CBSE Class 11 Mathematics Straight Lines Notes Set 01

Important Notes and Formulas

Distance Formula

The distance between the points A(x₁, y₁) and B(x₂, y₂) is \( \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \).

Section Formula

The point P(x, y) which divides the line joining A(x₁, y₁) and B(x₂, y₂) in the ratio m : n is given by
\( x = \frac{mx_2 + nx_1}{m + n};\ \ y = \frac{my_2 + ny_1}{m + n} \)
Note: If m/n is positive, the division is internal, but if m/n is negative, the division is external.
In particular, the mid-point of AB is \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).

Centroid of a Triangle

If A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) are the vertices of triangle ABC, then the co-ordinates of its centroid G are
\( G \equiv \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)

Area of Triangle

If A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) are the vertices of triangle ABC, then its area is equal to
\( \Delta ABC = \frac{1}{2} \left[ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right] \), provided the vertices are considered in the counter clockwise sense.
The above formula will give a (−)ve area if the vertices (x_i, y_i), i = 1, 2, 3 are placed in the clockwise sense. So, in practice, we take the absolute value of the expression.

Slope Formula

If θ is the angle at which a straight line is inclined to the positive direction of x-axis, and 0° ≤ θ < 180°, θ ≠ 90°, then the slope of the line, denoted by m, is defined by m = tan θ. If θ is 90°, m does not exist, but the line is parallel to the y-axis. If θ = 0, then m = 0 and the line is parallel to the x-axis.
If A(x₁, y₁) and B(x₂, y₂), x₁ ≠ x₂, are points on a straight line, then the slope of the line is given by
\( m = \frac{y_1 - y_2}{x_1 - x_2} \)

Condition of Collinearity of Three Points

Points A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) are collinear if:
(i) m_AB = m_BC = m_CA, i.e. \( \frac{y_1 - y_2}{x_1 - x_2} = \frac{y_2 - y_3}{x_2 - x_3} \)
(ii) ΔABC = 0, i.e. x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) = 0
(iii) AC = AB + BC or AB ~ BC (sum of two of the distances equals the third)

Equation of a Straight Line in Various Forms

(i) Point-Slope form: y − y₁ = m(x − x₁) is the equation of a straight line whose slope is m and which passes through the point (x₁, y₁).
(ii) Slope-Intercept form: y = mx + c is the equation of a straight line whose slope is m and which makes an intercept c on the y-axis.
(iii) Two-point form: \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \) is the equation of a straight line which passes through the points (x₁, y₁) and (x₂, y₂).
(iv) Intercept form: \( \frac{x}{a} + \frac{y}{b} = 1 \) is the equation of a straight line which makes intercepts a and b on OX and OY respectively.
(v) General form: ax + by + c = 0 is the equation of a straight line in the general form. In this case, slope of the line \( = -\frac{a}{b} \).

Length of Perpendicular from a Point on a Line

The length of perpendicular from P(x₁, y₁) on ax + by + c = 0 is \( \left| \frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}} \right| \)

Angle Between Two Straight Lines in Terms of Their Slopes

If m₁ and m₂ are the slopes of two intersecting straight lines (m₁m₂ ≠ −1) and θ is the acute angle between them, then
\( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \)

Parallel Lines

(i) When two straight lines are parallel their slopes are equal. Thus any line parallel to y = mx + c is of the type y = mx + d, where d is a parameter.
(ii) Two lines ax + by + c = 0 and a′x + b′y + c′ = 0 are parallel if \( \frac{a}{a'} = \frac{b}{b'} \neq \frac{c}{c'} \).
Thus any line parallel to ax + by + c = 0 is of the type ax + by + k = 0, where k is a parameter.
(iii) The distance between two parallel lines with equations ax + by + c₁ = 0 and ax + by + c₂ = 0 is \( \left| \frac{c_1 - c_2}{\sqrt{a^2 + b^2}} \right| \).
The coefficients of x and y in both the equations must be the same.

Perpendicular Lines

(i) When two lines of slopes m₁ and m₂ are at right angles, the product of their slopes is −1, i.e. m₁m₂ = −1. Thus any line perpendicular to y = mx + c is of the form \( y = -\frac{1}{m}x + d \), where d is any parameter.
(ii) Two lines ax + by + c = 0 and a′x + b′y + c′ = 0 are perpendicular if aa′ + bb′ = 0. Thus any line perpendicular to ax + by + c = 0 is of the form bx − ay + k = 0, where k is any parameter.

Straight Lines Making Angle α with a Given Line

The equations of the lines passing through the point (x₁, y₁) and making an angle α with the line y = mx + c are given by
(y − y₁) = tan(θ − α)(x − x₁) and (y − y₁) = tan(θ + α)(x − x₁), where tan θ = m.

Solved Examples

Example 1. Using distance formula show that the points (−1, 2), (5, 0) and (2, 1) are collinear.
Answer:
Let the given points (−1, 2), (5, 0) and (2, 1) be denoted by A, B and C, respectively.
Now \( AB = \sqrt{(5 - (-1))^2 + (0 - 2)^2} = \sqrt{36 + 4} = 2\sqrt{10} \)
\( BC = \sqrt{(2 - 5)^2 + (1 - 0)^2} = \sqrt{9 + 1} = \sqrt{10} \)
\( CA = \sqrt{(-1 - 2)^2 + (2 - 1)^2} = \sqrt{9 + 1} = \sqrt{10} \)
It is clear that BC + CA = AB.
Hence A, B and C are collinear.

Example 2. 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.
Answer:
(i) In this case x₁ = x₂.
Now \( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{0 + (y_2 - y_1)^2} = |y_2 - y_1| \)

(ii) In this case, y₁ = y₂.
Now \( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(x_2 - x_1)^2 + 0} = |x_2 - x_1| \)

Example 3. Find a point on the line-segment joining points (0, 4) and (2, 0) and dividing the line-segment (i) internally in ratio 2 : 3, (ii) externally in ratio 3 : 2.
Answer:
(i) Here the points are (0, 4) and (2, 0) and the division is internal in the given ratio 2 : 3. The coordinates of point P are
\( \left( \frac{2 \times 2 + 3 \times 0}{2 + 3}, \frac{2 \times 0 + 3 \times 4}{2 + 3} \right) \) or \( \left( \frac{4}{5}, \frac{12}{5} \right) \)
(ii) Here the ratio is 3 : 2 and the division is external, so the coordinates of the point are
\( \left( \frac{3 \times 2 - 2 \times 0}{3 - 2}, \frac{3 \times 0 - 2 \times 4}{3 - 2} \right) \) or (6, −8).

Example 4. Find the ratio in which the line joining (2, −3) and (5, 6) is divided by (i) the x-axis, (ii) the y-axis.
Answer:
Let P(x, y) divide the join of A(2, −3) and B(5, 6) in the ratio k : 1.
The co-ordinates of P are \( \left( \frac{5k + 2}{k + 1}, \frac{6k - 3}{k + 1} \right) \)
(i) Since P lies on the x-axis, its y-co-ordinate is zero.
i.e., \( \frac{6k - 3}{k + 1} = 0 \Rightarrow 6k - 3 = 0 \Rightarrow k = \frac{1}{2} \)
Hence P divides in the ratio \( \frac{1}{2} : 1 \), i.e., 1 : 2 internally.
(ii) Since P lies on the y-axis, its x-co-ordinate is zero.
i.e., \( \frac{5k + 2}{k + 1} = 0 \Rightarrow 5k + 2 = 0 \Rightarrow k = -\frac{2}{5} \)
Hence P divides externally in the ratio 2 : 5.

Example 5. If the points (1, 1), (a, 0), (0, b) are collinear, show that \( \frac{1}{a} + \frac{1}{b} = 1 \).
Answer:
As the points (1, 1), (a, 0) and (0, b) are collinear, the area of the triangle with these vertices is zero.
i.e., \( \frac{1}{2}[1(0 - b) + a(b - 1) + 0(1 - 0)] = 0 \)
or, −b + ab − a = 0
or, \( \frac{-b}{ab} + \frac{ab}{ab} - \frac{a}{ab} = 0 \) [Dividing both sides by ab]
or, \( -\frac{1}{a} + 1 - \frac{1}{b} = 0 \)
or, \( \frac{1}{a} + \frac{1}{b} = 1 \)

Example 6. Find the equation of the locus of a point equidistant from the points A(1, 3) and B(−2, 1).
Answer:
Let P(x, y) be any point on the locus and let A(1, 3) and B(−2, 1) be the given points.
By the given condition, PA = PB
or PA² = PB²
or (x − 1)² + (y − 3)² = (x + 2)² + (y − 1)²
or x² + 1 − 2x + y² − 6y + 9 = x² + 4 + 4x + y² + 1 − 2y
or 6x + 4y = 5,
which is the equation of the required locus, a straight line (the perpendicular bisector of AB).

Example 7. What are the slopes of the lines whose inclinations from the positive direction of X-axis are: (i) 60° (ii) 0° (iii) 150° (iv) 120°
Answer:
(i) Slope = tan 60° = √3
(ii) Slope = tan 0° = 0
(iii) Slope = tan 150° = − cot 60° \( = -\frac{1}{\sqrt{3}} \)
(iv) Slope = tan 120° = − cot 30° = −√3

Example 8. The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.
Answer:
Let the slopes of the two given lines be m and 2m and θ be the angle between them.
According to the question, \( \tan \theta = \frac{1}{3} \)
Now \( \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \)
or \( \frac{1}{3} = \frac{2m - m}{1 + 2m^2} \) or \( \frac{1}{3} = \frac{m - 2m}{1 + 2m^2} \)
or 1 + 2m² = 3m    or    1 + 2m² = −3m
or 2m² − 3m + 1 = 0    or    2m² + 3m + 1 = 0
or 2m(m − 1) − 1(m − 1) = 0    or    2m(m + 1) + 1(m + 1) = 0
or (m − 1)(2m − 1) = 0    or    (m + 1)(2m + 1) = 0
or \( m = 1, \frac{1}{2} \)    or    \( m = -1, -\frac{1}{2} \)
Hence, the slopes (m, 2m) of the lines are \( \left( \frac{1}{2}, 1 \right) \) or (1, 2) or \( \left( -\frac{1}{2}, -1 \right) \) or (−1, −2).

Example 9. If the angle between two lines is π/4 and slope of one of the lines is 1/2, find the slope of the other line.
Answer:
If θ is the acute angle between two lines with slopes m₁ and m₂, then
\( \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \) ...(i)
Now \( m_1 = \frac{1}{2}, m_2 = m \) and \( \theta = \frac{\pi}{4} \)
Now from (i), \( \tan \frac{\pi}{4} = \left| \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} \right| \Rightarrow 1 = \left| \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} \right| \)
\( \Rightarrow \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} = 1 \) or \( \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} = -1 \)
\( \Rightarrow m = 3 \) or \( m = -\frac{1}{3} \)
Hence, the slope of the other line is 3 or \( -\frac{1}{3} \).

Example 10. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (−1, −1) are the vertices of a right angled triangle.
Answer:
Let the vertices be A(4, 4), B(3, 5) and C(−1, −1).
Slope of AB, \( m_1 = \frac{5 - 4}{3 - 4} = \frac{1}{-1} = -1 \)
Slope of BC, \( m_2 = \frac{-1 - 5}{-1 - 3} = \frac{-6}{-4} = \frac{3}{2} \)
Slope of AC, \( m_3 = \frac{-1 - 4}{-1 - 4} = 1 \)
Now m₁m₃ = −1 × (1) = −1
Thus AB ⊥ AC or ∠A = 90°.
Hence ΔABC is a right angled triangle.

Example 11. Using the concept of slope, show that (−2, −1), (4, 0), (3, 3) and (−3, 2) are vertices of a parallelogram.
Answer:
Let A, B, C and D be the points (−2, −1), (4, 0), (3, 3) and (−3, 2), respectively. We have
Slope of AB \( = \frac{0 - (-1)}{4 - (-2)} = \frac{1}{6} \);   Slope of BC \( = \frac{3 - 0}{3 - 4} = -3 \);
Slope of CD \( = \frac{3 - 2}{3 - (-3)} = \frac{1}{6} \);   Slope of DA \( = \frac{2 - (-1)}{-3 - (-2)} = -3 \).
We note that the slopes of AB and BC are different and therefore points A, B and C are not collinear. Also, points A, D and C are not collinear. Thus, the given points form a quadrilateral.

Also, the slope of AB = Slope of CD, i.e., AB is parallel to CD and the slope of BC = Slope of DA, i.e., BC is parallel to DA. Thus, opposite sides of the quadrilateral ABCD are parallel.
Hence, the given four points constitute the vertices of a parallelogram.

Example 12. Determine the equation of the line passing through the point (−1, −2) and having slope 4/7.
Answer:
We are given that x₁ = −1, y₁ = −2 and m = 4/7.
Substituting these values in the point-slope form of the equation of a line, we get
\( y - (-2) = \frac{4}{7}[x - (-1)] \)
or 7(y + 2) = 4(x + 1) or 7y = 4x − 10, i.e., 4x − 7y − 10 = 0,
which is the required equation of the line.

Example 13. Find the equation of the line which satisfies the given conditions:
(i) Passing through (2, 2) and inclined to x-axis at 45°
(ii) Intersecting x-axis at a distance of 3 units to the left of the origin with slope −2.
(iii) Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with the positive direction of x-axis.
Answer:
(i) The slope of the given line = tan 45° = 1
The required equation of line is \( \frac{y - 2}{x - 2} = 1 \Rightarrow y - 2 = x - 2 \Rightarrow x - y = 0 \).
(ii) Since the line intersects the x-axis at a distance of 3 units to the left of the origin, i.e., it passes through (−3, 0). Also its slope = −2 [Given]
The required equation of line is \( \frac{y - 0}{x - (-3)} = -2 \Rightarrow y = -2(x + 3) \Rightarrow 2x + y + 6 = 0 \).
(iii) Since the line intersects the y-axis at a distance of 2 units above the origin, i.e., it passes through (0, 2). Also its slope = tan 30° \( = \frac{1}{\sqrt{3}} \).
The required equation of line is \( \frac{y - 2}{x - 0} = \frac{1}{\sqrt{3}} \Rightarrow \sqrt{3}y - 2\sqrt{3} = x \), i.e., \( x - \sqrt{3}y + 2\sqrt{3} = 0 \).

Example 14. Find the ratio in which the line segment joining the points (2, 3) and (4, 5) is divided by the line joining the points (6, 8) and (−3, −2).
Answer:
The equation of the line joining the points (6, 8) and (−3, −2) is
\( y - 8 = \frac{-2 - 8}{-3 - 6}(x - 6) \Rightarrow y - 8 = \frac{10}{9}(x - 6) \)
⇒ 9y − 72 = 10x − 60 ⇒ 10x − 9y + 12 = 0 ...(i)
Let this line divide the join of (2, 3) and (4, 5) at the point P in the ratio k : 1.
Then the coordinates of P are \( \left( \frac{4k + 2}{k + 1}, \frac{5k + 3}{k + 1} \right) \)
Now the point P lies on the line (i), i.e., \( 10\left( \frac{4k + 2}{k + 1} \right) - 9\left( \frac{5k + 3}{k + 1} \right) + 12 = 0 \)
⇒ 40k + 20 − 45k − 27 + 12k + 12 = 0 ⇒ 7k = −5 ⇒ k = −5/7
Since the value of k is negative, the line segment is divided externally (here P is the point (−3, −2) on AB produced).

The required ratio is 5 : 7 externally.

Example 15. Find the equation of the perpendicular bisector of the line segment joining the points A(2, 3) and B(6, −5).
Answer:
The slope of AB is given by \( m = \frac{-5 - 3}{6 - 2} = -2 \)   \( \left[ \text{Using } m = \frac{y_2 - y_1}{x_2 - x_1} \right] \)
The slope of a line ⊥ to AB \( = -\frac{1}{m} = \frac{1}{2} \)
Let P be the mid-point of AB. Then the coordinates of P are \( \left( \frac{2 + 6}{2}, \frac{3 - 5}{2} \right) \), i.e., (4, −1).
Thus, the required line passes through P(4, −1) and has slope 1/2. So its equation is
\( y + 1 = \frac{1}{2}(x - 4) \)   [Using y − y₁ = m(x − x₁)]
⇒ x − 2y − 6 = 0.

Example 16. Two opposite vertices of a square are (3, 4) and (1, −1). Find the coordinates of the other vertices.
Answer:
Let the square have the vertices A(3, 4), B(x₁, y₁), C(1, −1) and D(x₂, y₂).
Slope of AC \( = \frac{4 - (-1)}{3 - 1} = \frac{5}{2} \)
The sides AD and AB make an angle of 45° with the diagonal AC. Let m be the slope of a line making an angle of 45° with AC.
Hence, \( \tan 45^\circ = \left| \frac{m - \frac{5}{2}}{1 + m \cdot \frac{5}{2}} \right| \Rightarrow \frac{m - \frac{5}{2}}{1 + \frac{5m}{2}} = \pm 1 \)
⇒ 2m − 5 = ±(2 + 5m)
⇒ 2m − 5 = 2 + 5m ⇒ −3m = 7 ⇒ m = −7/3
and 2m − 5 = −(2 + 5m) ⇒ 7m = 3 ⇒ m = 3/7
Let the slope of AD be −7/3. Then, since CD ⊥ AD, the slope of CD = 3/7.
Equation of AD is y − 4 = −7/3 (x − 3) ⇒ 7x + 3y − 33 = 0 ...(i)
Equation of CD is y + 1 = 3/7 (x − 1) ⇒ 3x − 7y − 10 = 0 ...(ii)
Solving (i) and (ii), we get \( \frac{x}{-30 - 231} = \frac{y}{-99 + 70} = \frac{1}{-49 - 9} \)
\( \Rightarrow x = \frac{-261}{-58} = \frac{9}{2} \) and \( y = \frac{-29}{-58} = \frac{1}{2} \)
Coordinates of D are \( \left( \frac{9}{2}, \frac{1}{2} \right) \)
M is the mid-point of BD and AC both.
Hence \( \left( \frac{x_1 + 9/2}{2}, \frac{y_1 + 1/2}{2} \right) = \left( \frac{3 + 1}{2}, \frac{4 + (-1)}{2} \right) \)
\( \Rightarrow \frac{2x_1 + 9}{4} = 2, \frac{2y_1 + 1}{4} = \frac{3}{2} \Rightarrow x_1 = -\frac{1}{2}, y_1 = \frac{5}{2} \)
Coordinates of B are \( \left( -\frac{1}{2}, \frac{5}{2} \right) \)

Example 17. The vertices of triangle PQR are P(2, 1), Q(−2, 3) and R(4, 5). Find the equation of the median through the vertex R.
Answer:
Given: P(2, 1), Q(−2, 3) and R(4, 5).
Mid-point of PQ is \( \left( \frac{2 + (-2)}{2}, \frac{1 + 3}{2} \right) = (0, 2) \)
Now we have to find the equation of the median passing through R, i.e., of line AR, where A(0, 2).
Equation of the line in two-point form is \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \)
\( \Rightarrow y - 2 = \frac{5 - 2}{4 - 0}(x - 0) \Rightarrow y - 2 = \frac{3}{4}x \)
⇒ 4y − 8 = 3x ⇒ 3x − 4y + 8 = 0

Example 18. Find the equation of the line passing through (−3, 5) and perpendicular to the line through the points (2, 5) and (−3, 6).
Answer:
The required line passes through A(−3, 5) and is perpendicular to the line through the points B(2, 5) and C(−3, 6).
\( m_{BC} = \frac{6 - 5}{-3 - 2} = -\frac{1}{5} \)
Slope of required line = 5.
Now the equation of the line in point-slope form is y − y₁ = m(x − x₁)
y − 5 = 5(x + 3) ⇒ y − 5 = 5x + 15 ⇒ 5x − y + 20 = 0.

Example 19. Given the triangle with vertices A(10, 4), B(−4, 9), C(−2, −1). Find the equation of the altitude through B.
Answer:
The given vertices are A(10, 4), B(−4, 9), C(−2, −1).
Slope of AC \( = \frac{-1 - 4}{-2 - 10} = \frac{-5}{-12} = \frac{5}{12} \)
The altitude through B is perpendicular to AC. Let m be the slope of the altitude through B.
Hence \( m \times \frac{5}{12} = -1 \Rightarrow m = -\frac{12}{5} \)
The equation of the altitude through B(−4, 9) is \( y - 9 = -\frac{12}{5}(x - (-4)) \)   [Using y − y₁ = m(x − x₁)]
⇒ 5y − 45 = −12x − 48 ⇒ 12x + 5y + 3 = 0

Example 20. The perpendicular from the origin to a line meets it at the point (−2, 9). Find the equation of the line.
Answer:
The perpendicular from the origin O(0, 0) meets the line AB at A(−2, 9).
Hence, \( m_{OA} = \frac{9}{-2} = -\frac{9}{2} \Rightarrow m_{AB} = \frac{2}{9} \) (as AB ⊥ OA)
Now, the equation of the line will be \( y - 9 = \frac{2}{9}(x + 2) \)
⇒ 9y − 81 = 2x + 4 ⇒ 2x − 9y + 85 = 0

Example 21. Find the equations of the lines which cut off intercepts on the axes whose sum and product are 1 and −6, respectively.
Answer:
Let the intercepts on x-axis and y-axis be a and b, respectively, so that
a + b = 1 ...(i) and ab = −6 ...(ii)
Eliminating a from (i) and (ii), we get (1 − b)b = −6 ⇒ −b² + b = −6 ⇒ b² − b − 6 = 0
\( \Rightarrow b = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2} \), i.e. either b = 3 or b = −2
When b = −2, a = 3 and when b = 3, a = −2.
Thus, by the intercept form, the equations of the lines are \( \frac{x}{3} + \frac{y}{-2} = 1 \) or \( \frac{x}{-2} + \frac{y}{3} = 1 \)
i.e., 2x − 3y − 6 = 0 or 3x − 2y + 6 = 0

Example 22. Point R(h, k) divides a line-segment between the axes in the ratio 1 : 2. Find the equation of the line.
Answer:
Let AB be the required line, where A is (a, 0) on the x-axis and B is (0, b) on the y-axis, and R divides AB in the ratio 1 : 2.

Thus \( h = \frac{1 \times 0 + 2 \times a}{1 + 2} = \frac{2a}{3} \) and \( k = \frac{1 \times b + 2 \times 0}{1 + 2} = \frac{b}{3} \)
Therefore \( a = \frac{3h}{2} \) and b = 3k.
Now using the intercept form, the equation of the line AB is \( \frac{x}{a} + \frac{y}{b} = 1 \) or \( \frac{x}{\frac{3h}{2}} + \frac{y}{3k} = 1 \Rightarrow \frac{2x}{h} + \frac{y}{k} = 3 \)
⇒ 2kx + hy = 3kh, which is the required equation.

Example 23. Find the equation of a line which passes through the point (−2, 3) and makes angle 60° with the positive direction of x-axis.
Answer:
The slope of the line is m = tan 60° = √3.
Using the point-slope form y − y₁ = m(x − x₁) with x₁ = −2, y₁ = 3, we get
y − 3 = √3(x + 2)
or \( \sqrt{3}x - y + 3 + 2\sqrt{3} = 0 \).

This is the required equation of the line.

Example 24. Two sides of a square lie on the lines x + y = 1 and x + y + 2 = 0. What is its area?
Answer:
Clearly the length of the side of the square is equal to the distance between the parallel lines
x + y − 1 = 0 ...(i) and x + y + 2 = 0 ...(ii)
Putting x = 0 in (i), we get y = 1. So, (0, 1) is a point on line (i).
Now, distance between the parallel lines = length of the ⊥ from (0, 1) to x + y + 2 = 0
\( = \left| \frac{0 + 1 + 2}{\sqrt{1^2 + 1^2}} \right| = \frac{3}{\sqrt{2}} \)
Thus, the length of the side of the square is \( \frac{3}{\sqrt{2}} \) and hence its area \( = \left( \frac{3}{\sqrt{2}} \right)^2 = \frac{9}{2} \).

Example 25. Find the distance between parallel lines:
(i) 15x + 8y − 34 = 0 and 15x + 8y + 31 = 0
(ii) l(x + y) + p = 0 and lx + ly − r = 0
Answer:
(i) 15x + 8y − 34 = 0 and 15x + 8y + 31 = 0
Here A = 15, B = 8, C₁ = −34, C₂ = 31
Distance between parallel lines \( d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} = \frac{|-34 - 31|}{\sqrt{(15)^2 + 8^2}} = \frac{65}{17} \)
(ii) l(x + y) + p = 0, i.e. lx + ly + p = 0, and lx + ly − r = 0.
Here A = l, B = l, C₁ = p, C₂ = −r
Distance between parallel lines \( d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} = \frac{|p + r|}{\sqrt{l^2 + l^2}} = \frac{|p + r|}{\sqrt{2}\,|l|} = \frac{1}{\sqrt{2}} \left| \frac{p + r}{l} \right| \).

Example 26. Find the equations of the straight lines passing through the point (2, 3) and inclined at π/4 radians to the line 2x + 3y = 5.
Answer:
Let the line 2x + 3y = 5 make an angle θ with the positive x-axis.
Then \( \tan \theta = -\frac{2}{3} \)
The required lines make angles \( \theta + \frac{\pi}{4} \) and \( \theta - \frac{\pi}{4} \) with the x-axis, so the slopes of the required lines are \( \tan\left( \theta + \frac{\pi}{4} \right) \) and \( \tan\left( \theta - \frac{\pi}{4} \right) \).
\( \tan\left( \theta + \frac{\pi}{4} \right) = \frac{\tan \theta + \tan \frac{\pi}{4}}{1 - \tan \theta \tan \frac{\pi}{4}} = \frac{-\frac{2}{3} + 1}{1 - \left( -\frac{2}{3} \right)(1)} = \frac{1}{5} \)
and \( \tan\left( \theta - \frac{\pi}{4} \right) = \frac{\tan \theta - \tan \frac{\pi}{4}}{1 + \tan \theta \tan \frac{\pi}{4}} = \frac{-\frac{2}{3} - 1}{1 + \left( -\frac{2}{3} \right)(1)} = -5 \)
∴ Equations of the required lines are \( y - 3 = \frac{1}{5}(x - 2) \) and y − 3 = −5(x − 2)
i.e., x − 5y + 13 = 0 and 5x + y − 13 = 0

Example 27. Find the equations to the straight lines passing through the point (2, 3) and equally inclined to the lines 3x − 4y − 7 = 0 and 12x − 5y + 6 = 0.
Answer:
Let m be the slope of the required line. Then its equation is y − 3 = m(x − 2) ...(1)
The slope of 3x − 4y − 7 = 0 is 3/4 and the slope of 12x − 5y + 6 = 0 is 12/5.
It is given that line (1) is equally inclined to the lines 3x − 4y − 7 = 0 and 12x − 5y + 6 = 0, then
\( \frac{\frac{3}{4} - m}{1 + \frac{3}{4}m} = -\left( \frac{\frac{12}{5} - m}{1 + \frac{12}{5}m} \right) \Rightarrow \frac{3 - 4m}{4 + 3m} = -\left( \frac{12 - 5m}{5 + 12m} \right) \)
⇒ (3 − 4m)(5 + 12m) + (4 + 3m)(12 − 5m) = 0
⇒ 63m² − 32m − 63 = 0 ⇒ (7m − 9)(9m + 7) = 0
\( \therefore m = \frac{9}{7}, -\frac{7}{9} \)
Putting these values of m in (1) we obtain the equations of the required lines as 9x − 7y + 3 = 0 and 7x + 9y − 41 = 0.

Example 28. Find the equation of straight lines through point (2, 3) and having an intercept of length 2 units between the straight lines 2x + y = 3, 2x + y = 5.
Answer:
The given lines 2x + y = 3 ...(1) and 2x + y = 5 ...(2) are parallel, and the distance between them is \( \frac{2}{\sqrt{5}} < 2 \), so such lines exist.
Equation of any non-vertical line through (2, 3) is y − 3 = m(x − 2) or y = mx − 2m + 3 ...(3)
Let line (3) cut lines (1) and (2) at A and B respectively. Solving (1) and (3), we get \( A \equiv \left( \frac{2m}{m + 2}, \frac{6 - m}{m + 2} \right) \)
and solving (2) and (3), we get \( B \equiv \left( \frac{2m + 2}{m + 2}, \frac{m + 6}{m + 2} \right) \)
According to the question AB = 2 ⇒ (AB)² = 4
\( \Rightarrow \left( \frac{2}{m + 2} \right)^2 + \left( \frac{2m}{m + 2} \right)^2 = 4 \Rightarrow 1 + m^2 = m^2 + 4m + 4 \)
⇒ 1 = 4m + 4 ⇒ \( m = -\frac{3}{4} \), giving the line \( y - 3 = -\frac{3}{4}(x - 2) \), i.e., 3x + 4y = 18.
Now check the vertical line through (2, 3), i.e., x = 2. It meets (1) at (2, −1) and (2) at (2, 1); the length of the intercept is |1 − (−1)| = 2. So x = 2 also satisfies the condition.

Hence the equations of the required lines are 3x + 4y = 18 and x − 2 = 0.