Welcome to Class 11 Applied Maths Chapter 17, where we embark on an exciting journey into the world of advanced mathematical concepts tailored for Class 11 students.” Unlock the power of applied mathematics with expert solutions crafted by professionals at AppliedMath.com. Designed to propel students towards academic success, our meticulously curated ML Aggarwal Solutions for Applied Mathematics cater to Class 11 and class 12 students seeking mastery in their examinations. Every query from the CBSE ML Aggarwal Books finds a comprehensive answer on our platform, complete with detailed explanations and step-by-step solutions presented in an easily understandable language.
Dive into the world of applied mathematics and discover how our resources can elevate your understanding and performance. Keep reading to explore the wealth of ML Aggarwal Solutions for Class 11 and Class 12 Applied Mathematics.
Here we provide you with Class 11 Applied Maths Chapter 17, to help you gain a comprehensive understanding of the chapter and its concepts. https://appliedmathsolution.com/wp-admin/post.php?post=6&action=edit
Class 11 Applied Maths Chapter 17 Solutions
Straight Line
EXERCISE- 17.5
Q.1 Find the equation of the line which makes intercepts -3 and 2 on the x-axis and y-axis respectively.
Ans. Given, A line l1, whose x & y intercept are -3 & 2 respectively.
Using, x/a + y/b = 1, we get,
x/-3 + y/2 = 1
(-2x + 3y)/6 = 1
-2x + 3y = 6
0 = 2x – 3y + 6
Therefore, required equation :
2x – 3y + 6 = 0
Q.2 Find the equation of a straight line which passes through the point (1,-3) and makes an intercept on y-axis twice as long as on x-axis.
Ans. A straight line, l, passes through P(1, -3), whose y- intercept (b) twice of x- interceprt (a).
ATQ, b = 2a……..(i)
Using, x/a + y/b = 1, we get,
x/a + y/2a = 1,
(2x + y)/2 = 1
Therefore, equation of line (l) :
2x + y = 2a…….(ii)
ATQ, P(1, -3) lies on the line
Therefore, x & y coordinate of P would satisfy the equation.
2 x 1 + (-3) = 2a
2 – 3 = 2a
-1 = 2a
So, a = -1/2
Therefore, b = -1 (from 1)
On substituting value of ‘a’ in equation (ii), we get,
2x + y = 2 x -1/2
2x + y = -1
Therefore, required equation : 2x + y + 1 = 0
Q.3 Find the equation of the straight line which passes through the point (2, 3) and makes intercepts equal in magnitude but opposite in sign on the axes.
Ans. A straight line, l, passes through P(2, 3), whose y- intercept (b) and x- intercept (a) are equal in magnitude but opposite in sign.
ATQ, if x intercept is a, then y- intercept (b) would by ‘-a’.
Using, x/a + y/b = 1, we get,
x/a + y/-a = 1
(x – y)/a = 1
Equation of line (l) :
x – y = a…………..(i)
ATQ, P(2, 3) lies on the line.
Therefore, x & y coordinate of P would satisfy the equation,
So, 2 – 3 = a
-1 = a
Therefore, b = -a = -(-1) = 1
Substituting value of a in eq, (i), we get,
x – y = -1
Therefore, required equation : x – y + 1 = 0
Q.4 Find the equation of the straight line which passes through the point (3,-2) and cuts off positive intercepts on the x-axis and y-axis which are in the ratio 4: 3.
Ans. Given, A straight line, l, passes through P(3, -2), and has x & y intercept in the ratio 4 : 3.
ATQ, a/b = 4/3
Therefore, if a = 4t, b = 3t
Using, x/a + y/b = 1, we get,
x/4t + y/3t = 1
1/t[x/4 + y/3] = 1
x/4 + y/3 = t
(3x + 4y)/12 = t
Equation of line (l) :
3x + 4y = 12t…………..(i)
Now, ATQ, point P with x & y coordinate as 3 & -2 lies on the line.
3 x 3 + 4 x -2 = 12t
9 – 8 = 12t
1 = 12t
t = 1/12
Substituting value of ’12t’ in eq, (i), we get,
3x + 4y = 1
Therefore, required equation : 3x + 4y – 1 = 0
Q.5 Find the equation of the line whose length of perpendicular segment from the origin to the line is 4 units and the inclination of the perpendicular segment with positive direction of x-axis is 30°.
Ans. ATQ, Length of perpendicular (p) = 4
Angle of perpendicular (α) = 30°
Using, x cosα + y sinα = p, we get,
x cos 30° + y sin 30° = 4
x *3/2 + y*1/2 = 4
√3x/2 + y2 = 4
(√3x + y)/2 = 4
√3x + y = 8
Required equation : √3x + y – 8 = 0
Q.6 Prove that the straight line whose intercepts on the axes are 2 and -3 respectively passes through the point (4,3).
Ans. Given : A line (l), whose x & y intercept are 2 & -3.
To prove : Line passes through point P(4, 3)
Proof : Using, x/a + y/b = 1, we get,
x/ + y/-3 = 1
x/2 – y/2 = 1
3x/6 – 2y/6 = 1
(3x – 2y)/6 = 1
Therefore, equation of line (l) : 3x – 2y = 6
If P lies on the line, its x & y coordinate should satisfy the equation of line.
Now, for x = 4 & y = 3, we have
3x 4 – 2 x 3 = 6
12 – 6 = 6
6 = 6
LHS = RHS
Equation is satisfied & hence, (4, 3) lies on the line.
Q.7 A straight line passes through the point P(1, 2) and the portion of the line intercepted between the axes is bisected at this point, find its equation.
Ans. Given, A line, l, intercepts x & y axis at point A & B. Also, P(1, 2) which is mid point of AB lies on the line.
Let, coordinate of A & B be (c, 0) & (0, d) respectively,
ATQ, P is mid point of AB .
Therefore, using mid point formula, we get,
1 = (c + 0)/2 ; 2 = (0 + d)/2
So, c = 2 & d = 4
Therefore, x- intercept = 2 & y- intercept = 4
Using, x/a + y/b = 1, we get,
x/2 + y/4 = 1
(2x + y)/4 = 1
2x + y = 4
Therefore, required equation : 2x + y – 4 = 0
Q.8 If the intercept of a line between the coordinate axes is divided by the point (-5, 4) in the ratio 1:2, then find the equation of the line.
Ans. Given, A straight line (l), which intersects x & y axis at A & B respectively. Another point P(-5, 4) divides AB in the ratio of 1 : 2.
For AB, using section formula, we get,
-5 = (1 x 0 + 2 x c)/3, 4 = (1 x d + 2 x 0)/3
-15/2 = c , 12 = d
x- intercept = -15/2 (a), y- intercept = 12 (b)
Now, using x/a + y/b = 1, we get,
x/(-15/2) + y/12 = 1
-2x/15 + y/12 = 1
(-8x + 5y)/60 = 1
-8x + 5y = 60
Required equation : -8x + 5y – 60 = 0
Q.9 Find the equation of line(s) which passes through the point (22, -6) and whose intercept on the x-axis exceeds the intercept on y-axis by 5.
Ans. Given, A straight line, l1, passes through P(22, -6) x intercept of l1 exceeds y- intercept by 5.
ATQ, a = b + 5
Using, x/a + y/b = 1, we gte,
x/(b+5) + y/b = 1……..[Equation of line l1]
Since, P lies on l1 x & y coordinate of P would satisfy the equation of line.
22/(b+5) + -6/b = 1
[22b – 6(b + 5)][(b + 5)b] = 1
[22b – 6b -30]/[b^2 + 5b] = 1
16b – 30 = b^2 + 5b
0 = b^2 + 5b – 16b + 30
b^2 – 11b + 30 = 0
b(b – 6) – 5(b – 6 ) = 0
(b – 6) (b – 5) = 0
Therefore, b = 6 or b = 5
a = 11, a = 10 [From a = b + 5]
If b = 6, then a = 11
If b = 5, then a = 10
Using, x/a + y/b = 1, we get,
x/11 + y/6 = 1 ; x/10 + y/5 = 1
(6x + 11y)/66 = 1 ; (x + 2y)/10 = 1
6x + 11y = 66 ; x + 2y = 10
Required equation : 6x + 11y – 66 = 0, x + 2y – 10 = 0
Q.10 Find the equation of a line which passes through (2, 2) and sum of its intercepts on the axes is 9.
Ans. Given, line (l), sum of whose intercepts (a & b) is 9 passes through (2, 2)
ATQ, a + b = 9
Therefore, b = 9 – a
Using, x/a + y/b = 1, we get,
x/a + y/(9 – a) = 1 [Equation of line]
ATQ, Point P(2, 2) lies on the line.
2/a + 2/(9 -a) = 1
[2(9 – a) + 2a]/[a(9 – a)] = 1
(18 – 2a + 2a)/(9a – a^2) = 1
18 = 9a – a^2
a^2 – 9a + 18 = 0
a^2 – 6a – 3a + 18 = 0
a(a – 6) – 3(a – 6) = 0
(a – 6) (a – 3) = 0
Therefore, a = 6 or 3
b = 3, or 6
Q.11 If p is the length of perpendicular from origin to the line and a is the angle which this perpendicular makes with the positive direction of x-axis, then sketch and write the equation of the line when :
(i) p = 5, a = 135°
(ii) p = 1, α = 90°
Ans. (i) Using, x cos α + y sin α = p, we get,
x cos 135° + y sin 135° = 5
x cos (180 – 45) + y sin (180 – 45) = 5
x * -cos 45° + y sin 45° = 5
x * -1/√2 + y * 1/√2 = 5
-x/√2 + y/√2 = 5
(y – x)/√2 = 5
y – x = 5√2
(ii) Using, x cos α + y sin α = p, we get,
x cos 90° + y sin 90° = 1
x * 0 + y * 1 = 1
So, y = 1
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.
