Straight Lines | ISC Class 11 Maths Notes

This note covers coordinates, division of line segments, triangle area, shifting the origin, incentres, slopes, angles between lines, equations of straight lines, perpendicular distances, families of lines, angle bisectors and loci.

How do coordinates describe points, segments and triangle areas?

In a Cartesian plane, two perpendicular number lines are the x-axis and y-axis. Their intersection is the origin, written O(0, 0). A point P(x, y) has horizontal coordinate x, called its abscissa, and vertical coordinate y, called its ordinate.

Coordinates are signed. Positive x is measured to the right and positive y upwards. A point on the x-axis has y = 0; a point on the y-axis has x = 0. Thus (6, −4) lies right of and below the origin.

Result: Distance and division of a segment

For distinct points P(x₁, y₁) and Q(x₂, y₂), the subscripts identify the first and second points. The length PQ is √[(x₂ − x₁)² + (y₂ − y₁)²]. The symbol √ denotes the non-negative square root; a squared coordinate difference is non-negative.

Let R divide PQ in the positive ratio r:s, meaning PR:RQ = r:s. For internal division, R lies between P and Q. Its coordinates are ((sx₁ + rx₂)/(r + s), (sy₁ + ry₂)/(r + s)). Notice that the weight r multiplies Q's coordinates.

For external division, R lies on the line beyond the segment. Its coordinates are ((rx₂ − sx₁)/(r − s), (ry₂ − sy₁)/(r − s)), with r ≠ s. Equal positive ratios do not give a finite external division point.

The midpoint divides the segment into equal lengths. Taking equal weights in the internal formula gives ((x₁ + x₂)/2, (y₁ + y₂)/2). It is the average of corresponding coordinates, not the average of all four numbers.

Worked example 1. Find the point dividing A(1, −3) and B(−3, 9) internally in the ratio 1:3.

Answer: The x-coordinate is [1(−3) + 3(1)]/4 = 0. The y-coordinate is [1(9) + 3(−3)]/4 = 0. The required point is the origin (0, 0).

Result: Area and collinearity

Let a third point be T(x₃, y₃). The triangle's area is ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. Vertical bars denote absolute value, which removes the sign of a real number.

For vertices (4, 4), (3, −2) and (−3, 16), the area is ½|4(−18) + 3(12) − 3(6)| = 27 square units. Three points are collinear when they lie on one line; their triangle area is zero.

How do shifting the origin and the incentre use coordinates?

Shifting the origin without rotating the axes

A shift of origin changes the reference point from which coordinates are measured. Let the new origin O′ have old coordinates (h, k), where h and k are fixed real numbers. Keep the new axes parallel to, and directed like, the original axes.

Let the same point have old coordinates (x, y) and new coordinates (X, Y). Then x = X + h and y = Y + k. Equivalently, X = x − h and Y = y − k. Capital letters here distinguish the new coordinates.

To transform an equation, substitute X + h for every x and Y + k for every y. This changes its coordinate description. The geometrical points and their distances remain unchanged because the axes have been translated, not stretched or rotated.

For the line Ax + By + C = 0, let A, B and C be fixed real coefficients, with A and B not both zero. The transformed equation is AX + BY + (Ah + Bk + C) = 0.

The coefficients of the variable coordinates remain A and B. If the new origin lies on the original line, Ah + Bk + C = 0, and the transformed line passes through (0, 0). This is a useful check on the substitution signs.

Coordinates of the incentre

The incentre of a triangle is the common point of its internal angle bisectors. An angle bisector divides an angle into two equal angles. The incentre has equal perpendicular distances from all three sides and is the centre of the circle touching them internally.

For triangle vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃), let a, b and c denote the positive lengths BC, CA and AB respectively. These vertex labels are separate from the equation coefficients used above.

The incentre I has coordinates ((ax₁ + bx₂ + cx₃)/(a + b + c), (ay₁ + by₂ + cy₃)/(a + b + c)). Each vertex is weighted by the length of the opposite side. Calculate those lengths with the distance formula before substituting.

This is a weighted average, not generally the unweighted average of the vertices. State which side each letter represents: attaching a side length to the wrong vertex changes the point and loses the equal-distance property.

What do inclination and slope tell us about a line?

Definition: The inclination θ of a line is its angle with the positive x-direction, measured anticlockwise. Its slope or gradient m is tan θ, provided θ ≠ 90°. The abbreviation tan denotes the tangent trigonometric function.

The symbol ° denotes degrees. A horizontal line has inclination 0° and slope zero. A vertical line has inclination 90° and its slope is undefined. Undefined slope is not a numerical value that can be substituted into a slope formula.

Result: Slope from two points

For distinct points P(x₁, y₁) and Q(x₂, y₂) on a non-vertical line, m = (y₂ − y₁)/(x₂ − x₁), where x₂ ≠ x₁. The numerator measures signed vertical change; the denominator measures signed horizontal change.

Subtract coordinates in the same order. Reversing both differences leaves their ratio unchanged. Reversing just one difference changes the sign incorrectly. A positive slope rises as x increases; a negative slope falls as x increases.

The sloping line passes through P(x₁, y₁) and Q(x₂, y₂). A horizontal segment from P meets the vertical through Q at M. The right angle at M and the marked angle θ at P show the right triangle used for the slope ratio.

See Fig. 9.3(i) in your NCERT textbook

In this diagram, PM is the horizontal change and MQ is the vertical change. For the acute angle shown, their ratio is tan θ. Signed coordinate differences also give the same slope formula when the inclination is obtuse, meaning greater than 90° but less than 180°.

Worked example 2. Find the slope through (3, −2) and (−1, 4), and compare it with the line through (3, −2) and (3, 4).

Answer: The first slope is [4 − (−2)]/(−1 − 3) = 6/(−4) = −3/2. For the second line the horizontal difference is zero. It is vertical, so its slope is undefined.

For collinearity, compare slopes only when the relevant denominators are non-zero. When all three x-coordinates are equal, the points are on a vertical line. The zero-area test handles this case without dividing by a coordinate difference.

How are parallelism, perpendicularity and angles related to slopes?

Let two non-vertical lines have slopes m₁ and m₂. For distinct lines, parallelism means that they do not meet in the plane. They are parallel exactly when m₁ = m₂. Equal slopes describe equal inclinations.

Result: The perpendicularity condition

Two lines with defined slopes are perpendicular, meaning that they meet at a right angle, exactly when m₁m₂ = −1. Thus one slope is the negative reciprocal of the other: m₂ = −1/m₁. A reciprocal is one divided by a non-zero number.

The horizontal and vertical pair must be handled separately because the vertical slope is undefined. Also, equal slopes alone do not distinguish distinct parallel lines from coincident lines, which are two equations describing the same geometrical line.

Finding the smaller angle

For non-parallel, non-perpendicular lines with defined slopes, let θ be their acute angle, meaning the angle between 0° and 90°. Then tan θ = |(m₂ − m₁)/(1 + m₁m₂)|, provided 1 + m₁m₂ ≠ 0.

The other angle is 180° − θ because adjacent angles formed by intersecting lines are supplementary, meaning their sum is 180°. If the denominator is zero, use the perpendicularity condition to obtain 90° instead of attempting division.

Worked example 3. Two lines make an angle of 45°. One has slope 1/2. Find the possible slopes of the other.

Answer: Write its slope as m. Since tan 45° = 1, |(m − 1/2)/(1 + m/2)| = 1. The positive case gives 2m − 1 = 2 + m, so m = 3. The negative case gives 2m − 1 = −2 − m, so m = −1/3.

Both answers are needed: a line may make the specified acute angle on either side of the given direction. The absolute value is therefore a geometrical requirement, not merely a device for making a final numerical answer positive.

Note: Conditions involving slopes carry the condition that those slopes exist. Check vertical lines before applying the parallel, perpendicular or angle formula.

How do we write equations for axis-parallel lines and a given slope?

An equation of a line is a condition satisfied by the coordinates of every point on that line, and by no point outside it. A known point can therefore be checked by substituting its two coordinates into the equation.

Horizontal and vertical lines

A horizontal line through a point (u, v), where u and v are its fixed coordinates, has equation y = v. The x-coordinate is free to vary. A vertical line through the same point has equation x = u, while y varies.

The x-axis itself has equation y = 0, and the y-axis has equation x = 0. A positive distance from an axis does not by itself choose a side: points above and below the x-axis can have the same distance from it.

Worked example 4. Find the lines parallel to the coordinate axes through (−2, 3).

Answer: The horizontal line keeps the y-coordinate fixed at 3, so its equation is y = 3. The vertical line keeps the x-coordinate fixed at −2, so its equation is x = −2.

Point-slope form

Let P₀(x₀, y₀) be a fixed point on a non-vertical line with known slope m. A variable point P(x, y) on it satisfies y − y₀ = m(x − x₀). This is the point-slope form.

It follows by rearranging the slope ratio between P₀ and P. Although the ratio requires distinct x-coordinates, the rearranged equation also includes P₀ itself. At that point both sides are zero, as required.

Worked example 5. Find the line through (−2, 3) with slope −4.

Answer: Substitution gives y − 3 = −4[x − (−2)] = −4(x + 2). Expanding gives y − 3 = −4x − 8, hence 4x + y + 5 = 0. The given point makes the left side zero.

When should we use two-point and slope-intercept forms?

Two-point form

Two distinct points determine one straight line. For points (x₁, y₁) and (x₂, y₂) with x₁ ≠ x₂, first calculate their slope and substitute it into point-slope form. This gives y − y₁ = [(y₂ − y₁)/(x₂ − x₁)](x − x₁).

If x₁ = x₂, the points determine the vertical line x = x₁. Avoid dividing by zero. An equivalent equation without division is (x₂ − x₁)(y − y₁) = (y₂ − y₁)(x − x₁); the points must still be distinct.

Worked example 6. Find the equation through (1, −1) and (3, 5).

Answer: The slope is [5 − (−1)]/(3 − 1) = 3. Therefore y + 1 = 3(x − 1), giving −3x + y + 4 = 0. Substituting either supplied point gives zero.

Slope-intercept form

A y-intercept c is the signed y-coordinate where a line meets the y-axis, at (0, c). For a non-vertical line with slope m, point-slope form at this point gives y = mx + c.

Similarly, an x-intercept d is the signed x-coordinate of the point (d, 0). A line with defined slope m and this intercept satisfies y = m(x − d). Here d denotes an intercept, not a perpendicular distance.

Positive and negative intercepts specify opposite sides of the origin. The constant in y = mx + c is directly the y-intercept; an arbitrary constant in another arrangement of the equation need not be an intercept.

Worked example 7. A line has slope 1/2 and y-intercept −3/2. Find its equation.

Answer: Substitute into slope-intercept form to obtain y = x/2 − 3/2. Multiplying by 2 and rearranging gives x − 2y − 3 = 0. At x = 0 the equation gives y = −3/2, confirming the intercept.

Choose the form that uses the data directly. With two points, use their coordinate differences. With a slope and an intercept, begin with the corresponding intercept point. Rearrange only after the supplied conditions have been incorporated.

How do intercept form and the general equation describe a line?

Intercept form

Let a and b now denote the non-zero x-intercept and y-intercept of a line. Its intercept points are (a, 0) and (0, b). Substituting them into two-point form and simplifying gives x/a + y/b = 1.

These intercepts are signed coordinates. A line meeting a negative part of an axis has a negative intercept on that axis. The formula requires both intercepts to be finite and non-zero, so it is unsuitable for axis-parallel lines or lines through the origin.

A descending line meets the positive x-axis at (a, 0) and the positive y-axis at (0, b). The horizontal and vertical distances from O are labelled a and b. The drawing illustrates positive intercepts.

See Fig. 9.13 in your NCERT textbook

Worked example 8. Find the line with x-intercept −3 and y-intercept 2.

Answer: Intercept form gives x/(−3) + y/2 = 1. Multiplying by 6 gives −2x + 3y = 6, or 2x − 3y + 6 = 0. Its intercept points are (−3, 0) and (0, 2).

General form and its conditions

The general equation is Ax + By + C = 0, where A, B and C are real coefficients and A and B are not both zero. Unlike a slope form, it includes vertical lines.

For A and B both non-zero, setting y = 0 gives x-intercept −C/A, and setting x = 0 gives y-intercept −C/B. Multiplying the whole equation by a non-zero number leaves its solution points unchanged.

This last property explains why different-looking equations can describe one line. Preserve every coefficient's sign when rearranging or scaling. Changing just the constant, without changing the other coefficients, generally produces a different parallel line.

What is the normal form of a straight line?

A normal to a line is a line perpendicular to it. Let p be the positive perpendicular distance from the origin to a line not through the origin. Let α be the angle made anticlockwise by that directed perpendicular with the positive x-axis.

The line's normal form is x cos α + y sin α = p. Here cos and sin denote the cosine and sine trigonometric functions. The angle α describes the perpendicular from the origin, not the inclination of the original line.

Why does the form work?

Let H be the foot of this perpendicular, meaning its intersection with the line. Its coordinates are (p cos α, p sin α). The displacement along the perpendicular direction from the origin to any point on the line has the same component p.

For P(x, y) on the line, that component is x cos α + y sin α. Equating it to p gives the normal equation. The identity cos² α + sin² α = 1 expresses the unit length of the direction used in this projection.

Converting a general equation

Begin with Ax + By = −C. Divide the whole equation by one of the two numbers √(A² + B²) and −√(A² + B²), choosing the sign that makes the right side positive.

The resulting coefficients of x and y are cos α and sin α. Their squares sum to one. Use the signs of both coefficients to choose the angle correctly, rather than using a tangent ratio alone.

If C = 0, the distance is zero. An equation x cos α + y sin α = 0 still represents the line when α specifies a normal direction, but the zero-length perpendicular from the origin does not select a unique directed normal.

Note: In normal form, p is a non-negative distance. In intercept form, the intercepts are signed coordinates. Their sign conventions answer different geometrical questions.

How is the perpendicular distance from a point to a line calculated?

The distance from a point to a line means the length of the perpendicular segment joining them. For point P(x₁, y₁) and line Ax + By + C = 0, write this distance as d.

Result: Point-to-line distance

The formula is d = |Ax₁ + By₁ + C|/√(A² + B²), with A and B not both zero. Substitute the point's coordinates only in the numerator; the denominator depends on the line's coefficients.

The absolute value keeps distance non-negative. If the numerator is zero, the point satisfies the line's equation and lies on it. The denominator also ensures that multiplying the complete line equation by a non-zero number leaves the distance unchanged.

Line L meets the axes at R and Q. Point P lies away from the line. Segment PM meets it at a marked right angle at M and is labelled d. Dashed segments join P to R and Q.

See Fig. 9.14 in your NCERT textbook

When the line has distinct intercept points Q and R, its triangle with P has area ½ × QR × PM. Therefore PM is twice the triangle's area divided by QR. Substituting coordinate expressions for area and base leads to the distance formula.

The formula itself also applies to horizontal and vertical lines, even though that intercept-triangle construction is unavailable for them. Read coefficients after writing the equation with zero on one side, so that the constant is included with its correct sign.

Worked example 9. Find the distance of (3, −5) from 3x − 4y − 26 = 0.

Answer: Here A = 3, B = −4 and C = −26. Thus d = |3(3) + (−4)(−5) − 26|/√[3² + (−4)²] = |3|/5 = 3/5 unit.

A distance measured along a specified oblique line is a different quantity. Find where that line meets the given line, then use the distance between the two points. Use the perpendicular-distance formula only when the perpendicular length is wanted.

How do we find the distance between parallel lines?

The separation of two parallel lines is measured perpendicularly. Take any point on one line and find its perpendicular distance to the other. This gives a constant separation because the lines have the same direction.

Match the variable coefficients first

Write the lines as Ax + By + C₁ = 0 and Ax + By + C₂ = 0, with exactly the same A and B. Here C₁ and C₂ are their respective constant terms. Then d = |C₂ − C₁|/√(A² + B²).

To see why, let P(x₁, y₁) lie on the first line. It satisfies Ax₁ + By₁ = −C₁. Substitution into the distance formula for the second line leaves |−C₁ + C₂| in the numerator, independent of which point P was selected.

If the variable coefficients are merely proportional, multiply or divide an entire equation to make them identical. Subtracting the displayed constants before doing this would compare differently scaled equations and could give an incorrect distance.

Worked example 10. Find the distance between 3x − 4y + 7 = 0 and 3x − 4y + 5 = 0.

Answer: The variable coefficients already match. The distance is |7 − 5|/√[3² + (−4)²] = 2/5 unit. No point of intersection is needed because the lines are parallel.

For non-vertical lines y = mx + c₁ and y = mx + c₂, where c₁ and c₂ are their y-intercepts, the same rule becomes d = |c₂ − c₁|/√(1 + m²). The vertical difference of intercepts is not generally the perpendicular separation.

If the constants also agree after scaling, the equations describe the same line and the separation is zero. Check coincidence as well as parallelism before interpreting a calculated distance.

How do families of lines and concurrency problems work?

A family of lines is a collection described by an equation containing an adjustable parameter. A parameter is a quantity held fixed for one member but allowed to vary when choosing another member of the collection.

Lines through an intersection

Let L₁ = A₁x + B₁y + C₁ and L₂ = A₂x + B₂y + C₂ be two linear expressions. Their coefficients are fixed real numbers. Suppose L₁ = 0 and L₂ = 0 represent distinct intersecting lines.

Then L₁ + λL₂ = 0 gives lines through their intersection, where λ, pronounced lambda, is a real parameter. At that intersection both expressions vanish, so their linear combination also vanishes.

This form includes L₁ = 0 when λ = 0, but excludes L₂ = 0 for finite λ. The complete family is uL₁ + vL₂ = 0, where u and v are real parameters not both zero. Taking u = 0 includes L₂.

Use an extra condition to choose a member: substitute a required point, equate a slope, or impose parallelism or perpendicularity. Retain the condition that the generating lines intersect; a family generated by parallel lines requires a different interpretation.

Concurrency

Lines are concurrent if they pass through a common point. For three given equations, solve two suitable equations simultaneously and substitute their intersection into the third. The third equation then supplies any unknown coefficient.

Worked example 11. Find k if 2x + y − 3 = 0, 5x + ky − 3 = 0 and 3x − y − 2 = 0 are concurrent.

Answer: Adding the first and third equations gives 5x − 5 = 0, so x = 1. The first equation then gives y = 1. Substitution into the second gives 5 + k − 3 = 0, hence k = −2.

The family method and the intersection method express the same common-point requirement. Direct simultaneous solution is convenient when the intersection is simple. A family equation is useful when the extra condition can be applied without calculating that point first.

How do loci lead to equations of angle bisectors?

Definition: A locus is the set of all points satisfying a specified geometrical condition. Its equation translates that condition into a relation between the variable coordinates x and y.

To find a locus equation, let P(x, y) be an arbitrary point satisfying the condition. Express the required distances, slopes or ratios in coordinates, then simplify. Check that the resulting points satisfy the original condition, especially after squaring or cancelling expressions.

Equidistance and angle bisectors

For distinct intersecting lines L₁ = 0 and L₂ = 0, use the linear expressions and coefficients defined in the preceding section. Equidistance from the lines means |L₁|/√(A₁² + B₁²) = |L₂|/√(A₂² + B₂²).

The locus consists of their two angle bisectors. Removing the absolute values requires both cases: L₁/√(A₁² + B₁²) = ±L₂/√(A₂² + B₂²). The symbol ± means that the plus sign and the minus sign give separate equations.

Dividing by the coefficient lengths is essential. Equal values of unnormalised line expressions do not generally mean equal distances. Also, the plus sign does not by itself identify a particular internal bisector; reversing one line equation's sign exchanges the two labels.

Worked example 12. Find the locus of a point equidistant from 3x − 2y = 5 and 3x + 2y = 5.

Answer: Their distance denominators are both √13. Thus |3x − 2y − 5| = |3x + 2y − 5|. Equal signed expressions give −4y = 0, hence y = 0. Opposite signed expressions give 6x − 10 = 0, hence x = 5/3.

The full answer contains both straight lines, not just one selected branch. Every point on either line has equal perpendicular distances from the original pair. Their intersection is also the intersection of the original two lines.

For the internal bisector of a specified triangle angle, select the branch lying inside that angle. Its intersection with another internal bisector gives the incentre, connecting the locus method with the equal-distance property of the triangle's centre.

Glossary

• Cartesian coordinates — An ordered pair giving a point's signed horizontal and vertical positions relative to perpendicular axes.
• Origin — The intersection of the coordinate axes, having both coordinates equal to zero.
• Internal division — Division of a segment by a point lying between its two endpoints in a specified ratio.
• External division — Division in a specified ratio by a point on the line outside the segment.
• Collinear points — Points lying on one straight line, with zero area for their associated triangle.
• Inclination — The angle a line makes with the positive x-direction, measured anticlockwise.
• Slope — The tangent of a non-vertical line's inclination, also its ratio of signed coordinate changes.
• Intercept — The signed coordinate at which a line meets a specified coordinate axis.
• Normal — A line perpendicular to another line at their point of intersection.
• Perpendicular distance — The length of the perpendicular segment joining a point to a line.
• Concurrent lines — Lines that all pass through the same common point in the coordinate plane.
• Family of lines — A collection of lines represented by an equation with one or more adjustable parameters.
• Locus — The set of all points satisfying a given geometrical condition or collection of conditions.
• Incentre — The intersection of a triangle's internal angle bisectors, equally distant from its three sides.

Common errors and misconceptions

• Misconception: A vertical line has slope zero. Correct: Its slope is undefined because the horizontal coordinate difference is zero. A horizontal line has slope zero.
• Misconception: Reversing just one coordinate difference preserves a slope. Correct: Subtract corresponding coordinates in the same order in both numerator and denominator.
• Misconception: The slope-product test applies to every perpendicular pair. Correct: It requires both slopes to exist. Recognise the horizontal and vertical pair separately.
• Misconception: Intercepts are necessarily positive distances. Correct: Intercepts are signed coordinates, while perpendicular distances are non-negative lengths.
• Misconception: Constants can be subtracted immediately to find parallel-line separation. Correct: First make the coefficients of x and y identical in both equations.
• Misconception: Normal form uses the original line's inclination. Correct: Its angle describes the directed perpendicular from the origin to the line.
• Misconception: Equal absolute values give only equal signed expressions. Correct: The expressions may be equal or opposite, so an equidistance locus can contain two bisectors.
• Misconception: Shifting the origin to (h, k) means substituting x = X − h. Correct: Old and new coordinates satisfy x = X + h and y = Y + k.

Exam-style questions with model answers

Q1. Find the slope of the line through (3, −2) and (−1, 4). [2 marks]

1. The slope is the signed vertical change divided by the signed horizontal change: m = [4 − (−2)]/(−1 − 3).
2. Thus m = 6/(−4) = −3/2. The denominator is non-zero, so the slope is defined.

Q2. Two lines make an acute angle of 45°. One line has slope 1/2. Find both possible slopes of the other line. [3 marks]

1. Let m be the other slope. The angle formula gives 1 = |(m − 1/2)/(1 + m/2)| because tan 45° = 1.
2. For the positive case, m − 1/2 = 1 + m/2. Multiplying by 2 gives 2m − 1 = 2 + m, so m = 3.
3. For the negative case, m − 1/2 = −1 − m/2, giving 3m = −1. Hence the possible slopes are 3 and −1/3.

Q3. Find the equation of the line through (1, −1) and (3, 5), and verify that the second point lies on it. [3 marks]

1. The two x-coordinates differ, so use the slope formula. The slope m is [5 − (−1)]/(3 − 1) = 6/2 = 3.
2. Point-slope form through the first point gives y + 1 = 3(x − 1). Expansion and rearrangement give −3x + y + 4 = 0.
3. At the second point (3, 5), the left side is −3(3) + 5 + 4 = 0. Therefore that point satisfies the equation.

Q4. Find the equation of the line whose x-intercept is −3 and y-intercept is 2. State its intercept points. [3 marks]

1. The intercepts are signed coordinates. The line therefore meets the x-axis at (−3, 0) and the y-axis at (0, 2).
2. Both intercepts are non-zero, so intercept form applies: x/(−3) + y/2 = 1. The negative sign belongs to the x-intercept.
3. Multiplying the entire equation by 6 gives −2x + 3y = 6. Equivalently, the required equation is 2x − 3y + 6 = 0.

Q5. Find (a) the distance of (3, −5) from 3x − 4y − 26 = 0 and (b) the distance between 3x − 4y + 7 = 0 and 3x − 4y + 5 = 0. [4 marks]

1. For part (a), the point-to-line formula gives d = |3(3) + (−4)(−5) − 26|/√[3² + (−4)²], where d denotes perpendicular distance.
2. The numerator is |9 + 20 − 26| = 3 and the denominator is 5. Hence the first distance is 3/5 unit.
3. For part (b), the two equations already have identical variable coefficients, 3 and −4. Their constant terms are 7 and 5.
4. The parallel-line separation is therefore |7 − 5|/√[3² + (−4)²] = 2/5 unit. Both results are non-negative lengths.

Q6. Triangle PQR has vertices P(2, 1), Q(−2, 3) and R(4, 5). Find the equation of the median from R and verify that it contains the midpoint of PQ. [5 marks]

1. A median joins a triangle's vertex to the midpoint of the opposite side. Let M denote the midpoint of PQ.
2. By the midpoint formula, M = ((2 − 2)/2, (1 + 3)/2) = (0, 2). These coordinates average the corresponding coordinates of P and Q.
3. The median passes through R(4, 5) and M(0, 2). Its slope is (5 − 2)/(4 − 0) = 3/4.
4. Using point-slope form at M gives y − 2 = (3/4)x. Multiplying by 4 and rearranging gives 3x − 4y + 8 = 0.
5. At M, the left side is 0 − 8 + 8 = 0; at R, it is 12 − 20 + 8 = 0. Thus the line contains the required midpoint and vertex.

Q7. Determine k so that 2x + y − 3 = 0, 5x + ky − 3 = 0 and 3x − y − 2 = 0 are concurrent. Verify the result. [5 marks]

1. Concurrency means that all three lines pass through one common point. First find the intersection of the first and third lines, which contain no unknown coefficient.
2. Add 2x + y − 3 = 0 and 3x − y − 2 = 0. This gives 5x − 5 = 0, hence x = 1.
3. Substitute x = 1 into the first equation: 2 + y − 3 = 0. Thus y = 1 and the intersection is (1, 1).
4. For the second line to contain this point, 5(1) + k(1) − 3 = 0. Therefore k = −2.
5. With k = −2, substitution of (1, 1) gives zero in each equation: 2 + 1 − 3, 5 − 2 − 3 and 3 − 1 − 2. This verifies concurrency.

Q8. Find the complete locus of a point having equal perpendicular distances from 3x − 2y = 5 and 3x + 2y = 5. Explain its geometrical meaning. [5 marks]

1. Let P(x, y) be any point on the locus. Its distances from the two lines are |3x − 2y − 5|/√13 and |3x + 2y − 5|/√13.
2. Equating these distances and cancelling their common positive denominator gives |3x − 2y − 5| = |3x + 2y − 5|. Both sign possibilities must be retained.
3. If the expressions have the same sign and value, 3x − 2y − 5 = 3x + 2y − 5. Therefore y = 0.
4. If they are opposite, 3x − 2y − 5 = −3x − 2y + 5. Therefore 6x = 10, giving x = 5/3.
5. The complete locus is the pair y = 0 and x = 5/3. These are the two angle bisectors of the given intersecting lines, and every point on either has the required equal distances.

Key takeaways

• Use signed coordinate differences consistently when calculating slope; treat vertical lines separately because their slopes are undefined.
• Internal division uses sums of positive weights, while external division uses differences and requires unequal ratio terms.
• Shifting the origin changes coordinates through x = X + h and y = Y + k without changing geometrical distances.
• Choose a line equation from the given data: a point and slope, two points, intercepts or a perpendicular from the origin.
• Non-vertical parallel lines have equal slopes; perpendicular lines with defined slopes have product −1.
• Use absolute values in distance formulae, and match variable coefficients before subtracting constants for parallel-line separation.
• A family through two intersecting lines' common point follows from linear combinations of their equations.
• Equal perpendicular distances from intersecting lines give both angle bisectors, so retain both sign choices.

Test yourself

Why is the slope of a vertical line undefined?

Its points have equal x-coordinates, so the slope ratio would require division by zero.

What are the equations of the coordinate axes?

The x-axis has equation y = 0; the y-axis has equation x = 0.

How can triangle area test whether three points are collinear?

Calculate the area from their coordinates. Zero area means that the three points lie on one line.

What is the restriction on the coefficients in Ax + By + C = 0?

A, B and C are real, and A and B must not both be zero.

In normal form x cos α + y sin α = p, what do α and p describe?

For a line not through the origin, α gives the directed perpendicular's angle with the positive x-axis, and p is its positive length.

What must be checked before using the difference of constants for parallel-line distance?

The coefficients of x and y must match exactly in both equations, after scaling an entire equation if necessary.

Which side lengths weight the vertices in the incentre formula?

Each vertex is weighted by the length of its opposite side, with the perimeter, the sum of the three side lengths, as the common denominator.

Why do equal distances from two intersecting lines give two equations?

Equality of absolute values permits equal or opposite signed expressions. The two resulting lines are the angle bisectors.