Class 11 Applied Maths Chapter 17 (Ex – 17.7)

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Class 11 Applied Maths Chapter 17 Solutions

Straight Line

EXERCISE- 17.7

Q.1 Find the distance of the point P from the line l in the following cases:

(i) P(3, -5), line l is 3x – 4y – 26 = 0

(ii) P(-1, 1), line l is 12(x + 6) = 5(y – 2)

Ans. (i) Here, x1 = 3, y1 = -5, A = 3, B = -4, C = -26

Formula for Distance of a point from a line, d = |(Ax1 + By1 + C)|/(√A^2+B^2), we get,

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

= |[3(3) + (-4)(-5) – 26]|/[√(3)^2 + (-4)^2]

= |[9 + 20 – 26|/√25

= 3/5 unit

(ii) Here, line is 12(x + 6) = 5(y – 2)

So, 12x + 72 = 5y – 10

12x + 72 – 5y + 10

Therefore line, 12x – 5y + 82 = 0

Now, x1 = -1, y1 = 1, A = 12, B = -5, C = 82

Formula for Distance of a point from a line, d = |(Ax1 + By1 + C)|/(√A^2+B^2), we get,

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

= |[12(-1) + (-5)(1) + 82]|/[√(12)^2 + (-5)^2]

= |[-12 -5 + 82|/√169

= 65/13

= 5 units

Q.2 Find the value(s) of k, given that the distance of the point (4, 1) from the line 3x – 4y + k = 0 is 4 units.

Ans. Here, x1 = 4, y1 = 1, A = 3, B = -4, C = k, D = 4

Formula for Distance of a point from a line, d = |(Ax1 + By1 + C)|/(√A^2+B^2), we get,

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

4 = |[3(4) + (-4)(1) + k]|/[√(3)^2 + (-4)^2]

4 = |[8 + k|/√9 + 16

4 = |8 + k|/5

20 = |8 + k|

Taking ‘+’ sign,

20 = 8 + k

k = 12

Taking ‘-‘ sign,

20 = -(8 + k)

k = -8 – 20

k = -28

Therefore, k = 12, -28

Q.3 Find the distance between the following pairs of lines:

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

(ii) y = mx + c and y = mx + d

(iii) y = 2x + 4 and 6x = 3y + 5

(iv) l (x + y) + p = 0 and lx + ly – r = 0

Ans.(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

Formula for Distance between two parallel sides, d = |C1 – C2|/√A^2 + B^2, we get,

= |-35 – 31|/√(15^2 + 8^2)

= |-65|/√(225 + 64)

= 65/√289

= 65/17 units

(ii) y = mx + c and y = mx + d

Formula for Distance between two parallel sides, d = |C1 – C2|/√A^2 + B^2, we get,

= |c – d|/√(m^2 + (-1)^2)

= |c – d|/√(m^2 +

(iii) y = 2x + 4 and 6x = 3y + 5

2x – y + 4 = 0, 6x – 3y – 5 = 0

Multiplying [2x – y + 4 = 0] x 3 = 6x – 3y + 12 = 0

Formula for Distance between two parallel sides, d = |C1 – C2|/√A^2 + B^2, we get,

= |12 – (-5)|/√(6^2 + (-3)^2)

= |17|/√(36 + 9)

= 17/√45 units

(iv) l (x + y) + p = 0 and lx + ly – r = 0

lx + ly + p = 0, lx + ly – r = 0

d = |p – (-r)|/√l^2 + r^2

d = |p + r|/√2l^2

d = |p + r|/√2 l

d = 1/√2 |(p + r)/l|

Q.4 A vertex of a square is at the origin and its one side lies along the line 3x – 4y – 10 = 0. Find the area of the square.

Ans. Here, Point P(0, 0), Line 3x – 4y – 10 = 0

Formula for Distance of a point from a line, d = |(Ax1 + By1 + C)|/(√A^2+B^2), we get,

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

= |[3(0) + (-4)(0) + (-10)]|/[√(3)^2 + (-4)^2]

= |[-10|/√25

= 10/5

= 2 units

Area of square = Side x Side

= 2 x 2

= 4 sq units.

Q.5 If the equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, -1), then find the length of the side of the triangle.

Ans. Here, Vertex (2, -1) & equation x + y – 2 = 0, Let AD be altitude of △ABC,

Formula for Distance of a point from a line, d = |(Ax1 + By1 + C)|/(√A^2+B^2), we get,

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

= |[1(2) + (1)(-1) – 2]|/[√(1)^2 + (1)^2]

= |-1|/√2

= 1/√2

Using pythagoras theorem,

AB^2 = AD^2 + BD^2

x^2 = (1/√2)^2 + (x/2)^2

x^2 = 1/2 + x^2/4

x^2 – x^2/4 = 1/2

3x^2/4 = 1/2

x^2 = 2/3

x = √2/√3 = √2/3 units.

Q.6 Find the distance of the point (0, -1) from the line joining the points (1, 3) and (-2, 6).

Ans. The line passing through two points (x1, y1) and (x2,y2),

Equation of line, (y – y1) = [(y2 – y1)/(x2 – x1)] (x – x1), we get,

(y – 3) = [(6 – 3)/(-2 – 1)] (x – 1)

(y – 3) = 3/-3 (x – 1)

(y – 3) = -x + 1

x + y – 4 = 0

Formula for Distance of a point from a line, d = |(Ax1 + By1 + C)|/(√A^2+B^2), we get,

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

= |[1(0) + (1)(-1) – 4]|/[√(1)^2 + (1)^2]

= |-5|/√2

= 5/√2 units

Q.7 The points A(2, 3), B(4, -1) and C(-1, 2) are the vertices of a triangle. Find the length of perpendicular from A to BC and hence find the area of AABC.

Ans. In a △ABC, A(2, 3), B(4, -1), C(-1, 2) are vertex

Equation of line, (y – y1) = [(y2 – y1)/(x2 – x1)] (x – x1), we get,

Equation of BC,

(y + 1) = [(2 + 1)/(-1 – 4)] (x – 4)

(y + 1) = 3/-5 (x – 4)

-5y – 5 = 3x – 12

3x + 5y – 7 = 0

Formula for Distance of a point from a line, d = |(Ax1 + By1 + C)|/(√A^2+B^2), we get,

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

AD = |[3(2) + (5)(3) – 7]|/[√(3)^2 + (5)^2]

= |6 + 15 – 7|/√9 + 25

= 14/√34 units

BC = √(x2 – x1)^2+(y2 – y1)^2

=√(2 + 1)^2+(-1-4)^2

= √9+25 = √34

Area of △ = 1/2 x BC x AD

= 1/2 x √34 x 14/√34

= 7 sq units.

Q.8 Find the length of perpendicular from the point (4, -7) to the line joining the origin and the point of intersection of the lines 2x – 3y + 14 = 0 and 5x + 4y – 7 = 0.

Ans. Equation of l1, 2x – 3y + 14 = 0

Equation of l2, 5x + 4y – 7 = 0

From the first equation, 2x – 3y = -14

y = (2x + 14)/3

Substituting y = (2x + 14)/3 in second equation,

5x + 4[(2x + 14)/3] – 7 = 0

5x + (8x + 56)/3 – 7 = 0

(15x + 8x + 56 – 21)/3 = 0 [Multiplying by 3]

(23x + 35)/3 = 0

23x + 35 = 0, x = -35/23

Substitute x = -35/23 into y = (2x + 14)/3 :

y = [2(-35/23) + 14)/3]

= [(-70/23) + (322/23)]/3

= (252/23)/3 = 252/69 = 84/23

Thus, the point of intersection is (-35/23, 84/23).

Equation of line : y – 0 = [(84/23 – 0)]/[(-35/12)] (x – 0)

-35y = 84x

84x + 35y = 0

12x + 5y + 0 = 0

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

= |12 x 4 + 5 x (-7) + 0|/[√(12)^2 + (5)^2]

= |48 – 35|/√144+25

= 13/√169

= 13/13

= 1

Q.9 A vertex of a rectangle is at the origin and two of its sides lie along the lines 5x – 12y + 26 = 0 and 12x + 5y – 39 = 0. Find the area of the rectangle.

Ans. Using, point P(0, 0), line 12x + 5y – 39 = 0

Formula for Distance of a point from a line, d = |(Ax1 + By1 + C)|/(√A^2+B^2), we get,

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

= |[12(0) + (5)(0) – 39]|/[√(2)^2 + (5)^2]

= |-39|/√144 + 25

= 39/√169

= 39/13 = 3 units

Now, point P (0, 0) and Equation 5x – 12y + 26 = 0,

Formula for Distance of a point from a line, d = |(Ax1 + By1 + C)|/(√A^2+B^2), we get,

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

= |[5(0) + (-12)(0) + 26]|/[√(5)^2 + (-12)^2]

= |26|/13

= 2 units

Therefore, area of rectangle = l x b

= 3 x 2 = 6 sq. units

Q.10 Find the equations of lines with slope -1 and perpendicular distance from origin equal to 5 units.

Ans. General equation of line :

Ax + By + C = 0

Slope = -A/B

-1 = -A/B

B = A

Formula for Distance of a point from a line, d = |(Ax1 + By1 + C)|/(√A^2+B^2), we get,

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

= |[A(0) + (B)(0) + C]|/[√(A)^2 + (A)^2]

= |C|/√2A^2

5 = |C|/√2 A

Taking ‘+’ sign, C/√2 A = 5, C = 5√2 A

Taking ‘-‘ sign, C/√√2 A = -5, C = -5√2 A

Equation of line,

Ax + Ay + 5√2 A = 0

x + y + 5√2 = 0 or x + y – 5√2 = 0

Q.11 Find the equations of the lines through the point of intersection of the lines 3x + y + 3 = 0 and x – y – 3 = 0 and whose distance from the origin is 3 units.

Ans. Equation of l1, 3x + y + 3 = 0

Equation of l2, x – y – 3 = 0

From the second equation, x – y – 3 = 0

x – y = 3, y = x – 3

Substituting y = x – 3 in first equation,

3x + (x – 3) + 3 = 0,

4x = 0, x = 0

Substitute x = 0 into y = x – 3:

y = 0 – 3 = -3

Thus, the point of intersection is (0, -3).

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

= |C|/[√(A)^2 + (B)^2]……..(i)

Line passes through (0, -3)

= A(0) + B(-3) + C = 0

C = 3B

From eq (i)

3 = |3B|/√A^2+B^2

9 = 9B/(A^2+B^2)

A^2 + B^2 = B^2

A = 0

Equation of line,

= Ax + By + C = 0

= 0 + By + 3B = 0

= y + 3 = 0

Q.12 Find the equations of the two straight lines drawn through the point (0, 1) on which the perpendiculars dropped from the point (2, 2) are each of unit length.

Ans. y = mx + c

As it passes through (0, 1)

1 = m x 0 + c

So, c = 1

Therefore, equation of line, y = mx + 1

mx – y + 1 = 0…….(i)

Formula for Distance of a point from a line, d = |(Ax1 + By1 + C)|/(√A^2+B^2), we get,

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

1 = |[m(2) + (-1)(2) + 1]|/[√(m)^2 + (-1)^2]

1 = |2m – 1|/√m^2+1

Squaring both the sides,

1 = (2m – 1)^2/(m^2 + 1)

m^2 + 1 = 4m^2 – 4m + 1

3m^2 – 4m = 0

m(3m – 4) = 0

m = 0, m = 4/3

From equation (i),

If m = 0, then -y + 1 = 0, y – 1 = 0

If m = 4/3, then 4/3x – y + 1 = 0, 4x – 3y + 3 = 0

Q.13 Find the points on the x-axis whose perpendicular distance from the line x/3 + y/4 = 1 is 4 units.

Ans. Point of x-axis (k, 0)

Equation of line, x/3 + y/4 = 1

(4x + 3y)/12 = 1

4x + 3y = 12

4x + 3y – 12 = 0

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

4 = |[4(k) + (3)(0) – 12]|/[√(4)^2 + (3)^2]

4 = |4k – 12|/√25

1 = |k – 3|/5

|k – 3| = 5

Taking ‘+’ sign, k – 3 = 5, k = 8

Taking ‘-‘ sign, k – 3 = -5, k = -2

Therefore, points (8, 0) or (-2, 0)

Q.14 Find the points on the line y = x which are at a distance of 5 units from the line 4x + 3y – 1 = 0.

Ans. Here, Points (k, k) & equation of line 4x + 3y – 1 = 0, d = 5 units

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

5 = |[4(k) + (3)(0) – 12]|/[√(4)^2 + (3)^2]

5 = |7k – 1|/√25

5 = |7k – 1|/5

|7k – 1| = 25

Taking ‘+’ sign, 7k – 1 = 25,

7k = 26, k = 26/7

Taking ‘-‘ sign, 7k – 1 = -25,

7k = -24, k = -24/7

Therefore, points (26/7, 26/7) or (-24/7, -24/7).

Q.15 Show that the line 5x – 2y – 1 = 0 is mid-parallel to the lines 5x – 2y + 7 = 0 and 5x – 2y – 9 = 0.

Ans. Equation of l1, 5x – 2y – 1 = 0

Equation of l2, 5x – 2y + 7 = 0

Equation of l3, 5x – 2y – 9 = 0

Distance between l1 & l2,

d = |(C1 – C2)|/(√A^2+B^2)

= |1 – 7|/[√(5)^2 + (-2)^2]

= 8/√29

Now, distance between l1 & l3,

d = |(Ax1 + By1 + C)|/(√A^2+B^2)

= |1 – (-9)|/[√(5)^2 + (-2)^2]

= 8/√29

Both distance are equal

Therefore, l1 is mid parallel to l2 & l3.

Q.16 If two sides of a square lie on the lines 5x – 12y + 26 = 0 and 5x – 12y – 65 = , find its area.

Ans. Here, Equation of l1 5x – 12y + 26 = 0, Equation of l2, 5x – 12y – 65 = 0

Where both lines are parallel,

d = |(C1 – C2)|/(√A^2+B^2)

= |26 – (-65)|/[√(5)^2 + (-12)^2]

= |26 + 65|/√25+144

= 91/√169

= 91/13 = 7

Therefore, side of square = 7 units

Now, Area of square = Side x Side

= 7 x 7

= 49 sq. units

FAQ’s related to Class 11 Applied Maths Chapter 17 on Straight Line:

Q.1 What is the general equation of a straight line?

Ans. The general equation of a straight line is given by: Ax + By + C = 0 where A, B, and C are constants, and x and y are variables representing the coordinates of any point on the line.

Q.2 What is the slope-intercept form of a straight line?

Ans. The slope-intercept form of a straight line is: y = mx + c where m is the slope of the line, and c is the y-intercept (the point where the line crosses the y-axis).

Q.3 How do you find the slope of a line given two points?

Ans. The slope mmm of a line passing through two points (x1, y1) and (x2, y2) is calculated using the formula: m = (y2 – y1)/(x2 – x1)

Q.4 What is the equation of a horizontal line?

Ans. The equation of a horizontal line is : y = k where k is a constant representing the y-coordinate of all points on the line.

Q.5 What is the equation of a vertical line?

Ans. The equation of a vertical line is: x = k where k is a constant representing the x-coordinate of all points on the line.

These are a few Frequently Asked Questions relating to Class 11 Applied Maths Chapter 17

In Class 11 Applied Maths chapter 17, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Class 11 Applied Maths Chapter 17 promises an enriching learning experience that will set you on the path to success. Class 11 Applied Maths Chapter 17, we delve deep into advanced mathematical concepts that are crucial for understanding.