 0$: $m = \frac{1 \pm 5\sqrt{2}}{7}$.

Takeaway: "Equally inclined" means equal angle sizes on opposite sides — hence the minus sign linking the two tangents.

Example 30: A sum-of-distances locus

The sum of the distances of $P(x, y)$ from $x + y - 5 = 0$ and $3x - 2y + 7 = 0$ is always 10. Show that $P$ moves on a line.

Solution:

Step 1 — Write the condition. $\frac{|x + y - 5|}{\sqrt{2}} + \frac{|3x - 2y + 7|}{\sqrt{13}} = 10$.

Step 2 — Fix a region. Within any one of the four regions cut out by the two lines, each modulus opens with a definite sign.

Step 3 — Conclude. The condition becomes a single linear equation in $x, y$ — a straight line (segment) in each region. ∎

Takeaway: Moduli are piecewise-linear — argue region by region and the locus is linear in each.

Example 31: The midway parallel

Find the line equidistant from the parallel lines $9x + 6y - 7 = 0$ and $3x + 2y + 6 = 0$.

Solution:

Step 1 — Match the coefficients. Divide the first by 3: $3x + 2y - \frac{7}{3} = 0$.

Step 2 — Average the constants. The midway line keeps $A, B$ and takes $C = \frac{-\frac{7}{3} + 6}{2} = \frac{11}{6}$.

Step 3 — Write it. $3x + 2y + \frac{11}{6} = 0$, i.e. $18x + 12y + 11 = 0$.

Takeaway: Midway between parallels = average the constants — but only AFTER matching the coefficient pairs.

Example 32: Reflection off the $x$-axis

A ray through $(1, 2)$ reflects off the $x$-axis at $A$ and the reflected ray passes through $(5, 3)$. Find $A$.

Solution:

Step 1 — Use the mirror image. The reflected ray extended backwards passes through the image of $(1, 2)$ in the $x$-axis: $(1, -2)$.

Step 2 — Line through image and target. Through $(1, -2)$ and $(5, 3)$: slope $\frac{5}{4}$, $y + 2 = \frac{5}{4}(x - 1)$.

Step 3 — Intersect the mirror. At $y = 0$: $2 = \frac{5}{4}(x - 1)$, so $x = \frac{13}{5}$: $A = \left(\frac{13}{5}, 0\right)$.

Takeaway: Reflection problems straighten out through the image point — one line replaces two rays.

Example 33: Two classics

(i) Prove that the product of the perpendiculars from $\left(\pm\sqrt{a^2 - b^2}, 0\right)$ to $\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1$ is $b^2$. (ii) A person at the junction of $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$ wants to reach the path $6x - 7y + 8 = 0$ in least time. Find the path to follow.

Solution:

Step 1 — (i) Clear the line. $bx\cos\theta + ay\sin\theta - ab = 0$; write $c = \sqrt{a^2 - b^2}$.

Step 2 — (i) Multiply the two distances. $\frac{|bc\cos\theta - ab| \cdot |bc\cos\theta + ab|}{b^2\cos^2\theta + a^2\sin^2\theta} = \frac{|b^2c^2\cos^2\theta - a^2b^2|}{b^2\cos^2\theta + a^2\sin^2\theta}$.

Step 3 — (i) Collapse. $c^2\cos^2\theta - a^2 = -(a^2\sin^2\theta + b^2\cos^2\theta)$, so the fraction equals $b^2$. ∎

Step 4 — (ii) Junction and direction. The lines meet at $\left(-\frac{1}{17}, \frac{22}{17}\right)$; least time = walk perpendicular to the target path (slope $\frac{6}{7}$), so use slope $-\frac{7}{6}$: $y - \frac{22}{17} = -\frac{7}{6}\left(x + \frac{1}{17}\right)$, i.e. $119x + 102y = 125$.

Takeaway: "Least time to reach a line" always means the perpendicular from where you stand.