Straight lines for NDA

Straight lines is the entry point to coordinate geometry, and it is formula-dense rather than idea-dense. Once the dozen results below are automatic, most NDA questions become substitution. The questions that catch people are the ones with a small twist: two parallel lines written with different coefficients, a ratio that turns out to be external, or an angle that needs the absolute value.

Typical question types are:

• the point dividing a segment in a given ratio, and the centroid;
• the slope and inclination of a line, and the angle between two lines;
• the equation of a line through a point, parallel or perpendicular to another;
• the distance of a point from a line, or between two parallel lines;
• the foot of a perpendicular, or the image of a point in a line;
• collinearity of three points and concurrency of three lines;
• the area of a triangle from its vertices.

Points: distance, section and area

• Distance between (x₁, y₁) and (x₂, y₂): √[(x₂ − x₁)² + (y₂ − y₁)²].
• Section formula (internal, ratio m : n): ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)). For external division, replace n by −n.
• Midpoint: ((x₁ + x₂)/2, (y₁ + y₂)/2).
• Centroid of a triangle: ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3). It divides each median in the ratio 2 : 1 from the vertex.
• Area of a triangle: ½ × |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. Three points are collinear when this is 0.

Slope and angle

The slope of a line is m = tan θ, where θ is the angle the line makes with the positive x-axis. Through two points, m = (y₂ − y₁)/(x₂ − x₁). For ax + by + c = 0, m = −a/b. A horizontal line has slope 0, and a vertical line has no defined slope.

The acute angle θ between lines of slopes m₁ and m₂ satisfies tan θ = |(m₁ − m₂)/(1 + m₁m₂)|.

• Parallel: m₁ = m₂.
• Perpendicular: m₁m₂ = −1 (or one line vertical and the other horizontal).

Forms of the equation

Families. Every line parallel to ax + by + c = 0 has the form ax + by + k = 0. Every line perpendicular to it has the form bx − ay + k = 0. Find k by putting in the given point. This is faster than working out slopes.

Distances

• Point to line: the distance of (x₁, y₁) from ax + by + c = 0 is |ax₁ + by₁ + c|/√(a² + b²).
• Between parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0: |c₁ − c₂|/√(a² + b²). The x and y coefficients must be identical first.

Foot of the perpendicular from (x₁, y₁) to ax + by + c = 0 is the point (x, y) given by (x − x₁)/a = (y − y₁)/b = −(ax₁ + by₁ + c)/(a² + b²).

Image (mirror reflection) of (x₁, y₁) in the same line: (x − x₁)/a = (y − y₁)/b = −2(ax₁ + by₁ + c)/(a² + b²). The image is twice as far along the same perpendicular as the foot.

Which side? Two points lie on the same side of ax + by + c = 0 when ax + by + c has the same sign at both.

Concurrency

Three lines are concurrent (pass through one point) when the point of intersection of two of them satisfies the third. Equivalently, for a₁x + b₁y + c₁ = 0 and so on, the determinant |a₁ b₁ c₁; a₂ b₂ c₂; a₃ b₃ c₃| is 0, provided no two of the lines are parallel.

Worked NDA-style MCQs

Q1. The point dividing the join of (2, 3) and (7, 8) internally in the ratio 2 : 3 is:
(a) (5, 6) (b) (4, 5) (c) (3, 4) (d) (4.5, 5.5)

x = (2 × 7 + 3 × 2)/5 = 20/5 = 4 and y = (2 × 8 + 3 × 3)/5 = 25/5 = 5. Answer: (b).

Q2. The acute angle between x − 2y + 3 = 0 and 3x − y − 1 = 0 is:
(a) 30° (b) 45° (c) 60° (d) 90°

The slopes are 1/2 and 3. tan θ = |(3 − 1/2)/(1 + 3/2)| = (5/2)/(5/2) = 1, so θ = 45°. Answer: (b).

Q3. The line through (1, −2) perpendicular to 3x + 4y = 7 is:
(a) 4x − 3y = 10 (b) 4x + 3y = −2 (c) 3x + 4y = −5 (d) 4x − 3y = 2

Perpendicular family: 4x − 3y + k = 0. Through (1, −2): 4 + 6 + k = 0, so k = −10, giving 4x − 3y = 10. Answer: (a).

Q4. The distance between 3x + 4y − 7 = 0 and 6x + 8y + 1 = 0 is:
(a) 8/5 (b) 6/5 (c) 3/2 (d) 3/4

Divide the second equation by 2: 3x + 4y + 1/2 = 0. Distance = |−7 − 1/2|/5 = 7.5/5 = 3/2. Using the equations as given, with c = −7 and c = 1, would give the wrong answer 8/5. Answer: (c).

Q5. The image of (3, 4) in the line x + y − 1 = 0 is:
(a) (0, 1) (b) (−3, −2) (c) (−4, −3) (d) (2, −1)

(x − 3)/1 = (y − 4)/1 = −2(3 + 4 − 1)/2 = −6, so the image is (−3, −2). The foot of the perpendicular, using −3 instead, is (0, 1), which is option (a) as a distractor. Answer: (b).

Q6. The points (1, 2), (3, k) and (5, 8) are collinear when k is:
(a) 4 (b) 5 (c) 6 (d) 3

The slope from (1, 2) to (5, 8) is 6/4 = 3/2. The slope from (1, 2) to (3, k) is (k − 2)/2. Setting them equal gives k = 5. (Or notice that (3, 5) is the midpoint.) Answer: (b).

Common mistakes

• Unequal coefficients in the parallel-lines formula. Scale one equation first.
• Dropping the absolute value in the angle formula and reporting an obtuse angle's tangent.
• Slope of ax + by + c = 0 as a/b or −b/a. It is −a/b.
• Mixing internal and external division. External division uses m − n in the denominator.
• Foot versus image. The image uses the factor 2; the foot does not.

Practice set

1. The slope of 2x − 4y + 7 = 0 is: (a) 2 (b) −1/2 (c) 1/2 (d) −2
2. The intercepts of 3x + 2y = 6 on the axes are: (a) 3 and 2 (b) 2 and 3 (c) 6 and 6 (d) 1/2 and 1/3
3. The distance between (1, 1) and (4, 5) is: (a) 4 (b) 5 (c) 7 (d) √7
4. The distance between 3x + 4y = 5 and 3x + 4y = 15 is: (a) 2 (b) 4 (c) 10 (d) 1
5. The lines x + y = 3, 2x − y = 0 and kx + y = 5 are concurrent when k is: (a) 1 (b) 2 (c) 3 (d) 4
6. The inclination of √3x − y + 2 = 0 is: (a) 30° (b) 45° (c) 60° (d) 120°
7. The line through (2, 3) with equal intercepts on the axes is: (a) x + y = 5 (b) x − y = −1 (c) x + y = 6 (d) 3x − 2y = 0
8. The centroid of the triangle with vertices (1, 2), (3, −4) and (5, 8) is: (a) (3, 2) (b) (9, 6) (c) (3, 3) (d) (2, 3)

Answers:

1. (c). −a/b = −2/(−4) = 1/2.
2. (b). x/2 + y/3 = 1.
3. (b). √(9 + 16) = 5.
4. (a). |5 − 15|/5 = 2.
5. (c). The first two meet at (1, 2); then k + 2 = 5.
6. (c). Slope √3 = tan 60°.
7. (a). x/a + y/a = 1 through (2, 3) gives a = 5.
8. (a). (9/3, 6/3).

What to do next

• Rebuild the forms table and the distance formulas from memory, then test yourself on the foot-versus-image difference.
• Solve 30 old NDA questions from this chapter at about 60 seconds each.
• Carry the same tools into circles and conic sections, and see the determinant versions of area and concurrency in matrices and determinants.