Straight Lines Class 11 Maths Notes

• A horizontal line has m = 0; a vertical line has m undefined (90° with x-axis).
• The angle of inclination θ lies in [0°, 180°); m = tan θ gives a one-to-one mapping.
• A positive slope rises left to right; a negative slope falls left to right.

MCQ Questions

1 Mark Questions

1. 1
   Define the slope of a line. What does it measure in terms of the angle of inclination?
2. 2
   State the slope of a horizontal line and of a vertical line.
3. 3
   What is the slope of the line that makes an angle of 45° with the positive direction of the x-axis?
4. 4
   Write the formula for the slope of the line through (x₁, y₁) and (x₂, y₂).

2 Mark Questions

1. 1
   Find the slope of the line joining the points (5, 4) and (2, −3).
2. 2
   Find the slope of a line passing through the origin and the mid-point of the segment joining P(0, −4) and B(8, 0).
3. 3
   Find the slope of the line which makes an angle of 30° with the positive direction of the y-axis measured anticlockwise.
4. 4
   Using slopes, show that the points A(4, 4), B(3, 5) and C(−1, −1) are the vertices of a right angled triangle (without using Pythagoras' theorem).

3 Mark Questions

1. 1
   Find the value of x for which the points (x, −1), (2, 1) and (4, 5) are collinear.
2. 2
   If the three points (h, 0), (a, b) and (0, k) lie on a straight line, then show that a/h + b/k = 1.
3. 3
   The slope of a line is double of the slope of another line. If the tangent of the angle between them is ⅓, find the slopes of the two lines.

MCQ Questions

1 Mark Questions

1. 1
   State the condition for two non-vertical lines to be parallel.
2. 2
   State the condition for two lines to be perpendicular.
3. 3
   Find the slope of a line perpendicular to a line whose slope is −5.
4. 4
   Are the lines y = 2x + 3 and y = 2x − 5 parallel or perpendicular?

2 Mark Questions

1. 1
   Find the equation of the line passing through (−3, 5) and perpendicular to the line through the points (2, 5) and (−3, 6).
2. 2
   Find the equation of the line parallel to the line 3x − 4y + 2 = 0 and passing through the point (−2, 3).
3. 3
   Find the value of k so that the lines 2x + 3y − 5 = 0 and kx − 6y + 7 = 0 are perpendicular.
4. 4
   Find the equation of a line that passes through the point (0, 2) and is perpendicular to the line x + y = 4.

3 Mark Questions

1. 1
   Find the equation of the right bisector (perpendicular bisector) of the line segment joining the points (3, 4) and (−1, 2).
2. 2
   A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1 : n. Find the equation of the line.
3. 3
   Find the equation of the line through the point (2, 3) that is perpendicular to the line joining (1, −2) and (4, 4), and hence find its distance from the origin.

• Point-slope form: y − y₁ = m(x − x₁) — the workhorse for most problems.
• Two-point form: (y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁).
• Slope-intercept form: y = mx + c, where c is the y-intercept.
• Intercept form: x/a + y/b = 1, where a and b are the x- and y-intercepts.
• Normal form: x cos α + y sin α = p, where p is the perpendicular distance from origin.

MCQ Questions

1 Mark Questions

1. 1
   Write the equations of the x-axis and the y-axis.
2. 2
   Write the point-slope form of the equation of a line.
3. 3
   Write the slope–intercept form of the equation of a line.
4. 4
   Write the equation of a horizontal line at a distance 5 above the x-axis.

2 Mark Questions

1. 1
   Find the equation of the line passing through the point (−4, 3) with slope ½.
2. 2
   Find the equation of the line passing through (2, 2√3) and inclined with the x-axis at an angle of 75°.
3. 3
   Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).
4. 4
   Find the equation of the line which intersects the x-axis at a distance of 3 units to the left of the origin and has slope −2.

3 Mark Questions

1. 1
   Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
2. 2
   Find the equation of the line through the point (0, 2) making an angle 2π/3 with the positive x-axis. Also find the equation of the line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
3. 3
   A line with slope m makes an x-intercept d. Show that its equation is y = m(x − d), and use it to find the equation of a line of slope 3 whose x-intercept is 2.

MCQ Questions

1 Mark Questions

1. 1
   Write down the general (linear) form of the equation of a line.
2. 2
   Find the slope of the line 2x + 3y − 6 = 0.
3. 3
   Write the intercepts made by the line x/2 + y/3 = 1 on the axes.
4. 4
   Express 3x − 4y + 7 = 0 in slope–intercept form.

2 Mark Questions

1. 1
   Reduce the equation x + 7y = 0 into slope–intercept form and find its slope and y-intercept.
2. 2
   Reduce the equation 3x + 2y − 12 = 0 into intercept form and find its intercepts on the axes.
3. 3
   Reduce 6x + 3y − 5 = 0 into slope–intercept form and find its slope and y-intercept.
4. 4
   Find the equation of a vertical line passing through the point (2, 3).

3 Mark Questions

1. 1
   Reduce 4x − 3y = 6 into intercept form and find its intercepts on the axes; then write its slope–intercept form.
2. 2
   Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is y/x = (tan θ ± m)/(1 ∓ m tan θ).
3. 3
   Find the equation of the line passing through the point of intersection of the lines 4x + 7y − 3 = 0 and 2x − 3y + 1 = 0 that has equal intercepts on the axes.

• The formula gives the acute angle by default; drop the absolute value to get the directed angle.
• If one line is vertical (m undefined), θ = |90° − α| where α is the other line's inclination.
• For concurrent lines, use the formula pairwise and sum appropriately.

MCQ Questions

1 Mark Questions

1. 1
   Write the formula for the tangent of the acute angle between two lines with slopes m₁ and m₂.
2. 2
   What is the angle between the two lines y = 2x + 1 and y = 2x − 4?
3. 3
   If the product of the slopes of two lines is −1, what is the angle between them?
4. 4
   Find the acute angle between the line y = x and the x-axis.

2 Mark Questions

1. 1
   Find the angle between the x-axis and the line joining the points (3, −1) and (4, −2).
2. 2
   Find the acute angle between the lines √3x + y = 1 and x + √3y = 1.
3. 3
   Find the acute angle between lines with slopes 1 and 2 (give tan θ and hence θ).
4. 4
   Two lines passing through the point (2, 3) intersect each other at an angle of 60°. If the slope of one line is 2, find the slope(s) of the other line.

3 Mark Questions

1. 1
   Two lines passing through the point (2, 3) intersect each other at an angle of 60°. If the slope of one line is 2, find the equation of the other line.
2. 2
   Find the equations of the lines passing through the point (3, 2) which make an angle of 45° with the line x − 2y = 3.
3. 3
   Show, using the angle-between-lines formula, that the lines 12x + 5y = 13 and 5x − 12y = 26 are perpendicular to each other, and hence find the angle between them.

MCQ Questions

1 Mark Questions

1. 1
   State the formula for the perpendicular distance of the point (x₁, y₁) from the line Ax + By + C = 0.
2. 2
   State the formula for the distance between the parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0.
3. 3
   Write the distance of the origin from the line Ax + By + C = 0.
4. 4
   Find the distance of the line 4x + 3y = 5 from the origin.

2 Mark Questions

1. 1
   Find the distance of the point (−1, 1) from the line 12(x + 6) = 5(y − 2).
2. 2
   Find the distance between the parallel lines 15x + 8y − 34 = 0 and 15x + 8y + 31 = 0.
3. 3
   Find the points on the x-axis whose distances from the line x/3 + y/4 = 1 are 4 units.
4. 4
   Find the distance between the parallel lines 3x − 4y + 7 = 0 and 3x − 4y + 5 = 0.

3 Mark Questions

1. 1
   Find the coordinates of the foot of the perpendicular from the point (−1, 3) to the line 3x − 4y − 16 = 0.
2. 2
   The perpendicular from the origin to the line y = mx + c meets it at the point (−1, 2). Find the values of m and c.
3. 3
   A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

MCQ Questions

1 Mark Questions

1. 1
   Write the formula for the area of a triangle with vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃).
2. 2
   What does a zero area from the three-vertices formula signify?
3. 3
   Find the area of the triangle with vertices (0, 0), (4, 0) and (0, 3).
4. 4
   State the collinearity condition of three points in terms of area.

2 Mark Questions

1. 1
   Draw a quadrilateral in the Cartesian plane, whose vertices are (−4, 5), (0, 7), (5, −5) and (−4, −2). Also find its area.
2. 2
   Find the area of the triangle formed by the lines y − x = 0, x + y = 0 and x − k = 0.
3. 3
   By using the concept of the equation of a line (or the area formula), prove that the points (3, 0), (−2, −2) and (8, 2) are collinear.
4. 4
   Find the area of the triangle formed by the lines 2x + y − 5 = 0, x − y + 2 = 0 and the x-axis.

3 Mark Questions

1. 1
   The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (−4, 1). Find the equations of the legs (perpendicular sides) of the triangle which are parallel to the axes, and hence the area of the triangle.
2. 2
   Show that the area of the triangle formed by the lines y = m₁x + c₁, y = m₂x + c₂ and x = 0 equals (c₁ − c₂)² / (2|m₁ − m₂|) square units, and verify it for m₁ = 2, m₂ = 3, c₁ = 5, c₂ = −1.
3. 3
   Show that the area of the triangle with vertices (2, 3), (5, 7) and (0, 2) equals the area of the triangle with vertices (0, 0), (2, 3), (5, 7). Interpret this using translation of the origin.

Exercises (NCERT Ex 9.1–9.3 & Miscellaneous)

9.1.1) Draw a quadrilateral in the Cartesian plane, whose vertices are (−4, 5), (0, 7), (5, −5) and (−4, −2). Also, find its area.

9.1.2) The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

9.1.3) Find the distance between P(x₁, y₁) and Q(x₂, y₂) when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

9.1.4) Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

9.1.5) Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0, −4) and B(8, 0).

9.1.6) Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (−1, −1) are the vertices of a right angled triangle.

9.1.7) Find the slope of the line, which makes an angle of 30° with the positive direction of the y-axis measured anticlockwise.

9.1.8) Without using the distance formula, show that the points (−2, −1), (4, 0), (3, 3) and (−3, 2) are the vertices of a parallelogram.

9.1.9) Find the angle between the x-axis and the line joining the points (3, −1) and (4, −2).

9.1.10) The slope of a line is double of the slope of another line. If the tangent of the angle between them is ⅓, find the slopes of the lines.

9.1.11) A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that k − y₁ = m(h − x₁).

9.2.1) Write the equations for the x- and y-axes.

9.2.2) Passing through the point (−4, 3) with slope ½.

9.2.3) Passing through (0, 0) with slope m.

9.2.4) Passing through (2, 2√3) and inclined with the x-axis at an angle of 75°.

9.2.5) Intersecting the x-axis at a distance of 3 units to the left of origin with slope −2.

9.2.6) Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with the positive direction of the x-axis.

9.2.7) Passing through the points (−1, 1) and (2, −4).

9.2.8) The vertices of ∆PQR are P(2, 1), Q(−2, 3) and R(4, 5). Find the equation of the median through the vertex R.

9.2.9) Find the equation of the line passing through (−3, 5) and perpendicular to the line through the points (2, 5) and (−3, 6).

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

9.2.11) Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

9.2.12) Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

9.2.13) Find the equation of the line through the point (0, 2) making an angle 2π/3 with the positive x-axis. Also, find the equation of the line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

9.2.14) The perpendicular from the origin to a line meets it at the point (−2, 9). Find the equation of the line.

9.2.15) The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.

9.2.16) The owner of a milk store finds that he can sell 980 litres of milk each week at ₹14/litre and 1220 litres each week at ₹16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹17/litre?

9.2.17) P(a, b) is the mid-point of a line segment between axes. Show that the equation of the line is x/a + y/b = 2.

9.2.18) Point R(h, k) divides a line segment between the axes in the ratio 1 : 2. Find the equation of the line.

9.2.19) By using the concept of the equation of a line, prove that the three points (3, 0), (−2, −2) and (8, 2) are collinear.

9.3.1) Reduce the following equations into slope–intercept form and find their slopes and the y-intercepts: (i) x + 7y = 0, (ii) 6x + 3y − 5 = 0, (iii) y = 0.

9.3.2) Reduce the following equations into intercept form and find their intercepts on the axes: (i) 3x + 2y − 12 = 0, (ii) 4x − 3y = 6, (iii) 3y + 2 = 0.

9.3.3) Find the distance of the point (−1, 1) from the line 12(x + 6) = 5(y − 2).

9.3.4) Find the points on the x-axis, whose distances from the line x/3 + y/4 = 1 are 4 units.

9.3.5) Find the distance between the parallel lines (i) 15x + 8y − 34 = 0 and 15x + 8y + 31 = 0, (ii) l(x + y) + p = 0 and l(x + y) − r = 0.

9.3.6) Find the equation of the line parallel to the line 3x − 4y + 2 = 0 and passing through the point (−2, 3).

9.3.7) Find the equation of the line perpendicular to the line x − 7y + 5 = 0 and having x-intercept 3.

9.3.8) Find the angles between the lines √3x + y = 1 and x + √3y = 1.

9.3.9) The line through the points (h, 3) and (4, 1) intersects the line 7x − 9y − 19 = 0 at right angle. Find the value of h.

9.3.10) Prove that the line through the point (x₁, y₁) and parallel to the line Ax + By + C = 0 is A(x − x₁) + B(y − y₁) = 0.

9.3.11) Two lines passing through the point (2, 3) intersect each other at an angle of 60°. If the slope of one line is 2, find the equation of the other line.

9.3.12) Find the equation of the right bisector of the line segment joining the points (3, 4) and (−1, 2).

9.3.13) Find the coordinates of the foot of the perpendicular from the point (−1, 3) to the line 3x − 4y − 16 = 0.

9.3.14) The perpendicular from the origin to the line y = mx + c meets it at the point (−1, 2). Find the values of m and c.

9.3.15) If p and q are the lengths of perpendiculars from the origin to the lines x cos θ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k respectively, prove that p² + 4q² = k².

9.3.16) In the triangle ABC with vertices A(2, 3), B(4, −1) and C(1, 2), find the equation and length of the altitude from the vertex A.

9.3.17) If p is the length of the perpendicular from the origin to the line whose intercepts on the axes are a and b, show that 1/p² = 1/a² + 1/b².

9.M1) Find the values of k for which the line (k − 3)x − (4 − k²)y + k² − 7k + 6 = 0 is (a) parallel to the x-axis, (b) parallel to the y-axis, (c) passing through the origin.

9.M2) Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and −6, respectively.

9.M3) What are the points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units?

9.M4) Find the perpendicular distance from the origin to the line joining the points (cos θ, sin θ) and (cos φ, sin φ).

9.M5) Find the equation of the line parallel to the y-axis and drawn through the point of intersection of the lines x − 7y + 5 = 0 and 3x + y = 0.

9.M6) Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point where it meets the y-axis.

9.M7) Find the area of the triangle formed by the lines y − x = 0, x + y = 0 and x − k = 0.

9.M8) Find the value of p so that the three lines 3x + y − 2 = 0, px + 2y − 3 = 0 and 2x − y − 3 = 0 may intersect at one point.

9.M9) If three lines whose equations are y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ are concurrent, then show that m₁(c₂ − c₃) + m₂(c₃ − c₁) + m₃(c₁ − c₂) = 0.

9.M10) Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x − 2y = 3.

9.M11) Find the equation of the line passing through the point of intersection of the lines 4x + 7y − 3 = 0 and 2x − 3y + 1 = 0 that has equal intercepts on the axes.

9.M12) Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is y/x = (tan θ ± m)/(1 ∓ m tan θ).

9.M13) In what ratio, the line joining (−1, 1) and (5, 7) is divided by the line x + y = 4?

9.M14) Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x − y = 0.

9.M15) Find the direction in which a straight line must be drawn through the point (−1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.

9.M16) The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (−4, 1). Find the equations of the legs (perpendicular sides) of the triangle which are parallel to the axes.

9.M17) Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

9.M18) If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

9.M19) If the sum of the perpendicular distances of a variable point P(x, y) from the lines x + y − 5 = 0 and 3x − 2y + 7 = 0 is always 10, show that P must move on a line.

9.M20) Find the equation of the line which is equidistant from the parallel lines 9x + 6y − 7 = 0 and 3x + 2y + 6 = 0.

9.M21) A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

9.M22) Prove that the product of the lengths of the perpendiculars drawn from the points (√(a² − b²), 0) and (−√(a² − b²), 0) to the line x/a · cos θ + y/b · sin θ = 1 is b².

9.M23) A person standing at the junction (crossing) of two straight paths represented by the equations 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 wants to reach the path whose equation is 6x − 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.

Answers at a glance

1. 1
   9.1.1 — Shoelace: Σ xᵢyᵢ₊₁ = (−4)(7)+(0)(−5)+(5)(−2)+(−4)(5) = −58; Σ yᵢxᵢ₊₁ = (5)(0)+(7)(5)+(−5)(−4)+(−2)(−4) = 63. Area = ½|−58−63| = 121/2 = 60.5 sq units.
2. 2
   9.1.2 — Vertices are (0, a) and (0, −a) with apex at (±√3 a, 0) or (−√3 a, 0): the third vertex is (±√3a, 0).
3. 3
   9.1.3 — (i) PQ parallel to y-axis: distance = |y₂ − y₁|. (ii) Parallel to x-axis: distance = |x₂ − x₁|.
4. 4
   9.1.4 — Point on x-axis is (3, 0): distance to (7,6) = √(16+36), to (3,4) = 4; solve √((x−7)²+36) = √((x−3)²+16) → x = 3.
5. 5
   9.1.5 — Mid-point of P(0,−4) and B(8,0) is (4,−2); slope through origin = (−2−0)/(4−0) = −½.
6. 6
   9.1.6 — Slope AB = (5−4)/(3−4) = −1; slope BC = (−1−5)/(−1−3) = 3/2; slope AC = (−1−4)/(−1−4) = 1. Slope AB × slope AC = −1, so AB ⟂ AC.
7. 7
   9.1.7 — Angle with y-axis = 30°, so angle with x-axis = 60°; m = tan 60° = √3.
8. 8
   9.1.8 — Slope AB = (0+1)/(4+2) = 1/6; slope DC = (2−3)/(−3−3) = 1/6; slope BC = (3−0)/(3−4) = −3; slope AD = (2+1)/(−3+2) = −3. Opposite sides have equal slopes, so it is a parallelogram.
9. 9
   9.1.9 — Slope = (−2+1)/(4−3) = −1; angle with positive x-axis = 135°; acute angle with x-axis = 45°.
10. 10
    9.1.10 — Let the slopes be m and 2m. |(2m − m)/(1 + 2m²)| = 1/3 ⇒ 3m = ±(1 + 2m²). Plus: 2m² − 3m + 1 = 0 ⇒ m = 1/2 or 1. Minus: 2m² + 3m + 1 = 0 ⇒ m = −1/2 or −1. Slopes: (1/2, 1), (1, 2), (−1/2, −1) or (−1, −2).
11. 11
    9.1.11 — Slope of line through (x₁,y₁) and (h,k) is (k−y₁)/(h−x₁) = m ⇒ k − y₁ = m(h − x₁).
12. 12
    9.2.1 — x-axis: y = 0; y-axis: x = 0.
13. 13
    9.2.2 — y − 3 = ½(x + 4) ⇒ x − 2y + 10 = 0.
14. 14
    9.2.3 — y = mx.
15. 15
    9.2.4 — tan 75° = 2 + √3; y − 2√3 = (2+√3)(x − 2).
16. 16
    9.2.5 — y = −2(x + 3) ⇒ 2x + y + 6 = 0.
17. 17
    9.2.6 — tan 30° = 1/√3; y = (1/√3)x + 2 ⇒ x − √3y + 2√3 = 0.
18. 18
    9.2.7 — Slope = (−4−1)/(2+1) = −5/3; y − 1 = (−5/3)(x + 1) ⇒ 5x + 3y + 2 = 0.
19. 19
    9.2.8 — Median from R(4,5) meets PQ at its mid-point M((2−2)/2, (1+3)/2) = (0, 2); slope RM = (5−2)/(4−0) = 3/4; equation: 3x − 4y + 8 = 0.
20. 20
    9.2.9 — Slope of line through (2,5),(−3,6) = −1/5; perpendicular slope = 5; y − 5 = 5(x + 3) ⇒ 5x − y + 20 = 0.
21. 21
    9.2.10 — Slope of segment = 3; perpendicular slope = −1/3. Section formula (AP:PB = 1:n for A(1,0), B(2,3)): P = ((n·1+1·2)/(n+1), (n·0+1·3)/(n+1)) = ((n+2)/(n+1), 3/(n+1)). Line through P with slope −1/3: (n+1)x + 3(n+1)y = n + 11.
22. 22
    9.2.11 — Equal intercepts: x/a + y/a = 1 ⇒ x + y = a; through (2,3): a = 5 ⇒ x + y − 5 = 0.
23. 23
    9.2.12 — Let intercepts a, b with a + b = 9 and a·b: 2/a + 2/b = 1 ⇒ ab = 18. Factors of 18 with sum 9: a, b = 3, 6. Lines: x/3 + y/6 = 1 and x/6 + y/3 = 1.
24. 24
    9.2.13 — tan(2π/3) = −√3; y − 2 = −√3(x − 0) ⇒ √3x + y − 2 = 0. Parallel line crossing y-axis 2 units below origin: √3x + y + 2 = 0.
25. 25
    9.2.14 — Perpendicular from origin meets line at (−2, 9); slope of perpendicular = 9/(−2) = −9/2, so slope of line = 2/9; line through (−2,9): y − 9 = (2/9)(x + 2) ⇒ 2x − 9y + 85 = 0.
26. 26
    9.2.15 — Linear function L = mC + c. m = (125.134 − 124.942)/(110 − 20) = 0.192/90 = 8/375. Using C = 20: c = 124.942 − (8/375)(20) ≈ 124.515. So L = (8/375)C + 124.515 (cm).
27. 27
    9.2.16 — Linear demand: sale = 980 at ₹14, 1220 at ₹16. Slope = 240/2 = 120 litres per ₹; at ₹17 (₹3 above 14): 980 + 3×120 = 1340 litres/week.
28. 28
    9.2.17 — P(a, b) is mid-point between (2a, 0) and (0, 2b); line through them: x/a + y/b = 2.
29. 29
    9.2.18 — Let the segment join (α, 0) and (0, β). R(h, k) divides it in 1 : 2, so h = (1·0 + 2α)/3 = 2α/3 and k = (1β + 2·0)/3 = β/3, giving α = 3h/2, β = 3k. Intercept form: x/(3h/2) + y/(3k) = 1 ⇒ 2kx + hy = 3hk.
30. 30
    9.2.19 — Slope of (3,0):(−2,−2) = (−2−0)/(−2−3) = 2/5; slope of (−2,−2):(8,2) = (2+2)/(8+2) = 4/10 = 2/5; equal slopes with a common point ⇒ collinear.
31. 31
    9.3.1 — (i) x + 7y = 0 ⇒ y = −x/7: slope −1/7, y-intercept 0. (ii) 6x + 3y − 5 = 0 ⇒ y = −2x + 5/3: slope −2, y-intercept 5/3. (iii) y = 0: slope 0, y-intercept 0.
32. 32
    9.3.2 — (i) 3x + 2y = 12 ⇒ x/4 + y/6 = 1: intercepts 4, 6. (ii) 4x − 3y = 6 ⇒ x/(3/2) + y/(−2) = 1: intercepts 3/2, −2. (iii) 3y + 2 = 0 ⇒ y = −2/3, line parallel to x-axis.
33. 33
    9.3.3 — 12(x+6) = 5(y−2) ⇒ 12x − 5y + 82 = 0; distance = |12(−1) − 5(1) + 82|/√(144+25) = |−12−5+82|/13 = 65/13 = 5 units.
34. 34
    9.3.4 — x/3 + y/4 = 1 ⇒ 4x + 3y − 12 = 0. Point (x, 0): distance = |4x − 12|/5 = 4 ⇒ x = 8 or x = −2. Points: (8, 0) and (−2, 0).
35. 35
    9.3.5 — (i) d = |31 − (−34)|/√(15²+8²) = 65/17 units. (ii) d = |p + r|/(|l|√2).
36. 36
    9.3.6 — Parallel line: 3x − 4y + c = 0; through (−2, 3): −6 − 12 + c = 0 ⇒ c = 18 ⇒ 3x − 4y + 18 = 0.
37. 37
    9.3.7 — Perpendicular to x − 7y + 5 = 0 ⇒ 7x + y + k = 0; x-intercept 3 ⇒ at (3,0): 21 + k = 0 ⇒ k = −21 ⇒ 7x + y − 21 = 0.
38. 38
    9.3.8 — Slopes −√3 and −1/√3: tan θ = |(−1/√3 + √3)/(1+1)| = |(2/√3)/2| = 1/√3 ⇒ θ = 30°.
39. 39
    9.3.9 — Slope of line through (h,3),(4,1) = (1−3)/(4−h) = −2/(4−h). Perpendicular to 7x − 9y − 19 = 0 whose slope is 7/9 ⇒ (−2/(4−h))·(7/9) = −1 ⇒ 14 = 9(4−h) ⇒ h = 22/9.
40. 40
    9.3.10 — Required line has slope −A/B; line through (x₁,y₁): y − y₁ = (−A/B)(x − x₁) ⇒ A(x − x₁) + B(y − y₁) = 0.
41. 41
    9.3.11 — Other slope m: |(m−2)/(1+2m)| = tan 60° = √3 ⇒ m−2 = ±√3(1+2m). Case +: m = (2+√3)/(1−2√3); case −: m = (2−√3)/(1+2√3).
42. 42
    9.3.12 — Mid-point = (1, 3); slope of segment = (2−4)/(−1−3) = ½; perpendicular slope = −2 ⇒ y − 3 = −2(x − 1) ⇒ 2x + y − 5 = 0.
43. 43
    9.3.13 — Foot formula: foot = (x₁ − aL/(a²+b²), y₁ − bL/(a²+b²)) where L = ax₁ + by₁ + c. For (−1, 3) on 3x − 4y − 16 = 0: L = 3(−1) − 4(3) − 16 = −31, a²+b² = 25 ⇒ foot = (−1 − 3(−31)/25, 3 + 4(−31)/25) = (−1 + 93/25, 3 − 124/25) = (68/25, −49/25).
44. 44
    9.3.14 — Line through origin perpendicular to y = mx + c and meeting it at (−1, 2) ⇒ slope of perpendicular = −2 ⇒ m = ½; and (−1,2) lies on y = ½x + c ⇒ c = 5/2. So m = 1/2, c = 5/2.
45. 45
    9.3.15 — p = |k cos 2θ| (distance from origin to x cosθ − y sinθ = k cos2θ since √(cos²θ+sin²θ)=1) and q = |k|/√(sec²θ+cosec²θ) = |k| sinθcosθ ⇒ 4q² = 4k²sin²θcos²θ = k²sin²2θ; p² + 4q² = k²(cos²2θ + sin²2θ) = k².
46. 46
    9.3.16 — Altitude from A is perpendicular to BC. Slope BC = (2+1)/(1−4) = −1, so the altitude has slope 1; through A(2,3): y − 3 = x − 2, i.e. x − y + 1 = 0. Foot is where it meets BC: B(4,−1),C(1,2) ⇒ x + y = 3. Solve x − y = −1 and x + y = 3 ⇒ foot = (1, 2) = C. Length of the altitude = distance from A(2,3) to line x + y − 3 = 0 = |2 + 3 − 3|/√2 = 2/√2 = √2 units.
47. 47
    9.3.17 — Line x/a + y/b = 1 ⇒ bx + ay − ab = 0; p = |−ab|/√(a²+b²) ⇒ 1/p² = (a²+b²)/(a²b²) = 1/a² + 1/b².
48. 48
    9.M1 — (a) Parallel to x-axis: y coefficient zero ⇒ 4 − k² = 0 ⇒ k = ±2 (check k=3 gives 0x, invalid also). (b) Parallel to y-axis: x coefficient zero ⇒ k − 3 = 0 ⇒ k = 3. (c) Passing through origin: constant term zero ⇒ k² − 7k + 6 = 0 ⇒ k = 1 or 6.
49. 49
    9.M2 — Intercepts a, b with a + b = 1, ab = −6 ⇒ a,b = 3, −2 (roots of t² − t − 6 = 0). Lines: x/3 + y/(−2) = 1 and x/(−2) + y/3 = 1.
50. 50
    9.M3 — Line x/3 + y/4 = 1 ⇒ 4x + 3y − 12 = 0. Point (0, y): |3y − 12|/5 = 4 ⇒ 3y − 12 = ±20 ⇒ y = 32/3 or y = −8/3. Points (0, 32/3), (0, −8/3).
51. 51
    9.M4 — Distance from origin to line through P(cosθ, sinθ) and Q(cosφ, sinφ) = |x₁y₂ − y₁x₂| / √((x₂−x₁)² + (y₂−y₁)²) = |cosθ sinφ − sinθ cosφ| / √((cosφ−cosθ)² + (sinφ−sinθ)²) = |sin(φ−θ)| / √(2 − 2cos(φ−θ)) = |sin(φ−θ)| / (2|sin((φ−θ)/2)|) = |cos((θ−φ)/2)|.
52. 52
    9.M5 — Intersection of x − 7y + 5 = 0 and 3x + y = 0: y = −3x; x − 7(−3x) + 5 = 0 ⇒ 22x + 5 = 0 ⇒ x = −5/22, y = 15/22. A line parallel to the y-axis through it: x = −5/22 ⇒ 22x + 5 = 0.
53. 53
    9.M6 — x/4 + y/6 = 1 meets y-axis at (0, 6). Perpendicular slope: line slope = −6/4 = −3/2, perp slope = 2/3. Equation: y − 6 = (2/3)x ⇒ 2x − 3y + 18 = 0.
54. 54
    9.M7 — Lines y = x, y = −x, x = k form a right triangle with base on x = k from (k,k) to (k,−k) (length 2k) and altitude = distance of origin from x = k ⇒ k. Area = ½·(2k)·k = k².
55. 55
    9.M8 — Enter intersection of 3x+y−2=0 and 2x−y−3=0: solve: y = 2−3x; 2x − (2−3x) − 3 = 0 ⇒ 5x − 5 = 0 ⇒ x = 1, y = −1. Third line through (1,−1): px + 2y − 3 = 0 ⇒ p − 2 − 3 = 0 ⇒ p = 5.
56. 56
    9.M9 — Concurrent means the point of intersection of any two satisfies the third; equating determinants of the coefficients gives m₁(c₂−c₃) + m₂(c₃−c₁) + m₃(c₁−c₂) = 0.
57. 57
    9.M10 — Line x − 2y = 3 has slope ½. Lines making 45°: slopes m = (½ ± 1)/(1 ∓ ½) = (3/2)/(1/2) = 3, and (−1/2)/(3/2) = −1/3. Through (3,2): y − 2 = 3(x−3) and y − 2 = (−1/3)(x−3).
58. 58
    9.M11 — Family through intersection: (4x + 7y − 3) + k(2x − 3y + 1) = 0, slope = −(4 + 2k)/(7 − 3k). Equal intercepts on the axes ⇒ slope −1: (4 + 2k)/(7 − 3k) = 1 ⇒ k = 3/5. Line: (26/5)x + (26/5)y − 12/5 = 0 ⇒ 13x + 13y − 6 = 0 (intercepts 6/13, 6/13).
59. 59
    9.M12 — Slope of line through origin = y/x = m'. Angle with y = mx + c satisfies tan θ = |(m' − m)/(1 + m·m')| ⇒ (m' − m)/(1 + mm') = ± tan θ ⇒ m' = (m ± tan θ)/(1 ∓ m tan θ) ⇒ y/x = (tan θ ± m)/(1 ∓ m tan θ).
60. 60
    9.M13 — Point of division on x + y = 4: parametrise line (−1,1)→(5,7): (−1+6t, 1+6t); x+y = 4 ⇒ (−1+6t)+(1+6t) = 12t = 4 ⇒ t = 1/3. Ratio = t:(1−t) = 1/3:2/3 = 1:2.
61. 61
    9.M14 — Distance along 2x − y = 0: point (1,2) to intersection of 4x+7y+5=0 with y=2x: 4x+14x+5=0 ⇒ x=−5/18, y=−5/9. Distance = √((1+5/18)² + (2+5/9)²) = √((23/18)²+(23/9)²) = (23/18)√5.
62. 62
    9.M15 — Required line through (−1,2) meets x+y=4 at distance 3 from (−1,2). Point = (−1+3cosθ, 2+3sinθ) lies on x+y=4 ⇒ 1 + 3(cosθ+sinθ) = 4 ⇒ cosθ+sinθ = 1 ⇒ sin2θ = 0 ⇒ θ = 0 or 90° ⇒ directions: 0° (parallel to x-axis → y = 2) or 90° (x = −1).
63. 63
    9.M16 — Legs parallel to the axes: vertical through (1, 3): x = 1; horizontal through (−4, 1): y = 1. They meet at (1, 1), forming the right angle. Legs: x = 1 and y = 1.
64. 64
    9.M17 — Image of (3,8) in x + 3y = 7: foot H solves; image P' = 2H − P. Standard formula: image = ( (x₁ − 2a(ax₁+by₁+c)/(a²+b²)), (y₁ − 2b(ax₁+by₁+c)/(a²+b²)) ) ⇒ a=1,b=3,c=−7; val = 3+24−7 = 20; a²+b²=10 ⇒ P' = (3 − 2·20/10, 8 − 2·3·20/10) = (3−4, 8−12) = (−1, −4).
65. 65
    9.M18 — Slopes: line1 = 3, line2 = ½. Equally inclined to y = mx + 4 (slope m): |(m−3)/(1+3m)| = |(m−½)/(1+½m)|. Squaring and solving: (7/2)m² − m − 7/2 = 0 ⇒ 7m² − 2m − 7 = 0 ⇒ m = (2 ± √(4+196))/14 = (1 ± 5√2)/7.
66. 66
    9.M19 — Sum of perpendicular distances = |x+y−5|/√2 + |3x−2y+7|/√13 = 10. For P restricted to one region the absolute values open with fixed signs to give a linear equation in x, y — i.e., P moves on a straight line.
67. 67
    9.M20 — Equidistant line is the mid-parallel: rewrite both in same coefficients 3x + 2y: 9x+6y−7=0 ⇒ 3x+2y−7/3=0; 3x+2y+6=0. Mid value = (−7/3 + 6)/2 = (11/3)/2 = 11/6 ⇒ 3x + 2y + 11/6 = 0 ⇒ 18x + 12y + 11 = 0.
68. 68
    9.M21 — Reflection point A = (13/5, 0): image of P(1,2) in x-axis is (1,−2); line through (1,−2) and (5,3) cuts x-axis at A. Slope = 5/4; y+2 = (5/4)(x−1); y=0 ⇒ 2 = (5/4)(x−1) ⇒ x = 13/5. A = (13/5, 0).
69. 69
    9.M22 — Line: x/a cosθ + y/b sinθ = 1 ⇒ (cosθ/a)x + (sinθ/b)y − 1 = 0, with √(cos²θ/a² + sin²θ/b²) in each denominator. p₁p₂ = |((√(a²−b²)/a)cosθ − 1)(−(√(a²−b²)/a)cosθ − 1)| / (cos²θ/a² + sin²θ/b²) = |1 − ((a²−b²)/a²)cos²θ| / (cos²θ/a² + sin²θ/b²) = [(a²sin²θ + b²cos²θ)/a²]·[a²b²/(a²sin²θ + b²cos²θ)] = b².
70. 70
    9.M23 — Least-distance path is the line through the junction perpendicular to 6x − 7y + 8 = 0. Junction of 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0: from first x = (3y − 4)/2; into second: 3(3y−4)/2 + 4y − 5 = 0 ⇒ 9y − 12 + 8y − 10 = 0 ⇒ 17y = 22 ⇒ y = 22/17, x = (66/17 − 68/17)/2 = −1/17. Perpendicular to 6x − 7y + 8 = 0 has the form 7x + 6y + c = 0; through (−1/17, 22/17): −7/17 + 132/17 + c = 0 ⇒ c = −125/17. Line: 119x + 102y − 125 = 0.

MCQ Questions

1 Mark Questions

1. 1
   Write the equation of a line passing through the origin and having slope m.
2. 2
   State the condition for which the lines a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0 are perpendicular.
3. 3
   What is the area of the triangle with vertices (1,0), (0,1), (0,0)?
4. 4
   Find the slope of the line parallel to the x-axis.

2 Mark Questions

1. 1
   Find the equation of the line through (−1, 1) and (2, −4).
2. 2
   Find the value of h if the line through (h, 3) and (4, 1) is perpendicular to 7x − 9y − 19 = 0.
3. 3
   Reduce 2x − 3y + 6 = 0 to intercept form and state the intercepts.
4. 4
   Find the distance between 3x − 4y + 7 = 0 and 6x − 8y + 15 = 0.

3 Mark Questions

1. 1
   Find the equation of the right bisector of the segment joining (3, 4) and (−1, 2).
2. 2
   Two lines through (2, 3) make 60° with each other; one has slope 2. Find the equation of the other.
3. 3
   Find the foot of the perpendicular from (−1, 3) to 3x − 4y − 16 = 0.

5 Mark Questions

1. 1
   (a) Derive the normal form x cos ω + y sin ω = p of a line. (b) Find the normal form of the line √3x + y − 4 = 0 and hence its distance from the origin.
2. 2
   (a) Prove that the three lines y = m₁x+c₁, y = m₂x+c₂, y = m₃x+c₃ are concurrent iff m₁(c₂−c₃)+m₂(c₃−c₁)+m₃(c₁−c₂) = 0. (b) Find the area of the triangle formed by the lines y = 2x+3, y = 3x−1 and x = 0 using the concurrency of its sides' family.