CBSE Class 11 Mathematics Straight Lines Notes Set 02

Important Notes and Formulas

Definition

The straight line is a curve such that every point on the line segment joining any two points on it lies on it. The simplest locus of a point in a plane is a straight line. A line is determined uniquely by any one of the following:
(1) Two different points (because we know the axiom that one and only one straight line passes through two given points).
(2) A point and a given direction.

Thus, to determine a line uniquely, two geometrical conditions are required.

Slope (Gradient) of a Line

The trigonometrical tangent of the angle that a line makes with the positive direction of the x-axis in anticlockwise sense is called the slope or gradient of the line. The slope of a line is generally denoted by m. Thus, m = tan θ.

(1) Slope of a line parallel to x-axis is m = tan 0° = 0.
(2) Slope of a line parallel to y-axis is m = tan 90° = ∞ (not defined).
(3) Slope of the line equally inclined with the axes is 1 or −1.
(4) Slope of the line through the points A(x₁, y₁) and B(x₂, y₂) is \( \frac{y_2 - y_1}{x_2 - x_1} \), taken in the same order.
(5) Slope of the line ax + by + c = 0, b ≠ 0 is \( -\frac{a}{b} \).
(6) Slopes of two parallel lines are equal.
(7) If m₁ and m₂ are the slopes of two perpendicular lines, then m₁ · m₂ = −1.
Note: m can be defined as tan θ for 0 ≤ θ < π and θ ≠ \( \frac{\pi}{2} \).
If three points A, B, C are collinear, then slope of AB = slope of BC = slope of AC.

Equations of a Straight Line in Different Forms

(1) Slope form: Equation of a line through the origin and having slope m is y = mx.
(2) One point form or point-slope form: Equation of a line through the point (x₁, y₁) and having slope m is y − y₁ = m(x − x₁).
(3) Slope-intercept form: Equation of a line (non-vertical) with slope m and cutting off an intercept c on the y-axis is y = mx + c.
The equation of a line with slope m and the x-intercept d is y = m(x − d).

(4) Intercept form: If a straight line cuts the x-axis at A and the y-axis at B, then OA and OB are known as the intercepts of the line on the x-axis and y-axis respectively. The intercepts are positive or negative according as the line meets the positive or negative directions of the coordinate axes. In the figure, OA = x-intercept, OB = y-intercept.
Equation of a straight line cutting off intercepts a and b on the x-axis and y-axis respectively is \( \frac{x}{a} + \frac{y}{b} = 1 \).
Note: If the given line is parallel to the x-axis, then its x-intercept is undefined. If the given line is parallel to the y-axis, then its y-intercept is undefined.

(5) Two point form: Equation of the line through the points A(x₁, y₁) and B(x₂, y₂) is \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \).
In the determinant form it is given as \( \begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{vmatrix} = 0 \).

Lines parallel to the axes:
Equation of the x-axis ⇒ y = 0. Equation of a line parallel to the x-axis (or perpendicular to the y-axis) at a distance b from it ⇒ y = b.
Equation of the y-axis ⇒ x = 0. Equation of a line parallel to the y-axis (or perpendicular to the x-axis) at a distance a from it ⇒ x = a.

Equation of Parallel and Perpendicular Lines to a Given Line

(1) Equation of a line which is parallel to ax + by + c = 0 is ax + by + λ = 0.
(2) Equation of a line which is perpendicular to ax + by + c = 0 is bx − ay + λ = 0.
The value of λ in both cases is obtained with the help of the additional information given in the problem.

Point of Intersection of Two Lines

Let a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 be two non-parallel lines. If (x′, y′) is their point of intersection, then a₁x′ + b₁y′ + c₁ = 0 and a₂x′ + b₂y′ + c₂ = 0.
Solving these equations, we get \( (x', y') = \left( \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}, \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \right) \).
Note: Since the lines are not parallel, they have unequal slopes, so a₁b₂ − a₂b₁ ≠ 0.
In numerical questions we need not remember the coordinates (x′, y′) given above; we solve the two equations directly.

Angle between Two Non-parallel Lines

Let θ be the angle between the lines y = m₁x + c₁ and y = m₂x + c₂, intersecting at A, where m₁ = tan α and m₂ = tan β.

Since α = θ + β ⇒ θ = α − β
⇒ \( \tan\theta = \left| \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta} \right| \)
∴ \( \theta = \tan^{-1}\left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \)

(1) Angle between two straight lines when their equations are given: The angle θ between the lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 is given by \( \tan\theta = \left| \frac{a_2b_1 - a_1b_2}{a_1a_2 + b_1b_2} \right| \).
(i) Condition for the lines to be parallel: \( m_1 = m_2 \Rightarrow -\frac{a_1}{b_1} = -\frac{a_2}{b_2} \Rightarrow \frac{a_1}{a_2} = \frac{b_1}{b_2} \).
(ii) Condition for the lines to be perpendicular: \( m_1 m_2 = -1 \Rightarrow \left(-\frac{a_1}{b_1}\right)\left(-\frac{a_2}{b_2}\right) = -1 \Rightarrow a_1a_2 + b_1b_2 = 0 \).
(iii) Conditions for two lines to be coincident, parallel, perpendicular and intersecting: Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are
(a) Coincident, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
(b) Parallel, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
(c) Intersecting, if \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
(d) Perpendicular, if a₁a₂ + b₁b₂ = 0

Equation of a Straight Line through a Given Point Making a Given Angle with a Given Line

Let the given line y = mx + c make an angle θ with the x-axis, so that m = tan θ. A straight line L through (x₁, y₁) making an angle α with the given line makes an angle (θ + α) with the x-axis, so its equation is y − y₁ = tan(θ + α)(x − x₁); the other such line L′ makes an angle (θ − α) with the x-axis, so its equation is y − y₁ = tan(θ − α)(x − x₁).
Hence the equations of the straight lines which pass through a given point (x₁, y₁) and make a given angle α with the given straight line y = mx + c are
\( y - y_1 = \frac{m \pm \tan\alpha}{1 \mp m\tan\alpha}(x - x_1) \)

A Line Equally Inclined with Two Lines

Let the two lines with slopes m₁ and m₂ be equally inclined to a line with slope m. Then
\( \frac{m_1 - m}{1 + m_1 m} = -\left( \frac{m_2 - m}{1 + m_2 m} \right) \)
Note: The sign of m in both brackets is the same.

Length of Perpendicular

(1) Distance of a point from a line: The length p of the perpendicular from the point (x₁, y₁) to the line ax + by + c = 0 is given by \( p = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \).
Note: Length of the perpendicular from the origin to the line ax + by + c = 0 is \( \frac{|c|}{\sqrt{a^2 + b^2}} \).

(2) Distance between two parallel lines: Let the two parallel lines be ax + by + c₁ = 0 and ax + by + c₂ = 0.

First method: The distance between the lines is \( d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \).
Second method: The distance between the lines is \( d = \frac{\lambda}{\sqrt{a^2 + b^2}} \), where
(i) λ = |c₁ − c₂| if they are on the same side of the origin;
(ii) λ = |c₁| + |c₂| if the origin O lies between them.
Third method: Find the coordinates of any point on one of the given lines, preferably by putting x = 0 or y = 0. Then the perpendicular distance of this point from the other line is the required distance between the lines.
Note: Distance between the two parallel lines ax + by + c₁ = 0 and kax + kby + c₂ = 0 is \( \frac{\left| c_1 - \frac{c_2}{k} \right|}{\sqrt{a^2 + b^2}} \).
Distance between two non-parallel lines is always zero.

Some Important Results

(1) Area of the triangle made by the line ax + by + c = 0 with the coordinate axes is \( \frac{c^2}{2|ab|} \).
(2) Area of the rhombus formed by the lines ax ± by ± c = 0 is \( \left| \frac{2c^2}{ab} \right| \).
(3) The foot of the perpendicular (h, k) from (x₁, y₁) to the line ax + by + c = 0 is given by \( \frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{(ax_1 + by_1 + c)}{a^2 + b^2} \).
Hence, the coordinates of the foot of the perpendicular are \( \left( \frac{b^2x_1 - aby_1 - ac}{a^2 + b^2}, \frac{a^2y_1 - abx_1 - bc}{a^2 + b^2} \right) \).
(4) The equation of a line whose mid-point between the axes is (x₁, y₁) is \( \frac{x}{x_1} + \frac{y}{y_1} = 2 \).
(5) The equation of a straight line which makes a triangle with the axes having centroid (x₁, y₁) is \( \frac{x}{3x_1} + \frac{y}{3y_1} = 1 \).

Solved Examples

Example 1. The gradient of the line joining the points on the curve y = x² + 2x, whose abscissae are 1 and 3, is [MP PET 1997]
(a) 6
(b) 5
(c) 4
(d) 3
Answer:
Ans. (a)
The points are (1, 3) and (3, 15).
Hence gradient = \( \frac{y_2 - y_1}{x_2 - x_1} = \frac{12}{2} = 6 \).

Example 2. Slope of a line which cuts intercepts of equal lengths on the axes is [MP PET 1986]
(a) −1
(b) 0
(c) 2
(d) \( \sqrt{3} \)
Answer:
Ans. (a)
Equation of the line is \( \frac{x}{a} + \frac{y}{a} = 1 \)
⇒ x + y = a ⇒ y = −x + a. Hence slope of the line is −1.

Example 3. Equation to the straight line cutting off an intercept 2 from the negative direction of the axis of y and inclined at 30° to the positive direction of x, is [MP PET 2003]
(a) \( y + x - \sqrt{3} = 0 \)
(b) y − x + 2 = 0
(c) \( y - \sqrt{3}x - 2 = 0 \)
(d) \( \sqrt{3}y - x + 2\sqrt{3} = 0 \)
Answer:
Ans. (d)
Let the equation of the straight line be y = mx + c.
Here \( m = \tan 30^\circ = \frac{1}{\sqrt{3}} \) and c = −2.
Hence, the required equation is \( y = \frac{1}{\sqrt{3}}x - 2 \Rightarrow \sqrt{3}y - x + 2\sqrt{3} = 0 \).

Example 4. The equation of a straight line passing through (−3, 2) and cutting an intercept equal in magnitude but opposite in sign from the axes is given by [Rajasthan PET 1984; MP PET 1993]
(a) x − y + 5 = 0
(b) x + y − 5 = 0
(c) x − y − 5 = 0
(d) x + y + 5 = 0
Answer:
Ans. (a)
Let the equation be \( \frac{x}{a} + \frac{y}{-a} = 1 \Rightarrow x - y = a \).
But it passes through (−3, 2), hence a = −3 − 2 = −5. Hence the equation of the straight line is x − y + 5 = 0.

Example 5. The equation of the straight line passing through the point (4, 3) and making intercepts on the coordinate axes whose sum is −1, is [AIEEE 2004]
(a) \( \frac{x}{2} - \frac{y}{3} = -1 \) and \( \frac{x}{2} + \frac{y}{1} = 1 \)
(b) \( \frac{x}{2} - \frac{y}{3} = -1 \) and \( \frac{x}{-2} + \frac{y}{1} = -1 \)
(c) \( \frac{x}{2} - \frac{y}{3} = 1 \) and \( \frac{x}{2} + \frac{y}{1} = 1 \)
(d) \( \frac{x}{2} - \frac{y}{3} = 1 \) and \( \frac{x}{-2} + \frac{y}{1} = 1 \)
Answer:
Ans. (d)
Let the equation of the line be \( \frac{x}{a} + \frac{y}{-1 - a} = 1 \), which passes through (4, 3).
Then \( \frac{4}{a} + \frac{3}{-1 - a} = 1 \Rightarrow a^2 = 4 \Rightarrow a = \pm 2 \).
Hence the equations are \( \frac{x}{2} - \frac{y}{3} = 1 \) and \( \frac{x}{-2} + \frac{y}{1} = 1 \).

Example 6. Let PS be the median of the triangle with vertices P(2, 2), Q(6, −1) and R(7, 3). The equation of the line passing through (1, −1) and parallel to PS is [IIT Screening 2000]
(a) 2x − 9y − 7 = 0
(b) 2x − 9y − 11 = 0
(c) 2x + 9y − 11 = 0
(d) 2x + 9y + 7 = 0
Answer:
Ans. (d)
S = mid-point of QR = \( \left( \frac{6 + 7}{2}, \frac{-1 + 3}{2} \right) = \left( \frac{13}{2}, 1 \right) \)
∴ Slope of PS = \( \frac{2 - 1}{2 - \frac{13}{2}} = -\frac{2}{9} \)
∴ The required equation is \( y + 1 = -\frac{2}{9}(x - 1) \Rightarrow 2x + 9y + 7 = 0 \).

Example 7. The equation of the line which passes through (a, b) and is parallel to the line \( \frac{x}{a} + \frac{y}{b} = 1 \), is [Rajasthan PET 1986, 1995]
(a) \( \frac{x}{a} + \frac{y}{b} = 3 \)
(b) \( \frac{x}{a} + \frac{y}{b} = 2 \)
(c) \( \frac{x}{a} + \frac{y}{b} = 0 \)
(d) \( \frac{x}{a} + \frac{y}{b} + 2 = 0 \)
Answer:
Ans. (b)
The equation of a line parallel to the given line is \( \frac{x}{a} + \frac{y}{b} = \lambda \).
This line passes through the point (a, b). ∴ \( \frac{a}{a} + \frac{b}{b} = \lambda \Rightarrow \lambda = 2 \).
Hence, the required line is \( \frac{x}{a} + \frac{y}{b} = 2 \).

Example 8. A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is [IIT 1992]
(a) \( \frac{1}{3} \)
(b) \( \frac{2}{3} \)
(c) 1
(d) \( \frac{4}{3} \)
Answer:
Ans. (d)
The equation of a line passing through (2, 2) and perpendicular to 3x + y = 3 is \( y - 2 = \frac{1}{3}(x - 2) \) or x − 3y + 4 = 0.
Putting x = 0 in this equation, we obtain \( y = \frac{4}{3} \). So y-intercept = \( \frac{4}{3} \).

Example 9. The equation of the line bisecting perpendicularly the segment joining the points (−4, 6) and (8, 8) is [Karnataka CET 2003]
(a) 6x + y − 19 = 0
(b) y = 7
(c) 6x + 2y − 19 = 0
(d) x + 2y − 7 = 0
Answer:
Ans. (a)
Equation of the line passing through (−4, 6) and (8, 8) is \( y - 6 = \frac{8 - 6}{8 + 4}(x + 4) \Rightarrow y - 6 = \frac{2}{12}(x + 4) \Rightarrow 6y - x = 40 \) .....(i)
Now the equation of any line ⊥ to it is 6x + y + λ = 0 .....(ii)
This line passes through the mid-point of (−4, 6) and (8, 8), i.e., (2, 7).
∴ From (ii) 12 + 7 + λ = 0 ⇒ λ = −19. ∴ Equation of the line is 6x + y − 19 = 0.

Example 10. Equation of a line passing through the point of intersection of the lines 2x − 3y + 4 = 0, 3x + 4y − 5 = 0 and perpendicular to 6x − 7y + 3 = 0, then its equation is [Rajasthan PET 2000]
(a) 119x + 102y + 125 = 0
(b) 119x + 102y = 125
(c) 119x − 102y = 125
(d) None of these
Answer:
Ans. (b)
The point of intersection of the lines 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 is \( \left( \frac{-1}{17}, \frac{22}{17} \right) \).
The slope of the given line 6x − 7y + 3 = 0 is \( \frac{6}{7} \), so the slope of the required line = \( \frac{-7}{6} \).
Hence, the equation of the required line is \( y - \frac{22}{17} = \frac{-7}{6}\left( x + \frac{1}{17} \right) \Rightarrow 119x + 102y = 125 \).

Example 11. The equation of the straight line passing through the point of intersection of the straight lines 3x − y + 2 = 0 and 5x − 2y + 7 = 0 and having infinite slope is [UPSEAT 2001]
(a) x = 2
(b) x + y = 3
(c) x = 3
(d) x = 4
Answer:
Ans. (c)
From 3x − y + 2 = 0, y = 3x + 2. Substituting in 5x − 2y + 7 = 0: 5x − 6x − 4 + 7 = 0 ⇒ x = 3, y = 11.
So the point of intersection is (3, 11).
A line having infinite slope is parallel to the y-axis, i.e., of the form x = constant. Through (3, 11) it is x = 3.

Example 12. Angle between the lines 2x − y − 15 = 0 and 3x + y + 4 = 0 is [Rajasthan PET 2003]
(a) 90°
(b) 45°
(c) 180°
(d) 60°
Answer:
Ans. (b)
\( \tan\theta = \left| \frac{a_2b_1 - a_1b_2}{a_1a_2 + b_1b_2} \right| = \left| \frac{(3)(-1) - (2)(1)}{(3)(2) + (-1)(1)} \right| \Rightarrow \tan\theta = \left| \frac{-3 - 2}{6 - 1} \right| = \left| \frac{-5}{5} \right| = |-1| \)
\( \theta = \tan^{-1}|-1| = \tan^{-1}1 = 45^\circ \).

Example 13. To which of the following types do the straight lines represented by 2x + 3y − 7 = 0 and 2x + 3y − 5 = 0 belong [MP PET 1982]
(a) Parallel to each other
(b) Perpendicular to each other
(c) Inclined at 45° to each other
(d) Coincident pair of straight lines
Answer:
Ans. (a)
Here, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \); \( \frac{2}{2} = \frac{3}{3} \neq \frac{7}{5} \). Hence, the lines are parallel to each other.

Example 14. The equations of the lines which pass through the point (3, −2) and are inclined at 60° to the line \( \sqrt{3}x + y = 1 \) [IIT 1974; MP PET 1996]
(a) y + 2 = 0, \( \sqrt{3}x - y - 2 - 3\sqrt{3} = 0 \)
(b) x − 2 = 0, \( \sqrt{3}x - y + 2 + 3\sqrt{3} = 0 \)
(c) \( \sqrt{3}x - y - 2 - 3\sqrt{3} = 0 \)
(d) None of these
Answer:
Ans. (a)
The equation of the lines passing through (3, −2) is y + 2 = m(x − 3) .....(i)
The slope of the given line is \( -\sqrt{3} \). So, \( \tan 60^\circ = \pm \frac{m - (-\sqrt{3})}{1 + m(-\sqrt{3})} \). On solving, we get m = 0 or \( \sqrt{3} \).
Putting the values of m in (i), the required equations are y + 2 = 0 and \( \sqrt{3}x - y - 2 - 3\sqrt{3} = 0 \).

Example 15. In an isosceles triangle ABC, the coordinates of the points B and C on the base BC are respectively (1, 2) and (2, 1). If the equation of the line AB is y = 2x, then the equation of the line AC is [Roorkee 2000]
(a) \( y = \frac{1}{2}(x - 1) \)
(b) \( y = \frac{x}{2} \)
(c) y = x − 1
(d) 2y = x + 3

Answer:
Ans. (b)
Slope of BC = \( \frac{1 - 2}{2 - 1} = -1 \). Since AB = AC, ∠ABC = ∠ACB.
⇒ \( \left| \frac{2 + 1}{1 + 2(-1)} \right| = \left| \frac{m + 1}{1 + m(-1)} \right| \Rightarrow \left| \frac{m + 1}{1 - m} \right| = |-3| \Rightarrow \frac{m + 1}{1 - m} = \pm 3 \Rightarrow m = 2, \frac{1}{2} \)
But the slope of AB is 2; ∴ \( m = \frac{1}{2} \) (here m is the gradient of the line AC).
Equation of the line AC is \( y - 1 = \frac{1}{2}(x - 2) \Rightarrow x - 2y = 0 \) or \( y = \frac{x}{2} \).

Example 16. If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, then m = [ISM Dhanbad 1976]
(a) \( \frac{1 + 3\sqrt{2}}{7} \)
(b) \( \frac{1 - 3\sqrt{2}}{7} \)
(c) \( \frac{1 \pm 3\sqrt{2}}{7} \)
(d) \( \frac{1 \pm 5\sqrt{2}}{7} \)
Answer:
Ans. (d)
The line y = mx + 4 is equally inclined to the lines with slopes m₁ = 3 and \( m_2 = \frac{1}{2} \).
Then \( \frac{3 - m}{1 + 3m} = -\left( \frac{\frac{1}{2} - m}{1 + \frac{1}{2}m} \right) \Rightarrow 7m^2 - 2m - 7 = 0 \Rightarrow m = \frac{1 \pm 5\sqrt{2}}{7} \).

Example 17. The distance of the point (−2, 3) from the line x − y = 5 is [MP PET 2001]
(a) \( 5\sqrt{2} \)
(b) \( 2\sqrt{5} \)
(c) \( 3\sqrt{5} \)
(d) \( 5\sqrt{3} \)
Answer:
Ans. (a)
\( p = \left| \frac{x_1 - y_1 - 5}{\sqrt{1^2 + 1^2}} \right| = \left| \frac{-2 - 3 - 5}{\sqrt{2}} \right| = \left| \frac{-10}{\sqrt{2}} \right| = 5\sqrt{2} \)

Example 18. The distance between the lines 4x + 3y = 11 and 8x + 6y = 15 is [AMU 1979; MNR 1987; UPSEAT 2000; DCE 1999]
(a) \( \frac{7}{2} \)
(b) 4
(c) \( \frac{7}{10} \)
(d) None of these
Answer:
Ans. (c)
Given lines 4x + 3y = 11 and 8x + 6y = 15; distances from the origin to the two lines are \( \left| \frac{-11}{\sqrt{25}} \right| \) and \( \left| \frac{-15}{\sqrt{100}} \right| \Rightarrow \frac{11}{5}, \frac{15}{10} \).
Clearly both lines are on the same side of the origin.
Hence, the distance between the lines = \( \frac{11}{5} - \frac{15}{10} = \frac{7}{10} \).

Example 19. If the length of the perpendicular drawn from the origin to the line whose intercepts on the axes are a and b be p, then [Karnataka CET 2003]
(a) a² + b² = p²
(b) \( a^2 + b^2 = \frac{1}{p^2} \)
(c) \( \frac{1}{a^2} + \frac{1}{b^2} = \frac{2}{p^2} \)
(d) \( \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} \)
Answer:
Ans. (d)
Equation of the line is \( \frac{x}{a} + \frac{y}{b} = 1 \Rightarrow bx + ay - ab = 0 \).
Perpendicular distance from the origin to the line is \( p = \left| \frac{-ab}{\sqrt{a^2 + b^2}} \right| \Rightarrow \frac{\sqrt{a^2 + b^2}}{ab} = \frac{1}{p} \Rightarrow \frac{a^2 + b^2}{a^2b^2} = \frac{1}{p^2} \Rightarrow \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} \).

Example 20. The point on the x-axis whose perpendicular distance from the line \( \frac{x}{a} + \frac{y}{b} = 1 \) is a, is [Rajasthan PET 2001; MP PET 2003]
(a) \( \left[ \frac{a}{b}\left(b \pm \sqrt{a^2 + b^2}\right), 0 \right] \)
(b) \( \left[ \frac{b}{a}\left(b \pm \sqrt{a^2 + b^2}\right), 0 \right] \)
(c) \( \left[ \frac{a}{b}\left(a \pm \sqrt{a^2 + b^2}\right), 0 \right] \)
(d) None of these
Answer:
Ans. (a)
Let the point be (h, 0); then \( a = \pm \frac{bh + 0 - ab}{\sqrt{a^2 + b^2}} \Rightarrow bh = \pm a\sqrt{a^2 + b^2} + ab \Rightarrow h = \frac{a}{b}\left(b \pm \sqrt{a^2 + b^2}\right) \).
Hence the point is \( \left[ \frac{a}{b}\left(b \pm \sqrt{a^2 + b^2}\right), 0 \right] \).

Example 21. The vertex of an equilateral triangle is (2, −1) and the equation of its base is x + 2y = 1. The length of its sides is [UPSEAT 2003]
(a) \( \frac{4}{\sqrt{15}} \)
(b) \( \frac{2}{\sqrt{15}} \)
(c) \( \frac{4}{3\sqrt{3}} \)
(d) None of these

Answer:
Ans. (b)
\( |AD| = \left| \frac{2 - 2 - 1}{\sqrt{1^2 + 2^2}} \right| = \frac{1}{\sqrt{5}} \)
∵ \( \tan 60^\circ = \frac{AD}{BD} \Rightarrow \sqrt{3} = \frac{1/\sqrt{5}}{BD} \Rightarrow BD = \frac{1}{\sqrt{15}} \Rightarrow BC = 2BD = \frac{2}{\sqrt{15}} \).

Example 22. The coordinates of the foot of the perpendicular drawn from (2, 4) to the line x + y = 1 is [Roorkee 1995]
(a) \( \left( \frac{1}{3}, \frac{3}{2} \right) \)
(b) \( \left( -\frac{1}{2}, \frac{3}{2} \right) \)
(c) \( \left( \frac{4}{3}, \frac{1}{2} \right) \)
(d) \( \left( \frac{3}{4}, -\frac{1}{2} \right) \)
Answer:
Ans. (b)
Applying the formula for the foot of the perpendicular, the required coordinates are \( \left( \frac{1^2 \times 2 - 1 \times 1 \times 4 + 1}{1^2 + 1^2}, \frac{1^2 \times 4 - 1 \times 1 \times 2 + 1}{1^2 + 1^2} \right) = \left( \frac{-1}{2}, \frac{3}{2} \right) \).

Example 23. The area enclosed within the curve |x| + |y| = 1 is [Rajasthan PET 1990, 97; IIT 1981; UPSEAT 2003]
(a) \( \sqrt{2} \)
(b) 1
(c) \( \sqrt{3} \)
(d) 2
Answer:
Ans. (d)
The given lines are ±x ± y = 1, i.e., x + y = 1, x − y = 1, x + y = −1 and x − y = −1. These lines form a quadrilateral whose vertices are A(−1, 0), B(0, −1), C(1, 0) and D(0, 1).
Obviously ABCD is a square. Length of each side of this square is \( \sqrt{1^2 + 1^2} = \sqrt{2} \).
Hence, area of the square is \( \sqrt{2} \times \sqrt{2} = 2 \) sq. units.

Example 24. If x₁, x₂, x₃ and y₁, y₂, y₃ are both in G.P. with the same common ratio, then the points (x₁, y₁), (x₂, y₂) and (x₃, y₃) [AIEEE 2003]
(a) Lie on a straight line
(b) Lie on an ellipse
(c) Lie on a circle
(d) Are vertices of a triangle
Answer:
Ans. (a)
Taking the coordinates as \( \left( \frac{x}{r}, \frac{y}{r} \right) \), (x, y) and (xr, yr). The above coordinates satisfy the relation y = mx, where \( m = \frac{y}{x} \).
∴ The three points lie on a straight line.

Example 25. A square of side a lies above the x-axis and has one vertex at the origin. The side passing through the origin makes an angle \( \alpha \left( 0 \lt \alpha \lt \frac{\pi}{4} \right) \) with the positive direction of the x-axis. The equation of its diagonal not passing through the origin is [AIEEE 2003]
(a) y(cos α − sin α) − x(sin α − cos α) = a
(b) y(cos α + sin α) − x(sin α − cos α) = a
(c) y(cos α + sin α) + x(sin α + cos α) = a
(d) y(cos α + sin α) + x(sin α − cos α) = a

Answer:
Ans. (b)
Coordinates of A = (a cos α, a sin α); equation of OB is \( y = \tan\left( \frac{\pi}{4} + \alpha \right)x \).
∵ CA ⊥ OB; ∴ slope of CA = \( -\cot\left( \frac{\pi}{4} + \alpha \right) \).
Equation of CA: \( y - a\sin\alpha = -\cot\left( \frac{\pi}{4} + \alpha \right)(x - a\cos\alpha) \)
⇒ y(sin α + cos α) + x(cos α − sin α) = a, which is option (b).