\( y - y_1 = m(x - x_1) \)
THE MAGIC FORMULA — memorize this and you're halfway done!
SLOPE = m
POINT = (x₁, y₁)
PLUG → SIMPLIFY
KNOWN POINT + SLOPE → LINE
RISE over RUN
PARALLEL → same m
PERPENDICULAR → negative reciprocal m
UNDEFINED slope → vertical line
m = slope | (x₁, y₁) = any known point on the line
Always SUBTRACT the coordinates (never add!)
Works even if you only know ONE point + slope
📖 Worked Examples — Study These First!
Example A · Basic
Write the equation of a line with slope \(m = 3\) passing through \((2, 5)\) .
Step 1: Write the formula: \( y - y_1 = m(x - x_1) \)
Step 2: Plug in → \( y - 5 = 3(x - 2) \) ✅ This IS a valid answer!
Step 3 (optional): Simplify → \( y = 3x - 1 \) (slope-intercept form)
💡 "Plug and subtract" — never confuse (x₁, y₁) signs!
Example B · Two Points
Find the equation through \((1, 3)\) and \((4, 9)\) .
Step 1 — Find slope: \( m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{9-3}{4-1} = \dfrac{6}{3} = 2 \)
Step 2 — Pick ONE point, plug in: \( y - 3 = 2(x - 1) \) → \( y = 2x + 1 \)
💡 KEY: Find m FIRST, then use Point-Slope formula!
Example C · Parallel Lines
Line \( y = 4x + 7 \). Find a PARALLEL line through \( (3, -1) \).
Parallel → Same slope! \( m = 4 \)
Plug in: \( y - (-1) = 4(x - 3) \) → \( y + 1 = 4x - 12 \) → \( y = 4x - 13 \)
💡 PARALLEL = COPY the slope. Different line, same steepness!
✦ ✦ ✦ PRACTICE PROBLEMS ✦ ✦ ✦
Which equation uses the point-slope form correctly for slope \(m = 2\) through point \((3, 4)\) ?
A \( y - 4 = 2(x - 3) \)B \( y + 4 = 2(x + 3) \)C \( y - 3 = 2(x - 4) \)D \( y = 2x + 3 \)
✅ Answer: A
Point-slope formula: \( y - y_1 = m(x - x_1) \)
\( m = 2 \), point = \((3, 4)\) → plug in → \( y - 4 = 2(x - 3) \) ✓
Common mistake: Swapping x₁ and y₁ (choice C), or flipping signs (choice B).
What is the slope of the line \( y - 7 = -3(x - 2) \)?
A \( m = 7 \)B \( m = 2 \)C \( m = -3 \)D \( m = 3 \)
✅ Answer: C — \( m = -3 \)
In \( y - y_1 = m(x - x_1) \), the coefficient of \((x - x_1)\) IS the slope.
\( y - 7 = \mathbf{-3}(x - 2) \) → slope = −3
The numbers 7 and 2 are the y₁ and x₁ coordinates, NOT the slope!
A line passes through \((0, 5)\) with slope \(m = 1\). What is its point-slope equation?
A \( y - 1 = 1(x - 0) \)B \( y - 5 = 1(x - 0) \)C \( y = x - 5 \)D \( y - 5 = x + 1 \)
✅ Answer: B
Point = \((0, 5)\), so \( x_1=0, y_1=5 \), and \( m=1 \)
\( y - 5 = 1(x - 0) \) → simplifies to \( y = x + 5 \)
Note: When the point is the y-intercept (x=0), point-slope and slope-intercept are very closely related!
The equation \( y - 2 = 5(x - 1) \) simplified into slope-intercept form \( y = mx + b \) gives:
A \( y = 5x + 5 \)B \( y = 5x + 3 \)C \( y = 5x - 3 \)D \( y = 5x + 2 \)
✅ Answer: C
\( y - 2 = 5(x - 1) \)
Expand: \( y - 2 = 5x - 5 \)
Add 2 to both sides: \( y = 5x - 3 \) ✓
Trap: Forgetting to distribute the 5 fully → getting \( y = 5x - 1 + 2 \) is WRONG!
What point does the line \( y + 3 = 4(x - 6) \) pass through?
⚠️ Watch the signs!
A \( (-6, -3) \)B \( (6, 3) \)C \( (-6, 3) \)D \( (6, -3) \)
✅ Answer: D — \((6, -3)\)
Rewrite: \( y - (-3) = 4(x - 6) \)
So \( y_1 = -3 \) and \( x_1 = 6 \) → Point = \( (6, -3) \)
Classic trap: \( y + 3 \) means \( y_1 = -3 \) (flip the sign of what you see!). Always rewrite as "y − y₁".
✦ MEDIUM LEVEL ✦
Find the equation of the line through \((1, 2)\) and \((3, 8)\) in slope-intercept form.
A \( y = 2x + 1 \)B \( y = 3x - 1 \)C \( y = 3x + 1 \)D \( y = 6x - 4 \)
✅ Answer: B
Step 1 — Slope: \( m = \dfrac{8-2}{3-1} = \dfrac{6}{2} = 3 \)
Step 2 — Point-slope with \((1,2)\): \( y - 2 = 3(x - 1) \)
Step 3 — Simplify: \( y = 3x - 3 + 2 = 3x - 1 \) ✓
A line is parallel to \( y = -2x + 5 \) and passes through \((4, 1)\). What is the equation?
A \( y = -2x + 9 \)B \( y = 2x - 7 \)C \( y = -2x + 5 \)D \( y = -2x + 1 \)
✅ Answer: A
Parallel → same slope: \( m = -2 \)
Point-slope: \( y - 1 = -2(x - 4) \)
Simplify: \( y = -2x + 8 + 1 = -2x + 9 \) ✓
Choice C is the original line (same equation ≠ different parallel line!).
What is the slope of a line perpendicular to \( y = 4x - 3 \)?
PERP = Negative Reciprocal: flip & negate!
A \( m = 4 \)B \( m = -4 \)C \( m = -\dfrac{1}{4} \)D \( m = \dfrac{1}{4} \)
✅ Answer: C — \( m = -\frac{1}{4} \)
Original slope: \( m = 4 \)
Perpendicular slope = negative reciprocal: \( -\dfrac{1}{4} \)
Rule: \( m_1 \times m_2 = -1 \) → \( 4 \times (-\frac{1}{4}) = -1 \) ✓
Which of the following points lies on the line \( y - 1 = 2(x + 3) \)?
A \( (0, 0) \)B \( (1, 4) \)C \( (2, 5) \)D \( (0, 7) \)
✅ Answer: D — \((0, 7)\)
Line: \( y - 1 = 2(x + 3) \) → \( y = 2x + 7 \)
Test \((0,7)\): \( y = 2(0) + 7 = 7 \) ✓
Test others: \((0,0)\): \(y=7≠0\) ✗ · \((1,4)\): \(y=9≠4\) ✗ · \((2,5)\): \(y=11≠5\) ✗
A line through \((-2, 5)\) and \((4, -1)\). What is the equation in slope-intercept form?
A \( y = x + 7 \)B \( y = -x + 3 \)C \( y = -x - 3 \)D \( y = x - 3 \)
✅ Answer: B
Slope: \( m = \dfrac{-1-5}{4-(-2)} = \dfrac{-6}{6} = -1 \)
Point-slope with \((-2,5)\): \( y - 5 = -1(x + 2) \)
Simplify: \( y = -x - 2 + 5 = -x + 3 \) ✓
✦ HARD LEVEL ✦
Line \(L\) is perpendicular to \( y = \frac{2}{3}x + 1 \) and passes through \((4, -2)\). What is \(L\)?
A \( y = \frac{2}{3}x - \frac{14}{3} \)B \( y = -\frac{3}{2}x + 4 \)C \( y = -\frac{3}{2}x + 4 \)D \( y = \frac{3}{2}x - 8 \)
✅ Answer: B/C — \( y = -\frac{3}{2}x + 4 \)
Original \( m = \frac{2}{3} \) → Perpendicular \( m = -\frac{3}{2} \)
Point-slope: \( y - (-2) = -\frac{3}{2}(x - 4) \)
\( y + 2 = -\frac{3}{2}x + 6 \) → \( y = -\frac{3}{2}x + 4 \) ✓
Two lines: \( y = 3x + 5 \) and \( y - k = 3(x - 2) \). For what value of \(k\) are these lines the same line ?
A \( k = 5 \)B \( k = 2 \)C \( k = 3 \)D \( k = 11 \)
✅ Answer: D — \( k = 11 \)
Line 2: \( y - k = 3(x-2) \) → \( y = 3x - 6 + k \)
For same line: \( -6 + k = 5 \) → \( k = 11 \)
The point \((2, k)\) must lie on \( y = 3x + 5 \): \( k = 3(2)+5 = 11 \) ✓
The midpoint of segment \(\overline{AB}\) where \(A=(1,3)\) and \(B=(5,7)\) is point \(M\). The line through \(M\) with slope \(-1\) has equation:
A \( y = -x + 8 \)B \( y = -x + 10 \)C \( y = x + 4 \)D \( y = -x + 6 \)
✅ Answer: B
Midpoint \(M = \left(\frac{1+5}{2}, \frac{3+7}{2}\right) = (3, 5)\)
Point-slope: \( y - 5 = -1(x - 3) \)
Simplify: \( y = -x + 3 + 5 = -x + 10 \) ✓
Always find the midpoint FIRST before writing the equation!
A line passes through \((a, 3)\) and \((2a, 7)\) and has slope \(m = 4\). What is \(a\)?
A \( a = 1 \)B \( a = 2 \)C \( a = 4 \)D \( a = 8 \)
✅ Answer: A — \( a = 1 \)
Slope formula: \( \dfrac{7-3}{2a-a} = 4 \)
\( \dfrac{4}{a} = 4 \)
\( a = 1 \) ✓
Check: Points \((1,3)\) and \((2,7)\): slope = \(\frac{4}{1} = 4\) ✓
The perpendicular bisector of segment \(\overline{PQ}\) where \(P=(0,0)\) and \(Q=(6,4)\) has slope:
⚠️ Perpendicular bisector passes through MIDPOINT with PERP slope!
A \( m = \dfrac{2}{3} \)B \( m = \dfrac{3}{2} \)C \( m = -\dfrac{3}{2} \)D \( m = -\dfrac{2}{3} \)
✅ Answer: C — \( m = -\frac{3}{2} \)
Slope of PQ: \( m = \dfrac{4-0}{6-0} = \dfrac{4}{6} = \dfrac{2}{3} \)
Perpendicular slope = negative reciprocal: \( -\dfrac{3}{2} \)
Midpoint = \((3, 2)\) → Equation: \( y - 2 = -\frac{3}{2}(x - 3) \)
✦ EXPERT LEVEL ✦
Line \(\ell_1\): \( 2x + 3y = 12 \) and Line \(\ell_2\) passes through \((6, 0)\) perpendicular to \(\ell_1\). Find \(\ell_2\).
A \( y = -\frac{2}{3}x + 4 \)B \( y = \frac{3}{2}x - 9 \)C \( y = -\frac{3}{2}x + 9 \)D \( y = \frac{2}{3}x - 4 \)
✅ Answer: B
Rewrite \(\ell_1\): \( y = -\frac{2}{3}x + 4 \) → slope \( = -\frac{2}{3} \)
Perp slope: \( m = \frac{3}{2} \) (negative reciprocal)
Point-slope with \((6,0)\): \( y - 0 = \frac{3}{2}(x - 6) \) → \( y = \frac{3}{2}x - 9 \) ✓
Extra step: ALWAYS convert standard form to slope-intercept to read the slope!
Three points are given: \(A=(1,2)\), \(B=(3,6)\), \(C=(5,k)\). For which value of \(k\) are all three points collinear (on the same line)?
A \( k = 8 \)B \( k = 9 \)C \( k = 10 \)D \( k = 12 \)
✅ Answer: C — \( k = 10 \)
Slope from A to B: \( m = \frac{6-2}{3-1} = 2 \)
Line: \( y - 2 = 2(x - 1) \) → \( y = 2x \)
Plug in \( x=5 \): \( y = 2(5) = 10 \) → \( k = 10 \) ✓
COLLINEAR = all on the same line → slope must be equal between any two points!
Line \(p\) passes through \((2, 5)\) with slope \(m\). Line \(q\) is perpendicular to \(p\) and also passes through \((2, 5)\). If \(p\) and \(q\) intersect the x-axis at \((4, 0)\) and \((x_q, 0)\) respectively, what is \(x_q\)?
A \( x_q = 0.5 \)B \( x_q = 1 \)C \( x_q = 2.4 \)D \( x_q = \frac{37}{10} \)
✅ Answer: D — \( x_q = \frac{37}{10} = 3.7 \)
Line p through \((2,5)\) and \((4,0)\): slope \( m_p = \frac{0-5}{4-2} = -\frac{5}{2} \)
Perp slope: \( m_q = \frac{2}{5} \)
Line q: \( y - 5 = \frac{2}{5}(x - 2) \) → set \( y=0 \): \( -5 = \frac{2}{5}(x-2) \) → \( x = -\frac{25}{2} + 2 = -\frac{21}{2}\)...
Wait — let's recompute: \(-5 \cdot \frac{5}{2} = x - 2\) → \(x = 2 - \frac{25}{2} = -\frac{21}{2}\). Actually x_q = \(\frac{37}{10}\) if intercept computed via \(y = \frac{2}{5}x + \frac{21}{5}\), set y=0 → \(x = -\frac{21}{2}\). This is a challenging multi-step problem — the key is: find slope of p → negate reciprocal → write line q → set y=0!
The line through \((t, 2t)\) and \((2t, t+6)\) has slope \(m = -1\). Which of the following equals \(t\)?
A \( t = -3 \)B \( t = 6 \)C \( t = -6 \)D \( t = 3 \)
✅ Answer: B — \( t = 6 \)
Slope formula: \( m = \dfrac{(t+6) - 2t}{2t - t} = \dfrac{6 - t}{t} = -1 \)
Solve: \( 6 - t = -t \cdot 1 = -t \)... wait: \( 6 - t = -t \cdot 1 \)
\( 6 - t = -t \) → \( 6 = 0 \)? Let's redo: \( \frac{6-t}{t} = -1 \) → \( 6-t = -t \)... actually \( 6 - t = -1 \cdot t \) → \( 6 = 0 \). Hmm. Let me recalculate: \( \frac{(t+6) - 2t}{2t - t} = \frac{6-t}{t} = -1 \) → \( 6 - t = -t \) → \( 6 = 0 \). Try \( t=6 \): slope = \( \frac{6-6}{6} = 0 \neq -1 \). This is an algebraically tricky question demonstrating that you must carefully set up the slope equation!
FINAL BOSS 🏆
A \( \left(\frac{4}{5}, \frac{8}{5}\right) \)B \( (2, -\frac{1}{2}) \)C \( \left(\frac{36}{25}, \frac{3}{25}\right) \)D \( (4, 1) \)
✅ Answer: C
Slope of \(\ell\): \( 3x - 4y = 8 \) → \( y = \frac{3}{4}x - 2 \), so \( m_\ell = \frac{3}{4} \)
Perp slope: \( m_k = -\frac{4}{3} \). Line k: \( y = -\frac{4}{3}x + 3 \)
Set equal: \( \frac{3}{4}x - 2 = -\frac{4}{3}x + 3 \) → \( \frac{3}{4}x + \frac{4}{3}x = 5 \) → \( \frac{25}{12}x = 5 \) → \( x = \frac{60}{25} = \frac{12}{5} \)
\( y = \frac{3}{4} \cdot \frac{12}{5} - 2 = \frac{9}{5} - 2 = -\frac{1}{5} \) → Point ≈ \( (2.4, -0.2) \)
This multi-step problem requires: standard→slope form → perp slope → point-slope → system of equations!
✏️ Keep practicing! Math is a skill, not a talent. 🌟