Locus and shift of origin
If we mark on a map every place that is at the same distance from two poles, all those places fall on one fixed line. The path of all points that obey such a condition is a locus. Building on classes 9–10 : In classes 9–10 we learnt the distance formula and the midpoint formula; we use the same formulae to find the equation of a locus.
The set of all points that obey a given condition is the locus of that condition. The equation joining x and y that every such point satisfies is the equation of the locus. Keeping the directions of the axes the same and moving the origin to another point is called a shift of origin.
Rule 1: Three steps to find the equation of a locus — Step 1: take any point P(x, y) on the locus. Step 2: write the condition in x and y using the distance formula. Step 3: square both sides to remove square roots and simplify.
Rule 2: When the origin moves to (h, k) — x = X + h, y = Y + k that is X = x − h, Y = y − k. Here (x, y) are the old and (X, Y) the new coordinates.
In the figure : Points equidistant from A(1, 1) and B(5, 3) lie on the orange line; it is perpendicular to AB and passes through M.
In the figure : The new origin is O′(3, 2); the point P has old coordinates (5, 4) and new coordinates (2, 2).
How is the formula obtained? : P(x, y) is equidistant from A(1, 1) and B(5, 3): PA² = PB². So (x − 1)² + (y − 1)² = (x − 5)² + (y − 3)². After expanding, x² and y² are the same on both sides and cancel: −2x − 2y + 2 = −10x − 6y + 34, i.e. 8x + 4y = 32, so 2x + y − 8 = 0. For the shift: the new axis is h units from the old one, so a point is x from the old axis and x − h from the new axis; hence X = x − h.
- Easy : Locus of a point equidistant from A(1, 1) and B(5, 3): as above, 2x + y − 8 = 0.
- Medium : Locus of a point that is always 5 units from (2, −1): (x − 2)² + (y + 1)² = 25.
- Exam-type : New equation of 2x − 3y + 5 = 0 when the origin moves to (3, −2): put x = X + 3, y = Y − 2. 2(X + 3) − 3(Y − 2) + 5 = 2X + 6 − 3Y + 6 + 5 = 2X − 3Y + 17. So the new equation is 2X − 3Y + 17 = 0.
A locus is not one point but the whole set of points obeying the condition. A shift of origin does not change the shape, only the equation.
Writing X = x + h instead of X = x − h. If the new origin is (h, k), the new coordinate is the old one minus h.
When the origin goes to (h, k), an old point (x, y) becomes (x − h, y − k). In the other direction, to write an equation in the new coordinates put X + h in place of x. Memory clue : "new = old − home" (home means the new origin).
CET asks to remove the linear terms of x² + y² + 2gx + 2fy + c; complete the squares and take (−g, −f) as the new origin.
CET-style example : If the origin is shifted to (−1, 3), the new form of x² + y² + 2x − 6y + 1 = 0 is? (A) X² + Y² = 9 (B) X² + Y² = 1 (C) X² + Y² = 10 (D) X² + Y² = 11
Answer : (A). Put x = X − 1, y = Y + 3: (X − 1)² + (Y + 3)² + 2(X − 1) − 6(Y + 3) + 1 = X² + Y² − 9 = 0, so X² + Y² = 9. Time-saving method : Complete the squares: (x + 1)² + (y − 3)² = 9; with X = x + 1, Y = y − 3 you get X² + Y² = 9 at once.
Spot the difference
| Point | Locus | Shift of origin |
|---|---|---|
| What you get | an equation in x and y | new coordinates or a new equation |
| Start | condition and P(x, y) | new origin (h, k) |
| Main formula | distance formula | X = x − h, Y = y − k |
| Does the shape change | a new shape is formed | the shape stays the same |
Practice questions
- 1Find the locus of a point equidistant from A(0, 0) and B(4, 2).
- 2Write the locus of a point that is always 7 units from the origin.
- 3If the origin is shifted to (2, 3), find the new coordinates of (5, 7).
- 4Find the locus of a point whose distance from the X-axis is twice its distance from the Y-axis.
- 5If the origin is shifted to (−2, 1), find the new coordinates of (3, −4).
- 6If the origin is shifted to (1, −2), find the new equation of 3x + 2y − 7 = 0.
- 7Where should the origin be shifted to remove the linear terms of x² + y² − 6x + 4y − 3 = 0, and what is the new equation?
- 8Find the locus of a point equidistant from A(3, −1) and B(−1, 5).
Show answers (try them yourself first)
- 1From x² + y² = (x − 4)² + (y − 2)²: 8x + 4y = 20, so 2x + y − 5 = 0.
- 2x² + y² = 49.
- 3(5 − 2, 7 − 3) = (3, 4).
- 4|y| = 2|x|, so y² = 4x² (that is the two lines y = 2x and y = −2x).
- 5(3 + 2, −4 − 1) = (5, −5).
- 6x = X + 1, y = Y − 2: 3(X + 1) + 2(Y − 2) − 7 = 3X + 2Y − 8, so 3X + 2Y − 8 = 0.
- 7(x − 3)² + (y + 2)² = 16; the new origin is (3, −2) and the new equation is X² + Y² = 16.
- 8PA² = PB²: −8x + 12y − 16 = 0, i.e. 2x − 3y + 4 = 0.
Slope, inclination and angle between lines
To tell how steep a hill road is we say "10 metres up for every 100 metres forward". The slope of a line measures exactly this: rise ÷ run. Building on classes 9–10 : In classes 9–10 we learnt tan θ = opposite ÷ adjacent; the slope is that same tan θ.
The angle a line makes with the positive direction of the X-axis, measured anticlockwise, is its inclination θ (0° ≤ θ < 180°). The tan of the inclination is the slope m of the line. A vertical line has inclination 90°, so its slope is not defined.
Rule 1: Two ways to find a slope — m = tan θ and, from two points, m = (y₂ − y₁)/(x₂ − x₁).
In the figure : The line passes through (0, 0) and (3, 4): rise 4, run 3, so m = 4/3 and θ ≈ 53°.
Rule 2: Parallel and perpendicular lines — Parallel: m₁ = m₂. Perpendicular: m₁ × m₂ = −1.
Rule 3: Angle between two lines — tan θ = |(m₁ − m₂)/(1 + m₁m₂)| (for the acute angle). If 1 + m₁m₂ = 0 the lines are perpendicular.
In the figure : The angle θ between lines of slopes 1/2 and 3: tan θ = |(3 − 1/2)/(1 + 3/2)| = 1, so θ = 45°.
Rule 4: Condition for three points to be collinear — A, B, C are collinear if the slope of AB = slope of BC.
How is the formula obtained? : Take two points (x₁, y₁) and (x₂, y₂) on a line and draw a right triangle through them: the horizontal side is x₂ − x₁ and the vertical side y₂ − y₁. tan θ = vertical ÷ horizontal, so m = (y₂ − y₁)/(x₂ − x₁). If two lines have inclinations θ₁ and θ₂, the angle between them is θ = θ₂ − θ₁. tan(θ₂ − θ₁) = (tan θ₂ − tan θ₁)/(1 + tan θ₂ tan θ₁) = (m₂ − m₁)/(1 + m₁m₂). We want the acute angle, so we take the absolute value.
- Easy : Slope of the line through (2, 1) and (6, 9) = (9 − 1)/(6 − 2) = 8/4 = 2. If the slope is √3 then tan θ = √3, so θ = 60°.
- Medium : If (2, 3), (4, 7) and (6, k) are collinear find k: slope of AB = (7 − 3)/(4 − 2) = 2. Slope of BC = (k − 7)/(6 − 4) = 2, so k − 7 = 4 and k = 11.
- Exam-type : Acute angle between lines of slopes 2 and 1/3: tan θ = |(2 − 1/3)/(1 + 2 × 1/3)| = |(5/3)/(5/3)| = 1, so θ = 45°.
Inclination is measured between 0° and 180°. A horizontal line has slope 0 and a vertical line has no defined slope.
When the slope is negative the calculator gives a negative angle; take 180° − (acute angle) as the inclination. For example m = −1 gives θ = 135°, not −45°.
The slope of a perpendicular line is the given slope "flipped and with its sign changed". For example 2/3 → −3/2. A parallel line has the same slope. Memory clue : "perpendicular = flip and flip the sign".
CET rarely gives the slope directly; first bring the equation to y = mx + c or use −(coefficient of x)/(coefficient of y).
CET-style example : The line through A(1, 2) and B(3, k) is perpendicular to x + 2y − 5 = 0. Then k = ? (A) 4 (B) 6 (C) 3 (D) −2
Answer : (B). The slope of x + 2y − 5 = 0 is −1/2, so the slope of AB is 2. (k − 2)/(3 − 1) = 2, k − 2 = 4, k = 6. Time-saving method : Given slope −1/2 → perpendicular slope 2 → "rise = 2 × run 2 = 4", so k = 2 + 4.
Spot the difference
| Point | Parallel lines | Perpendicular lines | Collinear points |
|---|---|---|---|
| Condition | m₁ = m₂ | m₁m₂ = −1 | slope of AB = slope of BC |
| Angle between lines | 0° | 90° | it is one single line |
| Example | 2 and 2 | 2 and −1/2 | (2, 3), (4, 7), (6, 11) |
Practice questions
- 1Find the slope of the line through (1, −2) and (4, 7).
- 2What is the slope of a line with inclination 135°?
- 3Find the slope of a line with inclination 30°.
- 4The line through (1, 2) and (3, k) is parallel to the line through (0, 1) and (2, 7). Find k.
- 5The line through A(1, 1) and B(4, k) is perpendicular to the line through C(0, 2) and D(3, 0). Find k.
- 6If (−1, −5), (1, −1) and (4, k) are collinear, find k.
- 7Find the acute angle between lines of slopes 3 and −2.
- 8Show that the triangle A(1, 1), B(5, 3), C(3, 7) has a right angle at B.
Show answers (try them yourself first)
- 1(7 − (−2))/(4 − 1) = 9/3 = 3.
- 2m = tan 135° = −1.
- 3tan 30° = 1/√3.
- 4The given line has slope (7 − 1)/(2 − 0) = 3. (k − 2)/(3 − 1) = 3, so k = 8.
- 5Slope of CD = (0 − 2)/(3 − 0) = −2/3, so the slope of AB is 3/2. (k − 1)/(4 − 1) = 3/2, k − 1 = 9/2, k = 11/2.
- 6Slope of (−1, −5) and (1, −1) = 4/2 = 2. (k + 1)/(4 − 1) = 2, k + 1 = 6, k = 5.
- 7tan θ = |(3 − (−2))/(1 + 3 × (−2))| = |5/(−5)| = 1, so θ = 45°.
- 8Slope of AB = 2/4 = 1/2, slope of BC = (7 − 3)/(3 − 5) = −2. The product is −1, so AB ⊥ BC.
Standard forms of a line
Two pegs joined by a tight rope give a straight line. To fix the rope we must tell one peg and the direction, or two pegs, or the places where it meets the two axes; these give the different forms of a line. Building on classes 9–10 : In classes 9–10 we learnt y = mx + c; now we choose it or another form depending on the information given.
The equation of a line is the equation in x and y that is satisfied by every point (x, y) on the line. Depending on the information given we use the point-slope form, two-point form, slope-intercept form, double-intercept form or normal form.
Rule 1: Point-slope and two-point forms — y − y₁ = m(x − x₁) and (y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁).
Rule 2: Slope-intercept and double-intercept forms — y = mx + c (c = y-intercept) and x/a + y/b = 1 (a = x-intercept, b = y-intercept).
In the figure : The line with x-intercept 4 and y-intercept 3: x/4 + y/3 = 1, that is 3x + 4y − 12 = 0.
Rule 3: Normal form and special lines — x cos α + y sin α = p (p = perpendicular distance of the line from the origin, α = angle of that perpendicular). Horizontal line y = k, vertical line x = h.
How is the formula obtained? : Take P(x, y) on the line and the given point A(x₁, y₁). The slope of AP, (y − y₁)/(x − x₁), must equal the given slope m: (y − y₁)/(x − x₁) = m, so y − y₁ = m(x − x₁). This is the point-slope form. Putting x₁ = 0, y₁ = c in it gives y = mx + c. For the double-intercept form, the points (a, 0) and (0, b) give slope −b/a; using the point (a, 0) in the point-slope form gives y = −(b/a)(x − a), that is x/a + y/b = 1.
- Easy : Line through (3, −1) with slope 2: y + 1 = 2(x − 3), that is y + 1 = 2x − 6, so 2x − y − 7 = 0.
- Medium : Line through (2, 5) and (6, 1): slope = (1 − 5)/(6 − 2) = −1. y − 5 = −1(x − 2), so x + y − 7 = 0.
- Exam-type : Normal form of 3x + 4y = 10: divide by √(3² + 4²) = 5. (3/5)x + (4/5)y = 2. So cos α = 3/5, sin α = 4/5 and p = 2.
When the coordinates of a point on the line are put in its equation, both sides become equal; always check your answer by putting a given point in it.
Forgetting the sign when a or b is negative in x/a + y/b = 1; and forgetting that p is always positive in the normal form.
Choosing the form: one point + slope → point-slope; two points → find the slope first; places on the axes → x/a + y/b = 1; "perpendicular from the origin" → normal form. Memory clue : "point-slope = P-S", "intercepts = I-I": match the information given with the name.
CET often says "sum of intercepts" or "x-intercept twice the y-intercept": take a = 2b in x/a + y/b = 1 and put the given point to find b.
CET-style example : The x-intercept of a line through (4, 1) is twice its y-intercept. The line is? (A) x + 2y − 6 = 0 (B) 2x + y − 9 = 0 (C) x + y − 5 = 0 (D) x − 2y − 2 = 0
Answer : (A). Let the y-intercept be b and the x-intercept 2b: x/(2b) + y/b = 1. Put (4, 1): 2/b + 1/b = 1, b = 3. The line is x/6 + y/3 = 1, i.e. x + 2y − 6 = 0. Time-saving method : Test the options: putting (4, 1) in A gives 4 + 2 − 6 = 0 ✓. Then see the intercepts 6 and 3 (6 = 2 × 3).
Spot the difference
| Form | When to use | Formula |
|---|---|---|
| Point-slope | a point and the slope are given | y − y₁ = m(x − x₁) |
| Two-point | two points are given | find the slope, then point-slope |
| Slope-intercept | slope and y-intercept are given | y = mx + c |
| Double-intercept | both axis intercepts are given | x/a + y/b = 1 |
| Normal form | the perpendicular from the origin is given | x cos α + y sin α = p |
Practice questions
- 1Write the line with slope −2 passing through (1, 3).
- 2Write the equation of the line with x-intercept 5 and y-intercept 2.
- 3Write the equation of the line with y-intercept 4 and slope 3.
- 4Find the line through (−1, 2) and (3, −2).
- 5Find the line through (2, 1) parallel to the line joining (0, 0) and (3, 6).
- 6Find the line through (1, 1) perpendicular to x − 3y + 6 = 0.
- 7Find the line through (4, 1) that cuts equal intercepts on both axes.
- 8Write 4x + 3y − 15 = 0 in normal form and find p.
Show answers (try them yourself first)
- 1y − 3 = −2(x − 1), so 2x + y − 5 = 0.
- 2x/5 + y/2 = 1; multiply by 10: 2x + 5y = 10, so 2x + 5y − 10 = 0.
- 3y = 3x + 4, that is 3x − y + 4 = 0.
- 4Slope = (−2 − 2)/(3 + 1) = −1. y − 2 = −1(x + 1), so x + y − 1 = 0.
- 5Slope = 6/3 = 2. y − 1 = 2(x − 2), so 2x − y − 3 = 0.
- 6The given line has slope 1/3, so the perpendicular slope is −3. y − 1 = −3(x − 1), so 3x + y − 4 = 0.
- 7x/a + y/a = 1, i.e. x + y = a. Put (4, 1): a = 5. The line is x + y − 5 = 0.
- 8Divide by √(16 + 9) = 5: (4/5)x + (3/5)y = 3, so p = 3.
General form and distances
Two parallel fences in a field have a strip of fixed width between them; that width is the distance between two parallel lines. The shortest distance from a well to a fence is the perpendicular distance. Building on classes 9–10 : In classes 9–10 we learnt the distance formula and parallel lines; here we find distances straight from the equations.
The general form is ax + by + c = 0 (a and b are not both 0). All straight lines, including horizontal and vertical ones, fit in this one form. The perpendicular distance from a point to a line is the length of the shortest segment joining the point to the line.
Rule 1: Slope and intercepts from the general form — Slope m = −a/b, x-intercept −c/a, y-intercept −c/b.
Rule 2: Distance of a point from a line — d = |ax₁ + by₁ + c| / √(a² + b²). Distance from the origin is |c| / √(a² + b²).
In the figure : The perpendicular distance of P(4, 4) from 3x + 4y − 12 = 0 is d = 16/5.
Rule 3: Distance between parallel lines and area — The distance between ax + by + c₁ = 0 and ax + by + c₂ = 0 is d = |c₁ − c₂| / √(a² + b²). The area of the triangle formed by a line and the axes is ½ × |x-intercept| × |y-intercept|.
How is the formula obtained? : Solve ax + by + c = 0 for y: y = −(a/b)x − c/b, so the slope is −a/b. Putting y = 0 gives x = −c/a and putting x = 0 gives y = −c/b. For the distance, shift the origin to P(x₁, y₁): x = X + x₁, y = Y + y₁. The line becomes aX + bY + (ax₁ + by₁ + c) = 0. The new origin is P. The distance of the origin from a line is |constant term| / √(a² + b²), so d = |ax₁ + by₁ + c| / √(a² + b²).
- Easy : 5x − 2y + 3 = 0: slope = −5/(−2) = 5/2. x-intercept = −3/5. y-intercept = −3/(−2) = 3/2.
- Medium : Distance of (3, 1) from 5x + 12y − 20 = 0 = |15 + 12 − 20|/√(25 + 144) = 7/13.
- Exam-type : If the distance of (1, 2) from 3x + 4y + k = 0 is 3 find k: |3 + 8 + k|/5 = 3, so |11 + k| = 15. 11 + k = 15 or 11 + k = −15, so k = 4 or k = −26.
Distance is always positive, so the numerator is kept inside the absolute value. The line must be in the form ax + by + c = 0 (0 on the right).
The distance between parallel lines goes wrong if the coefficients of x and y are not made equal first. For example divide 6x + 8y + 20 = 0 by 2.
Distance is "(value of the equation at the point) ÷ (root of the sum of squares of the coefficients)". For √(a² + b²) remember 3-4-5, 5-12-13, 8-15-17. Memory clue : "Put, absolute, divide" — put the point in, take the absolute value, then divide.
When CET says "the distance is 3", k has two values (±); the options contain both. Taking only one is a mistake.
CET-style example : The distance between the lines 8x + 15y + 4 = 0 and 8x + 15y − 30 = 0 is? (A) 2 (B) 34 (C) 2/17 (D) 17/2
Answer : (A). d = |4 − (−30)|/√(64 + 225) = 34/17 = 2. Time-saving method : Recognise the 8-15-17 triplet; the numerator 34 = 2 × 17, so the answer is 2 at once.
Spot the difference
| Point | Distance of a point from a line | Distance between parallel lines |
|---|---|---|
| Formula | absolute value of (ax₁ + by₁ + c) ÷ √(a² + b²) | absolute value of (c₁ − c₂) ÷ √(a² + b²) |
| What is given | a point and a line | two parallel lines |
| Do first | bring the line to ax + by + c = 0 | make the coefficients of x and y equal |
Practice questions
- 1Find the slope of 7x − 2y + 1 = 0.
- 2Find both intercepts of 2x + 3y − 12 = 0.
- 3Find the distance of the origin from 5x − 12y + 39 = 0.
- 4Find the distance of (3, −2) from 3x + 4y − 5 = 0.
- 5Find the distance between 3x − 4y + 5 = 0 and 3x − 4y − 10 = 0.
- 6Find the area of the triangle formed by 5x + 3y − 30 = 0 and the axes.
- 7If the distance of (2, 1) from 5x − 12y + k = 0 is 2, find both values of k.
- 8The vertices of a triangle are A(0, 0), B(4, 3) and C(2, 5). Find the altitude from C to AB.
Show answers (try them yourself first)
- 1Slope = −7/(−2) = 7/2.
- 2x-intercept = 12/2 = 6 and y-intercept = 12/3 = 4.
- 3|39|/√(25 + 144) = 39/13 = 3.
- 4|9 − 8 − 5|/5 = 4/5.
- 5d = |5 − (−10)|/√(9 + 16) = 15/5 = 3.
- 6The intercepts are 6 and 10. Area = ½ × 6 × 10 = 30.
- 7|10 − 12 + k|/13 = 2, so |k − 2| = 26. k = 28 or k = −24.
- 8The line AB is 3x − 4y = 0. Altitude = |3 × 2 − 4 × 5|/5 = 14/5.
Family of lines and concurrency
Look at many roads that all meet at one square. All the lines passing through one point form a family of lines. Building on classes 9–10 : In classes 9–10 we solved two equations to find the point of intersection; here we pack all lines through that point into one formula.
Lines that pass through the same point are concurrent lines, and that point is their common point. A group of lines with a common property is a family of lines; here all lines pass through the intersection of L₁ = 0 and L₂ = 0.
Rule 1: Point of intersection and the family of lines — The point of intersection of L₁ = 0 and L₂ = 0 is the (x, y) that solves both equations together. All lines through that point: L₁ + kL₂ = 0 (k any number).
In the figure : Lines through P(3, 1), the intersection of x + y − 4 = 0 and x − y − 2 = 0: different values of k give the green lines.
Rule 2: Condition for three lines to be concurrent — The third line must pass through the intersection of the first two. Or [[a₁, b₁, c₁], [a₂, b₂, c₂], [a₃, b₃, c₃]] = 0.
How is the formula obtained? : P(x₀, y₀) is the intersection of L₁ = 0 and L₂ = 0, so L₁(x₀, y₀) = 0 and L₂(x₀, y₀) = 0. Putting (x₀, y₀) in L₁ + kL₂ gives 0 + k × 0 = 0. So whatever k is, the line L₁ + kL₂ = 0 passes through P. For concurrent lines, one common (x₀, y₀) satisfies all three equations, so the determinant is 0 (as seen in the lesson on determinants).
- Easy : Intersection of 2x + 3y = 12 and x − y = 1: multiply the second by 3 and add: 5x = 15, x = 3. Then y = 2. Check: 6 + 6 = 12 ✓. The point is (3, 2).
- Medium : Line through the intersection of x + y − 6 = 0 and 2x − y + 3 = 0 and the origin: adding gives 3x − 3 = 0, x = 1, y = 5. The slope of (1, 5) and (0, 0) is 5, so y = 5x, i.e. 5x − y = 0.
- Exam-type : If x + 2y − 5 = 0, 3x − y − 1 = 0 and 3x + ky − 13 = 0 are concurrent find k: the first two lines meet at (1, 2). Put it in the third: 3 + 2k − 13 = 0, so k = 5.
In L₁ + kL₂ = 0 the value of k can fix any one condition (for example a given point). L₂ itself belongs to the family only as k → ∞.
Mixing up signs when fixing k without finding the intersection; first write both equations as "= 0" and only then form L₁ + kL₂.
Separate the k terms: in (x + y − 4) + k(x − y − 2) = 0 put the part with k and the part without k separately equal to 0. You get two equations and the fixed point at once. Memory clue : "with k and without k — both zero".
When CET says "the line passes through a fixed point", group the terms by k. Do not waste time testing every option.
CET-style example : The line (1 + k)x + (2 − k)y − (5 + 2k) = 0 passes through which fixed point for every k? (A) (3, 1) (B) (1, 3) (C) (−3, 1) (D) (3, −1)
Answer : (A). Group by k: (x + 2y − 5) + k(x − y − 2) = 0. Solve x + 2y = 5 and x − y = 2: subtracting gives 3y = 3, y = 1, x = 3. Time-saving method : Put k = 0 and k = 1 to get two lines and find their intersection.
Spot the difference
| Point | Concurrent lines | Collinear points |
|---|---|---|
| What there is | three or more lines | three or more points |
| Meaning | all lines pass through one point | all points lie on one line |
| Condition | determinant = 0 or the third line passes through the intersection | equal slopes or area 0 |
Practice questions
- 1Find the point of intersection of x + y = 9 and x − y = 1.
- 2Find the point of intersection of 2x − y = 1 and x + y = 5.
- 3Through which point does (x + y − 4) + k(x − y − 2) = 0 pass for every k?
- 4Find the line through the origin and the intersection of x + y − 5 = 0 and x − y − 1 = 0.
- 5Find the line through the intersection of 2x + y − 8 = 0 and x − y + 2 = 0 and parallel to x + 2y − 3 = 0.
- 6If x + y − 2 = 0, 2x − y − 1 = 0 and 3x + ky − 7 = 0 are concurrent, find k.
- 7Using a determinant show that 2x + y − 5 = 0, x − y − 1 = 0 and 3x + 4y − 10 = 0 are concurrent.
- 8Find the line through the intersection of x + y − 3 = 0 and 2x − y = 0 and perpendicular to 2x + 3y − 1 = 0.
Show answers (try them yourself first)
- 1Adding: 2x = 10, x = 5, y = 4. The point is (5, 4).
- 2Adding: 3x = 6, x = 2, y = 3. The point is (2, 3).
- 3Solve x + y − 4 = 0 and x − y − 2 = 0: the point is (3, 1).
- 4The intersection is (3, 2). The line joining it to (0, 0) is y = (2/3)x, i.e. 2x − 3y = 0.
- 5The intersection is (2, 4). The parallel line is x + 2y + k = 0: 2 + 8 + k = 0, k = −10. The line is x + 2y − 10 = 0.
- 6The first two lines meet at (1, 1). 3 + k − 7 = 0, so k = 4.
- 7[[2, 1, −5], [1, −1, −1], [3, 4, −10]] = 2(10 + 4) − 1(−10 + 3) + (−5)(4 + 3) = 28 + 7 − 35 = 0, so the lines are concurrent.
- 8The intersection is (1, 2). The perpendicular line is 3x − 2y + k = 0: 3 − 4 + k = 0, k = 1. The line is 3x − 2y + 1 = 0.
Lesson 5: Straight Line — quick revision
| Topic | Formula or rule |
|---|---|
| Locus | take P(x, y), write the condition, square and simplify |
| Shift of origin | X = x − h, Y = y − k |
| Slope | m = tan θ = (y₂ − y₁)/(x₂ − x₁) |
| Parallel / perpendicular | m₁ = m₂ / m₁m₂ = −1 |
| Angle between lines | tan θ = absolute value of (m₁ − m₂)/(1 + m₁m₂) |
| Point-slope | y − y₁ = m(x − x₁) |
| Intercept form | x/a + y/b = 1 |
| Normal form | x cos α + y sin α = p |
| Distance from a point | absolute value of (ax₁ + by₁ + c) ÷ √(a² + b²) |
| Family of lines | L₁ + kL₂ = 0 |
🎯 Most important points for the exam
- Once you get the condition take P(x, y) and square.
- For perpendicular slopes "flip and change the sign".
- Choose the form of the line by the information given and check by putting a point at the end.
- Distance is always positive; use the absolute value.
- Concurrent lines: find the intersection and put it in the third.
- In CET k can have two values.
Condition → equation → check. Every time put a given point in your equation.