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Class 11 Applied Maths Chapter 17 Solutions
Straight Line
EXERCISE- 17.4
Q.1 Find the equation of a straight line parallel to x-axis at a distance
(i) 3 units above it
(ii) 3 units below it.
Ans. (i) The line is parallel to the x-axis and lies 3 units above it. Therefore, the equation of the line is:
y – 3 = 0
y = 3
(ii) The line is parallel to the x-axis and lies 3 units below it. Therefore, the equation of the line is:
y + 3 = 0
y = -3
Q.2 Find the equation of a straight line parallel to y-axis at a distance
(i) 2 units to the right
(ii) 2 units to the left of it.
Ans. (i) The line is parallel to the y-axis and lies 2 units to the right of it. Therefore, the equation of the line is: x – 2 = 0
x = 2
(ii) The line is parallel to the y-axis and lies 2 units to the left of it. Therefore, the equation of the line is:
x + 2 = 0
x = -2
Q.3 (i) Find the equation of a straight line parallel to the x-axis and having intercept -2 on the y-axis.
(ii) Find the equations of lines parallel to axes and passing through the point (-2, 3).
Ans. (i) A line parallel to the x-axis has a constant y-coordinate. The given intercept of −2 on the y-axis means the y-coordinate is −2 for all points on the line. Thus, the equation of the line is :
y + 2 = 0
y = -2
(ii) A line parallel to the x-axis passing through (−2,3) has a constant y-coordinate equal to 3. Its equation is: y – 3 = 0
y = 3
A line parallel to the y-axis passing through (−2,3) has a constant x-coordinate equal to −2. Its equation is:
x + 2 = 0
x = -2
Q.4 Find the equation of a horizontal line passing through the point (5,-2).
Ans. A horizontal line has a constant y-coordinate for all points on the line. Since the given point is (5,−2) the y-coordinate of the line is −2.
Thus, the equation of the horizontal line passing through (5,−2) is :
y = −2.
Q.5 (i) Find the equation of the line passing through (0, 0) with slope m.
(ii) Find the equation of the line passing through the point (-2, 3) with slope -4.
Ans. (i) We know that, y – y1 = m(x – x1)
Here, slope : m
x1 = 0, y1 = 0
Therefore, equation of line :
y – 0 = m (x – 0)
y = mx
(ii) ATQ, m = -4, x1 = -2, y1 = 3
Using, y – y1 = m (x – x1), we get,
y – 3= (-4) [x – (-2)]
y – 3 = -4 (x + 2)
y – 3 = -4x – 8
4x + y -3 + 8 = 0
Equation of line : 4x + y + 5 = 0
Q.6 Find the equation of the line passing through the point (-1, 1) and which is parallel to the line joining the points (2, 3) and (4, -1).
Ans. Given : A line (l1) passing through (-1, 1) is parallel to another line (l2) passing through (2, 3) & (4, -1).
Solution : For line l2, we have,
x1 = 2, y1 = 3
x2 = 4, y2 = -1
Using, m = (y2 – y1)/(x2 – x1), we get,
Slope of l2 (m2) = (-1 – 3)/(4 – 2) = -4/2 = -2
Therefore, m2 = -2
ATQ, l1 || l2
Therefore, m1 = m2 = -2
Now, for l1,
slope (m1) = -2, x1 = -1, y1 = 1
Using, y – y1 = m (x – x1), we get,
y – 1 = -2 [x – (-1)]
y – 1 = -2 (x + 1)
y – 1 = -2x – 2
2x + y- 1 + 2 = 0
Therefore, equation of l1 : 2x + y + 1 =0
Q.7 Find the equation of the straight line joining the points:
(i) (2, 3) and (4, 1)
(ii) (1,-1) and (3, 5).
Ans. (i) ATQ, x1 = 2, y1 = 3
x2 = 4, y2 = 1
Using, y – y1 = [(y2 – y1)/(x2 – x1)] (x – x1), we get,
y – 3 = [(1 – 3)/(4 – 2)] (x – 2)
y – 3 = -2/2 (x – 2)
y – 3 = -x + 2
x + y – 3 – 2 = 0
Equation of line : x + y – 5 = 0
(ii) ATQ, x1 = 3, y1 = 5
x2 = 1, y2 = -1
Using, y – y1 = [(y2 – y1)/(x2 – x1)] (x – x1), we get,
y – 5 = [(-1 – 5)/(1 – 3)] (x – 3)
y – 5 = -6/-2 (x – 3)
y – 5 = 3x – 9
0 = 3x – y – 9 + 5
Equation of line : 3x – y – 4 = 0
Q.8 Find the equation of a line whose
(i) slope = -2/7, y-intercept = -3
(ii) inclination = 135°, y-intercept = -5.
Ans. (i) ATQ, Slope (m) = -2/7
y – intercept (c) = -3
Using, y = mx + c (slope intercept form), we get,
y = -2/7x + (-3)y = -2x/7 – 3/1
y = (-2x – 21)/7
7xy = -2x – 21
7y = -2x – 21
Equation of line : 2x + 7y + 21 = 0
(ii) Inclination θ = 135° , y – intercept = -5
We know that, slope (m) tan θ
Here, m = tan 135°
m = tan (180° – 45°)
m = -tan 45°
m = -1
Using, y = mx + c, we get,
y = (-1)x + (-5)
y = -x – 5
Required equation : x + y + 5 = 0
Q.9 Find the equation of the line intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of x-axis.
Ans. ATQ, y – intercept (c) = 2
Inclination (θ) = 30°
We know that, m = tan θ
m = tan 30°
Therefore, m = 1/√3
Using, y = mx + c, we get,
y = 1/√3x + 2
y = (x + 2√3)/√3
√3y = x + 2√3
Therefore, required equation
x – √3y + 2√3 = 0
Q.10 Find the equation of the straight line passing through the point (5, 7) and inclined at 45° to x-axis. If it passes through the point P whose ordinate is -7, what is the abscissa of P?
Ans. Given, A line passes through (5, 7) & inclined at 45°
ATQ, θ = 45°
We know that, m = tan θ
m = tan 45°
Therefore, m = 1
Here, x1 = 5, y1 = 7 & m = 1
Using, y – y1 = m (x – x1), we get,
y – 7 = 1 (x – 5)
y – 7 = x – 5
0 = x – y – 5 + 7
Required equation : x – y + 2 = 0
Now, ordinate of P is -7 (y)
Since, P lies on the line, it would satisfy the equation of line.
On substituting value of y, we get,
x – (-7) + 2 = 0
x + 7 + 2 = 0
x + 9 = 0
x = -9
Q.11 Find the equation of the line passing through the point (2,-5) and making an intercept of −3 on the y-axis.
Ans. A line passes through (2, -5) & y – intercept (c) = -3
ATQ, y – intercept = -3 , line passes through (0, -3)
Here, x1 = 2, y1 = -5
x2 = 0, y2 = -3
Using, y – y1 = [(y2 – y1)/(x2 – x1)] (x – x1), we get,
y – (-5) = [(-3 – (-5))/(0 – 2)] (x – 2)
y + 5 = [(-3 + 5)/-2] (x – 2)
y + 5 = 2/-2 (x – 2)
(-1) (y + 5) = x – 2
-y – 5 = x – 2
0 = x + y + 5 – 2
Required equation = x + y + 3 = 0
Q.12 Find the equation of a straight line whose y-intercept is -5 and which is
(i) parallel to the line joining the points (3, 7) and (-2, 0)
(ii) perpendicular to the line joining the points (-1, 6) and (-2, -3).
Ans. (i) Given : A line (l1), with y – intercept (c) as -5, is parallel to line (l2), joining points (3, 7) & (-2, 0).
For l2, x1 = 3, y1 = 7, x2 = -2, y2 = 0
Using, m = [(y2 – y1)/(x2 – x1)], we get,
Slope of l2 (m2) = [(0 – 7)/(-2 – 3)] = -7/-5 = 7/5
ATQ, l1 || l2
Therefore, m1 = m2 = 7/5
For l2 : slope (m2) = 7/5
y – intercept (c) = -5
Using, y = mx + c, we get,
y = 7/5x + (-5)
y = (7x -25)/5
5y = 7x – 25
0 = 7x – 5y – 25
Therefore, required equation :
7x – 5y – 25 = 0
(ii) Given, A line (l1), having y – intercept as -, is perpendicular to another line (l2), joining points (1, 6) & (-2, -3).
For, l2, x1 = -1, y1 = 6, x2 = -2, y2 = -3
We know that, m = (y2 – y1)/(x2 – x1), we get,
Therefore, slope of l2 (m2) = (-3 – 6)/(-2 – (-1)) = -9/-1 = 9
Since, l1 ⊥ l2, therefore, m1 x m2 = -1
m1 x 9 = -1
So, m1 = -1/9
For l1 : slope (m1) = -1/9
c = -5
Using, y = mx + c, we get,
y = -1/9x + (-6)
y = (-x – 45)/9
9y = -x – 45
x + 9y + 45 = 0
Therefore, required equation : x + 9y + 45 = 0
Q.13 (i) Find the equation of the line through the point (-5, 1) and parallel to the line joining the points (7,-1) and (0,3).
(ii) Find the equation of the straight line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3,-1).
Ans. (i) Given, A line (l1), passing through (-5, 1) is parallel to another line (l2), joining points (7, -1) & (0, 3).
For, l2, x1 = 7, y1 = -1, x2 = 0, y2 = 3
We know that, m = (y2 – y1)/(x2 – x1), we get,
Therefore, slope of l2 (m2) = (3 – (-1))/(0 – 7) = 4/-7
Since, l1 || l2, therefore, m1 = m2 = 4/-7
For l1 : slope = 4/-7
Point P(-5, 1) [lies on l1]
Therefore, x = -5, y = 1
Using, y – y1 = m (x – x1), we get,
y – 1 = 4/-7 [x – (-5)]
-7(y – 1) = 4 (x + 5)
-7y + 7 = 4x + 20
0 = 4x + 7y + 20 – 7
Therefore, required equation : 4x + 7y + 13 = 0
(ii) Given, A line (l1) passing through (5, 2) is perpendicular to another line (l2), which is passing through (2, 3) & (3, -1).
For, l2, x1 = 2, y1 = 3, x2 = 3, y2 = -1
We know that, m = (y2 – y1)/(x2 – x1), we get,
Therefore, slope of l2 (m2) = (-1 – 3)/(3 – 2) = -4/1 = -4
Since, l1 ⊥ l2, therefore, m1 x m2 = -1
m1 x -4 = -1
So, m1 = 1/4
For l1 : slope (m1) = 1/4
Point P(5, 2) [lies on l1]
Therefore, x1 = 5, y1 = 2
Using, y – y1 = m(x – x1), we get,
y – 2 = 1/4 (x – 5)
4(y – 2) = x – 5
4y – 8 = x – 5
0 = x- 4y – 5 + 8
Therefore, required equation : x – 4y + 3 = 0
Q.14 If the points (1, 4), (3, -2) and (p, -5) lie on a straight line, find the value of p.
Ans. Given : Point A(1, 4), (B3, -2) and C(p, -5) lie on a straight line(l1).
Here, considering A & B we have,
x1 = 2, x2 = 3, y1 = 4, y2 = -2
Using, y – y1 = [(y2 – y1)/(x2 – x1)] (x – x1), we get,
y – 4 = [(-2 – 4)/(3 – 1)] (x – 1)
y – 4 = -6/2 (x – 1)
y – 4 = -3 (x – 1)
y – 4 = -3x + 3
Therefore, 3x + y – 7 = 0
Required equation : 3x + y – 7 = 0
Point c also lies on l1
So, values of x & y coordinate of point c should satisfy the equation.
3 X p + (-5) – 7 = 0
3p – 12 = 0
3p = 12
p = 4
Q.15 The vertices of a triangle PQR are P(2, 1), Q(-2, 3) and R(4, 5). Find the equation of the median through the vertex R.
Ans. Given : A △PQR with coordinate of vertices P(2, 1), Q(-2, 3) and R(4, 5)
Let, median pass through vertex R bisect PQ at Z.
Now, Z is mid point of PQ, (Median bisects the side)
Using, mid point formula, we get,
Coordinate of Z = [(2 – 2)/2], [(1 + 3)/2]
Coordinate of Z =(0, 2)
For line l1 : (median RZ)
x1 = 4, x2 = 0, y1 = 5, y2 = 2
Using, y – y1 = [(y2 – y1)/(x2 – x1)] (x – x1), we get,
y – 5 = [(2 -5)/(0 – 4)] (x – 4)
y – 5 = -3/-4 (x -4)
4 (y – 5) = 3(x – 4)
4y – 20 = 3x – 12
0 = 3x -4y + 20 – 12
Therefore, required equation : 3x – 4y + 8 = 0
Q.16 Write down the equation of the line whose slope is 3/2 and which passes through P where P divides the line segment joining A(-2, 6) and B (3,-4) in the ratio 2: 1.
Ans. Given, A line l1, with slope 3/2, point P lies on l1.
P divides line segment AB with coordinate (-2, 6) & (3, -4) respectively is ratio of 2:1.
For P dividing AB :
x1 = -2, x2 = 3, m1 = 2
y1 = 6, y2 = -4, m2 = 1
Using, section formula, we get,
Coordinate of P : x = (2 x 3 + 1 x -2)/(2 + 1), y = (2 x -4 + 1 x 6)/(2 + 1)
Coordinate of P : (6 – 2)/3, (-8 + 6)/3
Coordinate of P : 2/3, -2/3
For line (l1) :Slope (m) = 3/2
Point P lies on l1
Therefore, x1 = 2/3, y1 = -2/3
Using, y – y1 = m (x – x1), we get,
y – (-2/3) = 3/2 (x – 2/3)
(3y + 2)/3 = 3/2 [(3x – 2)/3]
2(3y + 2) = 3(3x – 4)
6y + 4 = 9x – 12
0 = 9x – 6y – 4 – 12
Required equation : 9x – 6y – 16 = 0
Q.17 Points A and B have coordinates (3, 2) and (7, 6) respectively. Find
(i) the equation of the right bisector of segment AB
(ii) the value of p if (-2, p) lies on it.
Ans. (i) Given : A line segment AB with coordinated (3, 2) & (7, 6)
Let, line l1 be the right bisector of AB.
Foe AB : x1 =3, x2 = 7, y1 = 2, y2 = 6
We know that, m = (y2 – y1)/(x2 – x1), we get,
m (slope of AB) = (6 – 2)/(7 – 3) = 4/4 = 1
Since, l1 is right bisector of AB .
Therefore, slope of AB x slope of l1 = -1
1 x slope of l1 = -1
So, slope of l1 = -1
Since, l1 is bisector of AB,
Therefore, mid point of AB (say P) would lie on l1.
Using, mid point formula, we get,
Coordinate of P : (3 + 7)/2, (2 + 6)/2
Coordinate of P : (5, 4)
For l1 : slope = -1
Point P lies on l1
Therefore, x = 5, y = 4
Using, y – y1 = m (x – x1), we get,
y – 4 = -1 (x – 5)
y – 4 = -x + 5
x + y – 4 – 5 = 0
x + y – 9 = 0
Required equation : x + y – 9 = 0
(ii) Equation of line : x + y – 9 = 0
Now, point Q (-2, p) lies on l1
Therefore, x & y coordinate of Q would satisfy the equation.
-2 + p – 9 = 0
p – 11 = 0
p = 11
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.
