Straight Lines Class 11 Notes AP Inter 1st Year Maths Chapter 9

Students can go through AP Inter 1st Year Maths Notes 9th Lesson Straight Lines will help students in revising the entire concepts quickly.

Straight Lines Class 11 Notes AP Inter 1st Year Maths 9th Lesson

→ Area of the triangle whose vertices are (x₁, y₁) (x₂, y₂) and (x₃, y₃) is \(\frac {1}{2}\)|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|

→ Area of the quadrilateral whose vertices are (x₁, y₁) (x₂, y₂) (x₃, y₃) and (x₄, y₄) is \(\frac {1}{2}\)|x₁(y₂ – y₄) + x₂(y₃ – y₁) + x₃(y₄ – y₂) + x₄(y₁ – y₃)|

→ Slope (m) of a non-vertical line passing through the points (x₁, y₁) and (x₂, y₂) is given by m = \(\frac{y_2-y_1}{x_2-x_1}=\frac{y_1-y_2}{x_1-x_2}\), x₁ ≠ x₂

→ If a line makes an angle a with the positive direction of the x-axis, then the slope of the line is given by m = tan α, α ≠ 90°

→ The slope of a horizontal line is zero, and the slope of a vertical line is undefined.

→ An acute angle (say θ) between lines L₁ and L₂ with slopes m₁ and m₂ is given by tan θ = \(\left|\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|\), 1 + m₁m₂ ≠ 0

→ Two lines are parallel if and only if their slopes are equal.

→ Two lines are perpendicular if and only if the product of their slopes is -1.

→ Three points A, B, and C are collinear if and only if the slope of AB = the slope of BC.

→ Equation of the horizontal line having distance a from the x-axis is either y = a or y = -a.

→ Equation of the vertical line having distance b from the y-axis is either x = b or x = -b.

→ The point (x, y) lies on the line with slope m and through the fixed point (x₀, y₀), if and only if its co-ordinates satisfy the equation y-y₀ = m(x-x₀).

→ Equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by y – y₁ = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\)(x – x₁)

→ The point (x, y) on the line with slope m and y-intercept c lies on the line if and only if y = mx + c.

→ If a line with slope m makes x-intercept d. The equation of the line is y = m(x – d).

→ Equation of a line making intercepts a and b on the x-and y-axis, respectively, is \(\frac{x}{a}+\frac{y}{b}\) = 1.

→ Any equation of the form Ax + By + C = 0, with A and B not zero, simultaneously, is called the general linear equation or general equation of a line.

→ The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by d = \(\frac{\left|\mathrm{Ax}_1+\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\)

→ Distance between the parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0, is given by d = \(\frac{\left|C_1-C_2\right|}{\sqrt{A^2+B^2}}\)

→ If Q(h, k) is the foot of the perpendicular from P(x₁, y₁) on the straight line ax + by + c = 0, then (h – x₁) : a = (k – y₁) : b = -(ax₁ + by₁ + c) : (a² + b²).

→ If Q(h, k) is the image of the point P(x₁, y₁) w.r.t. the straight line ax + by + c = 0, then (h – x₁) : a = (k – y₁) : b = -2(ax₁ + by₁ + c) : (a² + b²).

Example Problems

Question 1.
Find the slope of the lines:
(a) Passing through the points (3, -2) and (-1, 4)
(b) Passing through the points (3, -2) and (7, -2)
(c) Passing through the points (3, -2) and (3, 4)
(d) Making an inclination of 60° with the positive direction of the x-axis.
Solution:
(a) The slope of the line through (3, -2) and (-1, 4) is
m = \(\frac{4-(-2)}{-1-3}=\frac{6}{-4}=-\frac{3}{2}\)

(b) The slope of the line through the points (3, -2) and (7, -2) is
m = \(\frac{-2-(-2)}{7-3}=\frac{0}{4}\) = 0

(c) The slope of the line through the points (3, -2) and (3, 4) is
m = \(\frac{4-(-2)}{3-3}=\frac{6}{0}\) which is not defined.

(d) Here, inclination of the line α = 60°.
Therefore, the slope of the line is m = tan 60° = √3.

Question 2.
If the angle between two lines is \(\frac{\pi}{4}\) and the slope of one of the lines is \(\frac {1}{2}\), find the slope of the other line.
Solution:
We know that the acute angle θ between two lines with slopes m₁ and m₂ is given by
tan θ = \(\left|\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|\)
Let m₁ = \(\frac {1}{2}\), m₂ = m and θ = \(\frac{\pi}{4}\)
Now, putting these values in (1), we get

Therefore m = 3 or m = \(-\frac {1}{3}\)
Hence, the slope of the other line is 3 or \(-\frac {1}{3}\)

The figure explains the reason for the two answers.

Question 3.
The line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x.
Solution:
The slope of the line through the points (-2, 6) and (4, 8) is
m₁ = \(\frac{8-6}{4-(-2)}=\frac{2}{6}=\frac{1}{3}\)
The slope of the line through the points (8, 12) and (x, 24) is
m₂ = \(\frac{24-12}{x-8}=\frac{12}{x-8}\)
Since two lines are perpendicular, m₁m₂ = -1,
which gives \(\frac{1}{3} \times \frac{12}{x-8}\) or x = 4

Question 4.
Find the equations of the lines parallel to the coordinate axes and passing through (-2, 3).
Solution:
Position of the lines is shown in the Figure.
The y-coordinate of every point on the line parallel to the x-axis is 3.

Therefore, the equation of the line parallel to the x-axis and passing through (-2, 3) is y = 3.
Similarly equation of the line parallel to the y-axis and passing through (-2, 3) is x = -2.

Question 5.
Find the equation of the line through (-2, 3) with slope -4.
Solution:
Here m = -4 and the given point (x₀, y₀) is (-2, 3).
By the point-slope form formula (1) above,
the equation of the given line is y – 3 = -4(x + 2)
or 4x + y + 5 = 0, which is the required equation.

Question 6.
Write the equation of the line through the points (1, -1) and (3, 5).
Solution:
Here x₁ = 1, y₁ = -1, x₂ = 3, and y₂ = 5.
Using two-point form (2) above for the equation of the line,
we have y – (-1) = \(\frac{5-(-1)}{3-1}\)(x – 1)
or -3x + y + 4 = 0, which is the required equation.

Question 7.
Write the equation of the lines for which tan θ = \(\frac {1}{2}\), where θ is the inclination of the line and
(i) y-intercept is \(-\frac {3}{2}\)
(ii) x-intercept is 4.
Solution:
(i) Here, slope of the line is m = tan θ = \(\frac {1}{2}\) and y-intercept c = \(-\frac {3}{2}\).
Therefore, by slope-intercept form (3) above,
the equation of the line is y = \(\frac{1}{2} x-\frac{3}{2}\)
or 2y – x + 3 = 0, which is the required equation.

(ii) Here, we have m = tan θ = \(\frac {1}{2}\) and d = 4.
Therefore, by slope-intercept form (4) above,
the equation of the line is y = \(\frac {1}{2}\)(x – 4)
or 2y – x + 4 = 0, which is the required equation.

Question 8.
Find the equation of the line, which makes intercepts -3 and 2 on the x-axis and y-axis, respectively.
Solution:
Here, a = -3 and b = 2.
By intercept form above, equation of the line is \(\frac{x}{-3}+\frac{y}{2}\) = 1 or 2x – 3y + 6 = 0.
Any equation of the form Ax + By + C = 0, where A and B are not zero simultaneously, is called a general linear equation or a general equation of a line.

Question 9.
Find the distance of the point (3, -5) from the line 3x – 4y – 26 = 0.
Solution:
Give line is 3x – 4y – 26 = 0 ………(1)
Comparing (1) with the general equation of a line, Ax + By + C = 0,
we get A = 3, B = -4 and C = -26
Given point is (x₁, y₁) = (3, -5).
The distance of the given point from given line is d = \(\frac{\left|\mathrm{Ax}_1+\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\)
= \(\frac{|3.3+(-4)(-5)-26|}{\sqrt{3^2+(-4)^2}}\)
= \(\frac {3}{5}\)

Question 10.
Find the distance between the parallel lines 3x – 4y + 7 = 0 and 3x – 4y + 5 = 0.
Solution:
Here A = 3, B = -4, C₁ = 7 and C₂ = 5.
Therefore, the required distance is d = \(\frac{|7-5|}{\sqrt{3^2+(-4)^2}}=\frac{2}{5}\)

Question 11.
If the lines 2x + y – 3 = 0, 5x + ky – 3 = 0 and 3x – y – 2 = 0 are concurrent, find the value of k.
Solution:
Three lines are said to be concurrent if they pass through a common point,
i.e., the point of intersection of any two lines lies on the third line.
The lines given are
2x + y – 3 = 0 …….(1)
5x + ky – 3 = 0 …….(2)
3x – y – 2 = 0 ………..(3)
Solving (1) and (3) by the cross-multiplication method, we get
\(\frac{x}{-2-3}=\frac{y}{-9+4}=\frac{1}{-2-3}\)
or x = 1, y = 1
Therefore, the point of intersection of the two lines is (1, 1).
Since the above three lines are concurrent, the point (1, 1) will satisfy equation (2)
so that 5.1 + k.1 – 3 = 0 or k = -2.

Question 12.
Find the distance of the line 4x-y = 0 from the point P(4, 1) measured along the line making an angle of 135° with the positive x-axis.
Solution:
Given line is 4x – y = 0 ………..(1)
To find the distance of the line (1) from the point P(4, 1) along another line, we have to find the point of intersection of both lines.
For this purpose, we will first find the equation of the second line.
The slope of the second line is tan (135°) = -1.
Equation of the line with slope -1 through the point P(4, 1) is y – 1 = -1(x – 4)
or x + y – 5 = 0 ………(2)

Solving (1) and (2), we get x = 1 and y = 4, so that the point of intersection of the two lines is Q(1, 4).
Now, the distance of line (1) from the point P(4, 1) along the line (2) = the distance between the points P(4, 1) and Q(1, 4).
= \(\sqrt{(1-4)^2+(4-1)^2}\)
= 3√2 units.

Question 13.
Assuming that straight lines work as a plane mirror for a point, find the image of the point (1, 2) in the line x – 3y + 4 = 0.
Solution:
Let Q(h, k) is the image of the point P(1, 2) in the line x – 3y + 4 = 0 …….(1)
Therefore, the line (1) is the perpendicular bisector of the line segment PQ.

Hence slope of line PQ = \(\frac{-1}{\text { Slope of line } x-3 y+4=0}\)
So that \(\frac{k-2}{h-1}=\frac{-1}{\frac{1}{3}}\)
or 3h + k = 5 ………(2)
and the mid-point of PQ, i.e., point \(\left(\frac{\mathrm{h}+1}{2}, \frac{\mathrm{k}+2}{2}\right)\) will satisfy the equation (1)
so that \(\frac{h+1}{2}-3\left(\frac{k+2}{2}\right)+4=0\)
or h – 3k = -3 ………(3)
Solving (2) and (3), we get h = \(\frac {6}{5}\) and k = \(\frac {7}{5}\)
Hence, the image of the point (1, 2) in the line (1) is \(\left(\frac{6}{5}, \frac{7}{5}\right)\).

Question 14.
Show that the area of the triangle formed by the lines y = m₁x + c₁, y = m₂x + c_2, and x = 0 is \(\frac{\left(c_1-c_2\right)^2}{2\left|m_1-m_2\right|}\).
Solution:
Given lines are
y = m₁x + c₁ ……….(1)
y = m₂x + c₂ ………..(2)
x = 0 ……….(3)
We know that the line y = mx + c meets the line x = 0 (y-axis) at the point (0, c),
Therefore, two vertices of the triangle formed by lines (1) to (3) are P(0, c₁) and Q(0, c₂).
The third vertex can be obtained by solving equations (1) and (2).
Solving (1) and (2), we get

Question 15.
A line is such that its segment between the lines 5x – y + 4 = 0 and 3x + 4y – 4 = 0 is bisected at the point (1, 5). Obtain its equation.
Solution:
Given lines are
5x – y + 4 = 0 ……..(1)
3x + 4y – 4 = 0 ………(2)
Let the required line intersect the lines (1) and (2) at the points (α₁, β₁) and (α₂, β₂), respectively.

Therefore 5α₁ – β₁ + 4 = 0 and 3α₂ + 4β₂ – 4 = 0
or β₁ = 5α₁ + 4 and β₂ = \(\frac{4-3 \alpha_2}{4}\)
We are given that the midpoint of the segment of the required line between (α₁, β₁) and (α₂, β₂) is (1, 5).
Therefore \(\frac{\alpha_1+\alpha_2}{2}\) = 1 and \(\frac{\beta_1+\beta_2}{2}\) = 5
or α₁ + α₂ = 2 and \(\frac{5 \alpha_1+4+\frac{4-3 \alpha_2}{4}}{2}\) = 5
or α₁ + α₂ = 2 and 20α₁ – 3α₂ = 20 ………(3)
Solving equation (3) for α₁ and α₂, we get
α₁ = \(\frac {26}{23}\) and α₂ = \(\frac {20}{23}\)
and hence, β₁ = \(\text { 5. } \frac{26}{23}+4=\frac{222}{23}\)
Equation of the required line passing through (1, 5) and (α₁, β₁) is
y – 5 = \(\frac{\beta_1-5}{\alpha_1-1}\)(x – 1)
or y – 5 = \(\frac{\frac{222}{23}-5}{\frac{26}{23}-1}\)(x – 1)
or 107x – 3y – 92 = 0, which is the equation of the required line.

Question 16.
Show that the path of a moving point such that its distances from two lines 3x – 2y = 5 and 3x + 2y = 5 are equal is a straight line.
Solution:
Given lines are
3x – 2y = 5 ……..(1)
and 3x + 2y = 5 ………(2)
Let (h, k) be any point whose distances from the lines (1) and (2) are equal.
Therefore \(\frac{|3 \mathrm{~h}-2 \mathrm{k}-5|}{\sqrt{9+4}}=\frac{|3 \mathrm{~h}+2 \mathrm{k}-5|}{\sqrt{9+4}}\)
or |3h – 2k – 5| = |3h + 2k – 5|, which gives 3h – 2k – 5 = 3h + 2k – 5
or -(3h – 2k – 5) = 3h + 2k – 5.
Solving these two relations we get k = 0 or h = \(\frac {5}{3}\)
Thus, the point (h, k) satisfies the equations y = 0 or x = \(\frac {5}{3}\), which represent straight lines.
Hence, the path of the point equidistant from the lines (1) and (2) is a straight line.

Multiple Choice Questions

Question 1.
The equation of the straight line making an intercept of 3 units on the Y-axis and inclined at 45° to the X-axis is
(1) y = x – 1
(2) y = x + 3
(3) 45x + 3
(4) y = x + 45
Answer:
(2) y = x + 3

Question 2.
The angle made by the line segment joining (5, 2) (6, -15) at (0, 0) is
(1) \(\frac{\pi}{6}\)
(2) \(\frac{\pi}{4}\)
(3) \(\frac{\pi}{2}\)
(4) π
Answer:
(3) \(\frac{\pi}{2}\)

Question 3.
The lines 2x + 3y = 6, 2x + 3y = 8 cut the X-axis at A and B, respectively. A line ‘l’ drawn through the point (2, 2) meets the X-axis at ‘C’ in such a way that abscissae of A, B, C are in arithmetic progression, then the equation of the line ‘l’ is
(1) 2x + 3y = 10
(2) 3x + 2y = 10
(3) 2x – 3y = 10
(4) 3x – 2y = 10
Answer:
(1) 2x + 3y = 10

Question 4.
The area (in sq.m) of the triangle formed by the lines x = 0, y = 0, and 3x + 4y = 12 is
(1) 3
(2) 4
(3) 6
(4) 12
Answer:
(3) 6

Question 5.
The area of the triangle formed by the axes and the line x cos α + y sec α = 2
(1) 4
(2) 3
(3) 2
(4) 1
Answer:
(3) 2

Question 6.
If the area of the triangle formed by the lines x = 0; y = 0; 3x + 4y = a (a > 0) is 1; then a =
(1) √6
(2) 2√6
(3) 4√6
(4) 6√2
Answer:
(2) 2√6

Question 7.
Equation of the straight line passing through the image of (3, 6) in the line 2x – y = 5 and perpendicular to the line 3x + 4y = 15 is
(1) 4x – 3y = 10
(2) 4x – 3y = 7
(3) 4x – 3y = 5
(4) 4x – 3y = 16
Answer:
(4) 4x – 3y = 16

Question 8.
A straight line through the point A(3, 4) is such that its intercept between the axes bisects at A, and its equation is
(1) 4x + 3y = 24
(2) 3x + 4y = 25
(3) x + y = 7
(4) 3x – 4y + 7 = 0
Answer:
(1) 4x + 3y = 24

Question 9.
If the straight lines y = 4 – 3x, ay = x + 10, and 2y + bx + 9 = 0 represent the three consecutive sides of a rectangle, then ab =
(1) 18
(2) -3
(3) \(\frac {1}{2}\)
(4) \(-\frac {1}{3}\)
Answer:
(1) 18

Question 10.
The ends of a rod of length ‘l’ move on the coordinate axes. The locus of the point on the rod that divides it in the ratio 1 : 2 is
(1) 36x² + 9y² = 4l²
(2) 36x² + 9y² = l²
(3) 9x² + 36y² = 4l²
(4) 9x² + 36y² = l²
Answer:
(3) 9x² + 36y² = 4l²

Question 11.
Two equal sides of an isosceles triangle are 7x – y + 3 = 0, x + y – 3 = 0, and its third side passes through the point (1, -10). The equation of the third side is
(1) 3x + y + 7 = 0
(2) x – 3y + 29 = 0
(3) 3x + y + 3 = 0
(4) 3x + y – 3 = 0
Answer:
(1) 3x + y + 7 = 0

Question 12.
The perpendicular distance from (1, 2) to this straight line, 12x + 5y = 7, is
(1) \(\frac {15}{13}\)
(2) \(\frac {12}{13}\)
(3) \(\frac {5}{13}\)
(4) \(\frac {7}{13}\)
Answer:
(1) \(\frac {15}{13}\)

Question 13.
If p and q are perpendicular distances from the origin to the straight lines x sec θ + y cosec θ = a and x cos θ + y sin θ = a cos 2θ, then
(1) 4p² + q² = a²
(2) p² + q² = a²
(3) p² + 2q² = a²
(4) 4p² + q² = 2a²
Answer:
(1) 4p² + q² = a²

Question 14.
The equation of the line parallel to 5x + 12y = 1 and at a distance of 3 units from (1, 0) is
(1) 5x + 12y + 34 = 0
(2) 12x – 5y + 34 = 0
(3) 12x – 5y – 44 = 0
(4) 5x + 12y + 44 = 0
Answer:
(1) 5x + 12y + 34 = 0

Question 15.
The area (in sq. units) of the circle which touches the lines 4x + 3y = 15 and 4x – 3y = 5 is
(1) 4π
(2) 3π
(3) 2π
(4) π
Answer:
(4) π

Question 16.
If A(2, -1) and B(6, 5) are two points, the ratio in which the foot of the perpendicular from (4, 1) to AB divides it is
(1) 8 : 15
(2) 5 : 8
(3) -5 : 8
(4) -8 : 5
Answer:
(2) 5 : 8

Question 17.
A straight line which makes equal intercepts on positive X and Y axes and which is at a distance 1 unit from the origin intersects the straight line y = 2x + 3 + √2 at (x₀, y₀), then 2x₀ + y₀ =
(1) 3 + √2
(2) √2 – 1
(3) 1
(4) 0
Answer:
(2) √2 – 1

Question 18.
If a, b, c are in AP, the lines ax + by + c = 0 pass through the fixed point
(1) (1, 2)
(2) (-1, 2)
(3) (1, -2)
(4) (-1, -2)
Answer:
(3) (1, -2)

Question 19.
If a, b, c are in AP, then the lines ax + by + c = 0
(1) passes through a fixed point
(2) from an equilateral triangle
(3) from a rhombus
(4) from a square
Answer:
(1) passes through a fixed point

Question 20.
The point on the line 2x – 3y = 5, which is equidistant from (1, 2) and (3, 4), is
(1) (2, 3)
(2) (4, 1)
(3) (1, -1)
(4) (4, 6)
Answer:
(2) (4, 1)