Inter 1st Year Maths Straight Lines Solutions Exercise 9c

Practicing the AP Board Solutions Class 11 Maths and Chapter 9 Inter 1st Year Maths Straight Lines Solutions Exercise 9c Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Straight Lines Solutions Exercise 9c

Straight Lines Exercise 9c Solutions

Straight Lines Class 11 Exercise 9c Solutions – Straight Lines 9c Exercise Solutions

I.

Question 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
Solution:
(i) x + 7y = 0
Given The equation is x + 7y = 0
The slope-intercept form is represented in the form ‘y = mx + c’,
where m is the slope and c is the intercept.
So, the above equation can be expressed as y = \(\frac {-1}{7}\)x + 0
∴ The above equation is of the form y = mx + c,
Where m = \(\frac {-1}{7}\) and c = 0

(ii) 6x + 3y – 5 = 0
Given, The equation is 6x + 3y – 5 = 0
The slope-intercept form is represented in the form ‘y = mx + c’,
where m is the slope and c is the y-intercept.
So, the above equation can be expressed as
3y = -6x + 5
y = \(\frac{-6 x}{3}+\frac{5}{3}\)
y = -2x + \(\frac {5}{3}\)
∴ The above equation is of the form y = mx + c,
where m = -2 and c = \(\frac {5}{3}\)

(iii) y = 0
Given the equation is y = 0
The slope-intercept form is given by ‘y = mx + c’,
where m is the slope and c is the y-intercept.
y = 0 × x + 0
The above equation is of the form y = mx + c,
where m = 0 and c = 0.

Question 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
Solution:
(i) 3x + 2y – 12 = 0
Given The equation is 3x + 2y – 12 = 0
The equation of the line in intercept form is given by \(\frac{x}{a}+\frac{y}{b}\) = 1,
where ‘a’ and ‘b’ are intercepted on the x-axis and the y-axis, respectively.
So, 4x – 3y = 6
Now, let us divide both sides by 6; we get

∴ The above equation is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1
Where a = \(\frac {3}{2}\), b = -2
The intercept on the x-axis is \(\frac {3}{2}\)
The intercept on the y-axis is -2.

(ii) 4x – 3y = 6

This equation is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1
Where a = \(\frac {3}{2}\) and b = -2
Therefore, equation (1) is in the intercept form
and x-intercept = \(\frac {3}{2}\), y-intercept = -2.

(iii) 3y + 2 = 0
Given The equation is 3y + 2 = 0
The equation of the line in intercept form is given by \(\frac{x}{a}+\frac{y}{b}\) = 1,
Where ‘a’ and ‘b’ are intercepted on the x-axis and the y-axis, respectively.
So, 3y = -2
Now, let us divide both sides by -2;
We get \(\frac{3 y}{-2}=\frac{-2}{-2}\)
\(\frac{3 y}{-2}\) = 1
\(\frac{y}{\left(\frac{-2}{3}\right)}\) = 1
∴ The above equation is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1
Where a = 0, b = \(\frac {-2}{3}\)
The intercept on the x-axis is 0.
The intercept on the y-axis is \(\frac {-2}{3}\).

Question 3.
Find the distance of the point (-1, 1) from the line 12(x + 6) = 5(y – 2).
Solution:
Given The equation of the line is 12(x + 6) = 5(y – 2).
12x + 72 = 5y – 10
12x – 5y + 82 = 0 ………(1)
Now, compare equation (1) with the general equation of a line, Ax + By + C = 0
where A = 12, B = -5 and C = 82,
perpendicular distance (d) of a line Ax + By + C = 0 From a point (x₁, y₁) is given by
d = \(\frac{\left|\mathrm{Ax}_1+\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\)
Given point (x₁, y₁) = (-1, 1)
∴ The distance of the point (-1, 1) from the given line is

∴ The distance is 5 units.

Question 4.
Find the points on the x-axis, whose distances from the line \(\frac{x}{3}+\frac{y}{4}\) = 1 are 4 units.
Solution:
The equation of the line is \(\frac{x}{3}+\frac{y}{4}\) = 1
⇒ 4x + 3y -12 = 0 ………(1)
Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.
It is known that the perpendicular distance ‘d’ of a line Ax + By + C = 0 from a point (x, y) is given by
d = \(\frac{\left|\mathrm{Ax}_1+\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\)
Therefore, 4 = \(\frac{|4 a+3 \times 0-12|}{\sqrt{4^2+3^2}}\)
⇒ 4 = \(\frac{|4 a-12|}{5}\)
⇒ |4a – 12| = 20
⇒ ±(4a – 12) = 20
⇒ (4a – 12) = 20 or -(4a – 12) = 20
⇒ 4a = 20 + 12 or 4a = -20 + 12
⇒ a = 8 or a = -2
Thus, the required points on the x-axis are (-2, 0) and (8, 0).

Question 5.
Find the distance between parallel lines.
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
(ii) l(x + y) + p = 0 and l(x + y) – r = 0
Solution:
(i) The given parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
Therefore, the distance between them is

(ii) The given parallel lines are l(x + y) + p = 0 and l(x + y) – r = 0
⇒ lx + ly + p = 0 and lx + ly – r = 0
Therefore, the distance between them is

Question 6.
Find the equation of the line parallel to the line 3x – 4y + 2 – 0 and passing through the point (-2, 3).
Solution:
Given The line is 3x – 4y + 2 = 0
So y = \(\frac{3 x}{4}+\frac{2}{4}\)
y = \(\frac{3 x}{4}+\frac{1}{2}\)
Which is of the form y = mx + c, where m is the slope of the given line.
The slope of the given line is
We know that parallel lines have the same slope.
∴ Slope of other line = m = \(\frac {3}{4}\)
The equation of a line having slope m and passing through (x1, y1) is given by y – y₁ = m(x – x₁)
∴ The equation of the line having slope \(\frac {3}{4}\) and passing through (-2, 3) is
y – 3 = \(\frac {3}{4}\)(x – (-2))
⇒ 4y – 3 × 4 = 3x + 3 × 2
⇒ 3x – 4y = -18
∴ The equation is 3x – 4y +18 = 0.

Question 7.
Find the equation of the line perpendicular to the line x – 7y + 5 = 0 and having x-intercept 3.
Solution:
Given The equation of line is x – 7y + 5 = 0
So, y = \(\frac{1}{7} x+\frac{5}{7}\)
Which is of the form y = mx + c, where m is the slope of the given line
The slope of the given line is \(\frac {1}{7}\)
The slope of the line perpendicular to the line having slope m is \(\frac {-1}{m}\)
The slope of the line perpendicular to the line having a slope of \(\frac {1}{7}\) is \(\frac{-1}{\left(\frac{1}{7}\right)}\) = -7
So, the equation of the line with slope -7 and the x-intercept 3 is given by y = m(x – d)
⇒ y = -7(x – 3)
⇒ y = -7x + 21
⇒ 7x + y = 21
∴ The equation is 7x + y = 21.

Question 8.
Find the angles between the lines √3x + y = 1 and x + √3y = 1.
Solution:
Given The lines are √3x + y = 1 and x + √3y = 1
So, y = -√3x + 1 ……..(1)
y = \(\frac{-1}{\sqrt{3} x}+\frac{1}{\sqrt{3}}\) …….. (2)
The slope of the line (1) is m₁ = -√3,
while the slope of the line (2) is m₂ = \(\frac{-1}{\sqrt{3}}\)
Let θ be the angle between two lines
So, tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)

∴ The angle between the given lines is either 30° or 180° – 30° = 150°

Question 9.
The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at a right angle. Find the value of h.
Solution:
Let the slope of the line passing through (h, 3) and (4, 1) be m₁, then
m₁ = \(\frac{(1-3)}{(4-h)}=\frac{-2}{(4-h)}\)
Let the slope of line 7x – 9y – 19 = 0 be m₂
7x – 9y – 19 = 0
So, y = \(\frac{7 x}{9}-\frac{19}{9}\)
m₂ = \(\frac {7}{9}\)
Since the given lines are perpendicular,
m₁ × m₂ = -1
⇒ \(\frac{-2}{(4-h)} \times \frac{7}{9}\) = -1
⇒ \(\frac{-14}{(36-9 h)}\) = -1
⇒ -14 = -1 × (36 – 9h)
⇒ 36 – 9h = 14
⇒ 9h = 36 – 14
⇒ h = \(\frac {22}{9}\)
∴ The value of h is \(\frac {22}{9}\)

Question 10.
Find the ratio in which straight line 3x + 4y = 6 divides the line segment joining the points (2, -1) and (1, 1).
Solution:
L₁₁ = 3(2) + 4(-1) – 6 = -4 < 0
L₂₂ = 3(1) + 4(1) – 6 = 1 > 0
Required ratio = -(-4) : 1 = 4 : 1
∴ Given points lie on opposite sides of the line.

Question 11.
Find the point of intersection of the lines 7x + y + 3 = 0, x + y = 0
Solution:
Given 7x + y + 3 = 0 ……….(1)
and x + y = 0 ………(2)
⇒ y = -x
Substitute in equation (1) is 7x + (-x) + 3 = 0
⇒ 6x + 3 = 0
⇒ 6x = -3
⇒ x = \(\frac {-1}{2}\)
y = -x = \(-\left(\frac{-1}{2}\right)=\frac{1}{2}\)
∴ \(\left(\frac{-1}{2}, \frac{1}{2}\right)\)

Question 12.
Show that the straight lines (a – b)x + (b – c)y = c – a, (b – c)x + (c – a)y = a – b and (c – a)x + (a – b)y = b – c are concurrent.
Solution:
Given (a – b)x + (b – c)y = (c – a) ……….(1)
(b – c)x + (c – a)y = (a – b) …………(2)
(c – a)x + (a – b)y = (b – c) ………..(2)
Solving (1) and (2)


∴ The given lines are concurrent.

Question 13.
Find the value of p, if the straight lines x + p = 0, y + 2 = 0, and 3x + 2y + 5 = 0 are concurrent.
Solution:
Let (α, β) be the point of concurrence
∴ α + β = 0 ……..(1)
β + 2 = 0 ……..(2)
3α + 2β + 5 = 0 ………..(3)
(2) ⇒ β = -2,
(3) ⇒ 3α + 2(-2) + 5 = 0
⇒ 3α = -1
⇒ α = \(\frac {-1}{3}\)
(1) ⇒ \(\frac {-1}{3}\) + p = 0
⇒ p = \(\frac {1}{3}\)

Question 14.
Find the area of the triangle formed by the following straight lines and the coordinate axes.
(i) x – 4y + 2 = 0
(ii) 3x – 4y + 12 = 0
Solution:
(i) Area of the triangle = \(\frac{C^2}{2|a b|}=\frac{(2)^2}{2|(1)(-4)|}=\frac{1}{2}\) sq. unit
(ii) Area of the triangle = \(\frac{c^2}{2|a b|}=\frac{(12)^2}{2|(3)(-4)|}\) = 6 sq. unit

II.

Question 1.
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.
Solution:
Let the slope of line Ax + By + C = 0 be m
Ax + By + C = 0
So, y = \(-\frac{A}{B} x-\frac{C}{B}\)
m = \(\frac {-A}{B}\)
By using the formula,
Equation of the line passing through point (x₁, y₁) and having slope m = \(\frac {-A}{B}\) is y – y₁ = m(x – x₁)
⇒ y – y₁ = \(\frac {-A}{B}\)(x – x₁)
⇒ B(y – y₁) = -A(x – x₁)
⇒ A(x – x₁) + B(y – y₁) = 0
So, the line through point (x₁, y₁) and parallel to the line Ax + By + C = 0 is A(x – x₁) + B(y – y₁) = 0
Hence proved.

Question 2.
Two lines passing through the point (2, 3) intersect each other at an angle of 60°. If slope of one line is 2, find equation of the other line.
Solution:
Given m₁ = 2
Let the slope of the first line be m₁
And let the slope of the other line be m₂
The angle between the two lines is 60°.
So, tan θ = \(\left|\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|\)

So now let us consider
When m₂ = \(\frac{2-\sqrt{3}}{(2 \sqrt{3}+1)}\)
The equation of the line passing through point (2, 3) and having a slope m₂ is y = \(\frac{2-\sqrt{3}}{(2 \sqrt{3}+1)}\)x + c.

Question 3.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).
Solution:
Given the right bisector of a line segment bisects the line segment at 90°.
End-points of the line segment AB are given as A(3, 4) and B(-1, 2)
Let the midpoint of AB be (x, y)
x = \(\frac{(3-1)}{2}=\frac{2}{2}\) = 1
y = \(\frac{(4+2)}{2}=\frac{6}{2}\) = 3
∴ (x, y) = (1, 3)
Let the slope of line AB be m₁
m₁ = \(\frac{(2-4)}{(-1-3)}\)
= \(\frac{-2}{(-4)}\)
= \(\frac {1}{2}\)
And let the slope of the line perpendicular to AB be m₂
m₂ = \(\frac{-1}{\left(\frac{1}{2}\right)}\) = -2
The equation of the line passing through (1, 3) and having a slope of -2 is
y – 3 = \(\left(\frac{2-\sqrt{3}}{2 \sqrt{3}+1}\right)\) (x – 2)
⇒ (2√3 + 1)y – 3(2√3 + 1) = (2 – √3)x – 2(2 – √3)
⇒ (√3 – 2)x + (2√3 + 1)y = -4 + 2√3 + 6√3 + 3
⇒ (√3 – 2)x + (2√3 + 1)y = 8√3 – 1
∴ Equation of the other line is (√3 – 2)x + (2√3 + 1)y = 8√3 – 1
When m₂ = \(\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}\)
The equation of the line passing through the point (2, 3) and having a slope m₂ is
y – 3 = \(\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}\) (x – 2)
⇒ (2√3 – 1)y – 3(2√3 – 1) = -(2 + √3)x + 2(2 + √3)
⇒ (2√3 – 1)y + (2 + √3)x = 4 + 2√3 + 6√3 – 3
⇒ (2√3 – 1)y + (2 + √3)x = 8√3 + 1
∴ Equation of the other line is (2√3 – 1)y + (2 + √3)x = 8√3 + 1
⇒ (y – 3) = -2(x – 1)
⇒ y – 3 = -2x + 2
⇒ 2x + y = 5
∴ The required equation of the line is 2x + y = 5.

Question 4.
Find the coordinates of the foot of the perpendicular from the point (-1, 3) to the line 3x – 4y – 16 = 0.
Solution:

Let us consider the coordinates of the foot, of the perpendicular from (-1, 3) to the line 3x – 4y – 16 = 0, be (a, b)
So, let the slope of the line joining (-1, 3) and (a, b) be m₁
m₁ = \(\frac{(b-3)}{(a+1)}\)
And let the slope of the line 3x – 4y – 16 = 0 be m₂
y = \(\frac {3}{4}\)x – 4
m₂ = \(\frac {3}{4}\)
Since these two lines are perpendicular,
m₁ × m₂ = -1
⇒ \(\frac{(b-3)}{(a+1)} \times\left(\frac{3}{4}\right)\) = -1
⇒ \(\frac{(3 b-9)}{(4 a+4)}\) = -1
⇒ 3b – 9 = -4a – 4
⇒ 4a + 3b = 5 ……….(1)
point (a, b) lies on the line 3x – 4y = 16
⇒ 3a – 4b = 16 ……….(2)
Solving equations (1) and (2), we get
a = \(\frac {68}{25}\) and b = \(\frac {-49}{25}\)
∴ The co-ordinates of the foot of the perpendicular are \(\left(\frac{68}{25}, \frac{-49}{25}\right)\).

Question 5.
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.
Solution:
Given that the perpendicular from the origin meets the given line at (-1, 2).
The equation of the line is y = mx + c
The line joining the points (0, 0) and (-1, 2) is perpendicular to the given line.
So, the slope of the line joining (0, 0) and the slope of the given line is m
m × (-2) = -1
m = \(\frac {1}{2}\)
Since point (-1, 2) lies on the given line,
y = mx + c
2 = \(\frac {1}{2}\) × (-1) + c
c = 2 + \(\frac {1}{2}\) = \(\frac {5}{2}\)
∴ The values of m and c are \(\frac {1}{2}\) and \(\frac {5}{2}\) respectively.

Question 6.
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, then prove that p² + 4q² = k².
Solution:
The equations of the given lines are
x cos θ – y sin θ = k cos 2θ …….(1)
x sec θ + y cosec θ = k ……….(2)
Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by
d = \(\frac{\left|\mathrm{Ax}_1+\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\)
So now, compare equation (1) to the general equation of line, Ax + By + C = 0
we get, A = cos θ, B = -sin θ, and C = -k cos 2θ
It is given that p is the length of the perpendicular from (0, 0) to line (1).
P = \(\frac{|A \times 0+B \times 0+C|}{\sqrt{A^2+B^2}}=\frac{|c|}{\sqrt{A^2+B^2}}\)
= \(\frac{|-\mathrm{k} \cos 2 \theta|}{\sqrt{\cos ^2 \theta}+\sin ^2 \theta}\)
= k cos 2θ
p = k cos 2θ
Let us square on both sides, we get,
p² = k² cos² 2θ ………..(3)
Now, compare equation (2) to the general equation of line
i.e., Ax + By + C = 0, we get
A = sec θ, B = cosec θ and c = -k
It is given that q is the length of the perpendicular from (0, 0) to line (2)

= k cos θ sin θ
q = k cos θ sin θ
Multiply both sides by 2, and we get
2q = 2k cos θ sin θ = k × 2 sin θ cos θ
2q = k sin 2θ
Squaring both sides, we get
4q² = k² sin² 2θ ………(4)
Now add (3) and (4), we get
p² + 4q² = k² cos² 2θ + k² sin² 2θ
⇒ p² + 4q² = k² (cos² 2θ + sin² 2θ) [since, cos² 2θ + sin² 2θ = 1]
⇒ p² + 4q² = k²
Hence proved.

Question 7.
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.
Solution:
Let AD be the altitude of triangle ABC from vertex A
So, AD is perpendicular to BC.
Given, Vertices A(2, 3) B(4, -1) and C(1, 2)
Let the slope of the line BC = m₁
m₁ = \(\frac{(-1-2)}{(4-1)}\) = -1
Let the slope of the line BO = m₂
AD is perpendicular to BC
m₁ × m₂ = -1
⇒ -1 × m₂ = -1
⇒ m₂ = 1
The equation of the line passing through the point (2, 3) and having a slope of 1 is
y – 3 = 1 × (x – 2)
⇒ y – 3 = x – 2
⇒ y – x = 1
Equation of the altitude from the vertex
A = y – x = 1
Length of AD = Length of the perpendicular from A(2, 3) to BC
The equation of BC is y + 1 = -1 × (x – 4)
y + 1 = -x + 4
x + y – 3 = 0 ……….(1)
Perpendicular distance (d) of a line Ax + By + C = 0
From a point (x₁, y₁) is given by
d = \(\frac{\left|\mathrm{Ax}_1+\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\)
Now, compare equation (1) to the general equation of the line
i.e., Ax + By + C = 0, we get
Length of AB = \(\frac{|1 \times 2+1 \times 3-3|}{\sqrt{1^2+1^2}}=\frac{|2|}{\sqrt{2}}\) = √2 Units
[Where A = 1, B = 1 and C = 3]
∴ The equation and the length of the altitude from vertex A are y – x = 1 and √2 units, respectively.

Question 8.
If p is the length of a perpendicular from the origin to the line whose intercepts on the axes are a and b,
then show that \(\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\).
Solution:
The equation of line of a line whose intercepts on the axes are a and b is \(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ bx + ay = ab
⇒ bx + ay – ab = 0 ……….(1)
Perpendicular distance (d) of a line Ax + By + C = 0 From a point (x₁, y₁) is given by
d = \(\frac{\left|\mathrm{Ax}_1+\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\)
Now compare equation (1) to be general equation of a line
i.e., Ax + By + C = 0, We get
A = b, B = a and C = -ab
If p is the length of the perpendicular from point (x₁, y₁) = (0, 0) to line (1), we get

Hence proved.

Question 9.
A straight line meets the coordinate axes at A and B. Find the equation of the straight line when
(i) \(\overline{\mathbf{A B}}\) is divided in the ratio 2 : 3 at (-5, 2)
(ii) \(\overline{\mathbf{A B}}\) is divided in the ratio 1 : 2 at (-5, 4)
(iii) (p, q) bisects \(\overline{\mathbf{A B}}\)
Solution:

(i) Let a, b be the intercepts of the line:
A = (a, 0), B = (0, b)
The point which divides \(\overline{\mathbf{A B}}\) in the ratio 2 : 3 is \(\left(\frac{3 a}{5}, \frac{2 b}{5}\right)\)

(ii) Let a, b be the intercepts of the line:
A = (a, 0), B = (0, b)
The point which divides \(\overline{\mathbf{A B}}\) in the ratio 1 : 2 is \(\left(\frac{2 \mathrm{a}}{3}, \frac{\mathrm{~b}}{3}\right)\)

(iii) Let a, b be the intercepts of the line:
A = (a, 0), B = (r, b)

Question 10.
A triangle of area 24 sq. units is formed by a straight line and the coordinate axes in the first quadrant. Find the equation of the straight line, if it passes through (3, 4).
Solution:
If a line intersects the x-axis at (a, 0) and the y-axis at (0, b), the triangle formed with the coordinate axes has
area = \(\frac {1}{2}\) ab = 24
⇒ ab = 48 ……(1)
The general intercept form of line is \(\frac{x}{a}+\frac{y}{b}\) = 1
We substitute the point (3, 4) into this equation
\(\frac{3}{a}+\frac{4}{b}\) = 1 …….(2)
(1) ⇒ ab = 48
⇒ b = \(\frac {48}{a}\)
Substitute in equation (2)

Question 11.
Find the set of values of ‘a’ if the points (1, 2) and (3, 4) lie on the same side of the straight line 3x – 5y + a = 0.
Solution:
(1, 2), (3, 4) lie to the same side of L = 3x – 5y + a = 0
⇒ L(1, 2), L(3, 4) have the same sign
⇒ L(1, 2) L(3, 4) > 0
⇒ (3 – 10 + a) (9 – 20 + a) > 0
⇒ (a – 7) (a – 11) > 0
⇒ a < 7 or a > 11
∴ The set of values of a is (-∞, 7) ∪ (11, ∞).

Question 12.
Show that the lines 2x + y – 3 = 0, 3x + 2y – 2 – 0, and 2x – 3y – 23 = 0 are concurrent and find the point of concurrency.
Solution:
Let p(α, β) be the point of intersection of the lines 2x + y – 3 = 0, 3x + 2y – 2 = 0.
∴ 2α + β – 3 = 0, 3α + 2β – 2 = 0
By the method of cross multiplication:
\(\frac{\alpha}{-2+6}=\frac{\beta}{-9+4}=\frac{1}{4-3}\)
⇒ α = 4, β = -5
Now, 2α – 3β – 23 = 2(4) – 3(-5) – 23
= 8 + 15 – 23
= 0
P lies on the line 2x – 3y – 23 = 0
Given lines are concurrent, and the point of concurrence is p(4, -5).

Question 13.
Find the value of p, if the following lines are concurrent.
3x + 4y = 5, 2x + 3y = 4, px + 4y = 6
Solution:
Let (α, β) be the point of concurrence
3α + 4β = 5 ………..(1)
2α + 3β – 4 = 0 ………(2)
pα + 4β – 6 = 0 ……….(3)
From (1) and (2), we get,
\(\frac{\alpha}{-16+15}=\frac{\beta}{-10+12}=\frac{1}{9-8}\)
⇒ α = -1, β = 2
Equation (3) ⇒ -p + 8 – 6 = 0
⇒ p = 2
∴ Point of concurrence = (-1, 2)

Question 14.
If 3a + 2b + 4c = 0, then show that the equation ax + by + c = 0 represents a family of concurrent straight lines and find the point of concurrency.
Solution:
3a + 2b + 4c = 0
⇒ 3a + 2b = -4c
⇒ c = \(\frac{-(3 a+2 b)}{4}\)
a + by + c = 0
⇒ ax + by – \(\frac{(3 a+2 b)}{4}\) = 0
⇒ \(a\left(x-\frac{3}{4}\right)+b\left(y-\frac{1}{2}\right)=0\)
∴ ax + by + c = 0 represents a set of lines passing through the points of intersection of x – \(\frac {3}{4}\) = 0, y – \(\frac {1}{2}\) = 0
∴ ax + by + c = 0 represents a set of concurrent lines.
The point of concurrence is \(\left(\frac{3}{4}, \frac{1}{2}\right)\).

III.

Question 1.
Find the point on the straight line 3x + y + 4 = 0 which is equidistant from the points (-5, 6) and (3, 2).
Solution:
Let A(-5, 6) and B(3, 2)
Midpoint of \(\overline{\mathrm{AB}}\) is (-1, 4)
Slope of \(\overline{\mathrm{AB}}=\frac{2-6}{3+5}=\frac{-4}{8}=\frac{-1}{2}\)
∴ Slope of the perpendicular bisector of \(\overline{\mathrm{AB}}\) is 2.
Equation to the perpendicular bisector of \(\overline{\mathrm{AB}}\) is
y – 4 = 2(x + 1)
⇒ 2x – y + 6 = 0 ………..(1)
Given line equation in 3x + y + 4 = 0 ……..(2)
Solving (1) and (2), the required point is (-2, 2).

Question 2.
A straight line through P(3, 4) makes an angle of 60° with the positive direction of the X-axis. Find the co-ordinates of the points on the line which are 5 units away from P.
Solution:
Given P(3, 4), θ = 60°,
|r| = 5
⇒ r = ±5
The required point is (x, y) = (x1 + r cos θ, y1 + r sin θ)
If r = 5, then (x, y) = (3 + 5 cos 60°, 4 + 5 sin 60°)
= \(\left(3+\frac{5}{2}, 4+5 \frac{\sqrt{3}}{2}\right)\)
= \(\left(\frac{11}{2}, \frac{8+5 \sqrt{3}}{2}\right)\)
If r = -5 then (x, y) = (3 – 5 cos 60°, 4 – 5 sin 60°)
= \(\left(3-\frac{-5}{2}, 4-5 \frac{\sqrt{3}}{2}\right)\)
= \(\left(\frac{1}{2}, \frac{8-5 \sqrt{3}}{2}\right)\)

Question 3.
A straight line through Q(√3, 2) makes an angle \(\frac{\pi}{6}\) with the positive direction of the X-axis. If the straight line intersects the line √3x – 4y + 8 = 0 at P, find the distance PQ.
Solution:
m = tan \(\frac{\pi}{6}\) = \(\frac{1}{\sqrt{3}}\)
The equation of a line passing through Q(√3, 2) is y – 2 = \(\frac{1}{\sqrt{3}}\)(x – √3)
⇒ √3y – 2√3 = x – √3
⇒ x – √3y + √3 = 0 ……….(1)
Given equation is √3x – 4y + 8 = 0 ……..(2)
Now, P is the point of intersection of (1) and (2)

Question 4.
A straight line through Q(2, 3) makes an angle \(\frac{3 \pi}{4}\) with the negative direction of the X-axis. If the straight line intersects the line x + y – 7 = 0 at P, find the distance PQ.
Solution:
Slope of the line, m = \(\tan \left(\pi-\frac{3 \pi}{4}\right)\) = 1
Equation of the line passing through Q(2, 3) having Slope is y – 3 = 1(x – 2)
⇒ x – y + 1 = 0 ………..(1)
Given line is x + y – 7 = 0 ………..(2)

The point of intersection of (1) and (2) in (3, 4)
⇒ P = (3, 4)
∴ Distance PQ = \(\sqrt{(3-2)^2+(4-3)^2}=\sqrt{2}\)

Question 5.
Show that the straight lines x + y = 0, 3x + y – 4 = 0 and x + 3y – 4 = 0 form an isosceles triangle.
Solution:
Given lines are x + y = 0 ……..(1)
3x + y – 4 = 0 ………(2)
x + 3y – 4 = 0 ………..(3)
Slope of (1) is -1
Slope of (2) is -3
Slope of (3) is \(\frac {-1}{3}\)
If α is an angle between (1) and (2), then
tan α = \(\left|\frac{-1+3}{1+3}\right|=\frac{2}{4}=\frac{1}{2}\)
⇒ α = \(\tan ^{-1}\left(\frac{1}{2}\right)\)
If β is an angle between (1) and (3), then
tan β = \(\left|\frac{-1+\frac{1}{3}}{1+\frac{1}{3}}\right|=\frac{2}{4}=\frac{1}{2}\)
⇒ β = \(\tan ^{-1}\left(\frac{1}{2}\right)\)
∴ The triangle formed by (1), (2), and (3) is isosceles.