1st PUC Basic Maths Question Bank Chapter 17 Straight Lines

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Karnataka 1st PUC Basic Maths Question Bank Chapter 17 Straight Lines

Question 1.
Find the slope of the line joining the points (3, -4) and (-7,3).
Answer:
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 1

Question 2.
Show that the line joining points (2, -3) and (-5,1) is parallel to the line joining the points (7, -1) and (0,3) and perpendicular to the line joining the points (4,5) and (0, -2).
Answer:
Let m1 be the slope of the lien l1 joining points (2, -3) and (0, 3)
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 2
Let m2 be the slope of the line l2 going (7, -1) and (0,3)
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 3
⇒ m1 = m2
∴ L1 and L2 are parallel to each other.
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 4

1st PUC Basic Maths Question Bank Chapter 17 Straight Lines

Question 3.
L1 and L3 or l2 l3 are perpendicular to each other. The slope of a line is double the
slope of another line. If the tangent of the angle between the is \(\frac { 1 }{ 3 }\). Find the slope of the lines.
Answer:
If slope of one line is m, then the slope of the other is 2m.
Let the angle between them θ then tan θ = \(\frac { 1 }{ 3 }\)
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 5
⇒ \(\frac { 1 }{ 3 }\) = \(\frac{m}{1+2 m^{2}}\)
⇒ 1 + 2m2 = 3m
⇒ 2m2 – 3m + x = 0
⇒ (2m – 1)(m – 1 ) = 0
∴ 2m – 1 = 0 and m – 1 = 0
m = \(\frac { 1 }{ 2 }\) m = 1
∴ Slope are \(\frac { 1 }{ 2 }\), and 1.

On standard form of straight lines:

Question 1.
Find the equation of line passing through (-3, 5) with slope \(\frac { -1 }{ 5 }\)
Answer:
Let (x1 y1) = (-3, 5). slope m = \(\frac { -1 }{ 5 }\)
The equation of line is y – y1 = m(x – x1)
= y – 5 = \(\frac { -1 }{ 5 }\) (x + 3)
⇒ 5y – 25 = -x – 3
⇒ x + 5y – 22 = 0

1st PUC Basic Maths Question Bank Chapter 17 Straight Lines

Question 2.
Find the equation line passing through (-3, 2) and (11, -1).
Answer:
Let (x1, y1) = (-3,2), (x2, y2) = (11,-1)
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 6
⇒ \(\frac{y-2}{x+3}=\frac{-3}{14}\)
⇒ 14y – 28 = -3x – 9
⇒ 3x + 14y – 19 = 0 is the equation

Question 3.
Find the ratio in which the line join (1, 2) and (4, 3) is divided by the line joining the points (2,3) and (4, 1).
Answer:
Let A = (2,3) B = (4, 1)
∴ the equation of line is a given by \(\frac{y-3}{x-2}=\frac{1-3}{4-2}\)
⇒ \(\frac{y-3}{x-2}=\frac{-2}{2}\) = -1
⇒ y – 3 = -x + 2
∴ x + y – 5 = 0 is the equation of line
Let c = (1,2) and D(4,3) cut the line AB at P(x ,y) is the ratio 2 : 1 then
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 7
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 8
⇒ 4x + 1 + 3r + 2 – 5(r + 1) = 0
⇒ 2r – 2 = 0
⇒ r = 1
∴ the required ratio is 1 : 1

Question 4.
Express 9x – 4y – 13 = 0 in the intercept from
Answer:
Consider 9x – 4y = 13 = \(\frac{9}{13} x-\frac{4}{13} y\) = 1
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 18
∴ x – intercept = \(\frac{13}{9}\)
y – intercept = \(\frac{-13}{9}\)

1st PUC Basic Maths Question Bank Chapter 17 Straight Lines

Question 5.
If P ¡s the length of the perpendicular from the origin on a line, whose x and y intercepts are respectively a and b then show that = \(\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\)
Answer:
Let the line any the axis at A and y – axis at B
Then by data OA = a and OB = b from the figure
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 9
Area of OAB = \(\frac { 1 }{ 2 }\) (O.A) (O.B) = \(\frac { 1 }{ 2 }\) ab
or = \(\frac { 1 }{ 2 }\) (AB).P
⇒ \(\frac { 1 }{ 2 }\) ab = \(\frac { 1 }{ 2 }\) (AB).P ⇒ a2b2 = (AB)2. p2
Also AB2 = OA2 + OB2 ⇒ AB2 = a2 + b2 (∴ because a2b2 = (a2 + b2) . p2
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 10

Intersection of two lines and Concurrency of Lines

Question 1.
Find the equation of line parallel to the y-axis and drawn through the point of intersection of x – 7y + 5 = 0 and 3x + y – 7 = 0.
Answer:
The equation of line through the point of intersection of the given lines is of the form x – 7y + 5 + k(3x + y – 7) = 0
∴ (1 + 3k)x + (k -7)y + 5 – 7k = 0 (1)
Since the lines are parallel to y-axis, coefficient of y = 0
i.e., k – 7 =0
∴ k = 7 substitute in (1) we get
22x – 44 = 0
∴ x – 2 = 0 is required equation

Question 2.
Find the equation of the line through the intersection of 5x – 3y = 1 and 2x + 3y – 23 = 0 and perpendicular to the line whose equation is 5x – 3y -1 = 0 and perpendicular to the line whose equation is 5x – 3y – 1 = 0.
Answer:
We have 5x – 3y – 1 = 0 ….. (1)
and 2x + 3y – 23 = 0 ….. (2) .
∴ the equation will be in the form
(5x – 3y – 1) + k(2x + 3y – 23) = 0
⇒ (5 + 2k)x + 3(k – 1)y – (23k + 1) = 0 …… (3)
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 11
∴ Slope of line = \(\frac{5+2 k}{3(k-1)}\)
Also 5x – 3y – 1 = 0
⇒ y = \(\frac{5}{3} x \frac{1}{3}\) …….. (1)
∴slope = \(\frac{5}{3}\)
Given (4) and (5) are perpendicular
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 12
⇒ k = -34. Substituting the value of k in (3), we get
(5x – 3y – 1) -34 (2x + 3y – 25) = 0
⇒ 5x – 68x – 3y – 102y – 1 + 782 = 0
⇒ 63x + 105y – 781 = 0

1st PUC Basic Maths Question Bank Chapter 17 Straight Lines

Question 3.
If lines whose equation are y = m1 + c1y = m2x + c2 and y = m5x + c3 meet is a point then prove that m1( (c2 – c3) + m2 (c3 – c1) + m3(c1 – c2) = 0
Answer:
The equation of the given lines are
m1x – y + c1 = 0 ……. (1)
m2x – y + c2 = 0 …… (2)
m3x – y + c3 = 0 .. (3)
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 13
The three lines will be concurrent if the point of intersection of (1) and (2) lies on (3)
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 14
⇒ m3(c1 – c2) – (m2c1 – m1c2) + c3 (m2 – m1) =0, simplify,
⇒ m1(c2 – c3) + m2(c3 – c1) + m3(c1 – c2) =0

Question 4.
For what values of k are the lines x – 2y + 1 = 0, 2x – 5y + 3 = 0 and 5x – 9y + k = 0 are concurrent.
Answer:
consider
x – 2y + 1 = 0 ……. (1)
2x – 5y + 3 = 0 ….. (2)
5x – 9y + k = 0 …….. (3)
Solve (1) and (2) \(\frac{x}{-6+5}=\frac{-y}{3-2}=1\)
⇒ \(\frac{x}{-1}=\frac{-y}{-1}=\frac{1}{-1}\)
⇒ x = 1, y = 1
point of intersection of (1) and (2) is (1,1),
⇒ (1, 1) lies on (3) = 5 – 9 + k = 0 = k = 4

Angle between the lines, Length of perpendicular

Question 1.
Find the angle between the lines 2x + 3y – 4 = 0 and 3x – 2y + 5 =0
Answer:
By data
2x + 3y – 4 = 0 … (1)
3x – 2y + 5 = 0 …… (2)
∴ Slope of (1) = \(\frac{-a}{b}=\frac{-2}{3}\) = m1
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 15
∴ the angle between the lines is 90°

1st PUC Basic Maths Question Bank Chapter 17 Straight Lines

Question 2.
Determine the position of the points (2, 1) and (-1, 1) w.r.t the line 4x – ly + 1 = 0.
Answer:
4(2) -7(1) = -7 + 1 = 2 > 0
4( -1) – 7(1) = – 4 – 7 + 2 = -10 < 0
Since the two points are opposite in sign the two points lie an either sides of the given line 4x – 7y + 1 = 0

Question 3.
If points (9,8) and (-3,3) are equidistant from the line 5x + 2y + 7 = 0 find ‘a’.
Answer:
Distance from (a, 8) to the line 5x + 2y + 7 = 0
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 16
Given that the distance are equal
1st PUC Basic Maths Question Bank Chapter 17 Straight Lines - 17
⇒ 5a + 23 = 2
or
5a + 23 = -2
⇒ a = \(\frac{-21}{5}\) and a = -5