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Class 11 Applied Maths Chapter 17 Solutions
Straight Line
EXERCISE- 17.3
Q.1 Show that the line joining (2,-3) and (-5, 1) is
(i) parallel to the line joining (7,-1) and (0,3).
(ii) perpendicular to the line joining (4, 5) and (0,-2).
Ans. (i) Given : A line, l1, passing through points (2, -3) & (-5, 1)
Another line, l2, passing through points (7, -1) & (0, 3)
Using, m (y2 – y1)/(x2 – x1), we get,
m1 (slope of l1) = (1 – (-3))/(-5 – 2) = 4/-7
Similarly, m2 (slope of l2) = (3 – (-1))/(0 – 7) = 4/-7
Since, m1 = m2 = 4/-7
Therefore, l1 || l2.
(ii) Given : line, l3, passing through points (4, 5) & (0, -2)
m3 (slope of l3) = (-2 – 5)/(0 – 4) = -7/-4
= 7/4
Now, m1 x m3 = 4/-7 x 7/4 = -1
Since, m1 x m3 = -1
Therefore, l1 ⊥ l3.
Q.2 Find the value of x so that the line through (x, 9) and (2, 7) is parallel to the line through (2,-2) and (6, 4).
Ans. Given, A line, passing through points (x, 9) & (2, 7) is parallel to line, l2, passing through (2, -2) & (6, 4)
Using, m = (y2 – y1)/(x2 – y1), we get,
m1 (slope of l1) = (7 – 9)/(2 – x) = -2/(2 – x)
m2 (slope of l2) = (4 – (-2))/(6 – 2) = 6/4 = 3/2
ATQ, l1 || l2
Therefore, m1 = m2
-2/(2 – x) = 3/2
-2 x 2 = 3 (2 – x)
-4 = 6 – 3x
3x = 6 + 4
x = 10/3
Q.3 The line joining (-5, 7) and (0,-2) is perpendicular to the line joining (1,-3) and (4, y). Find y.
Ans. A line, l1, passing through points (-5, 7) & (0, -2) is perpendicular to line, l2, passing through (1, -3) & (4, y).
Using, m = (y2 – y1)/(x2 – x1), we get,
m1 (slope of l1) = (-2 – 7)/(0 – (-5)) = -9/5
m2 (slope of l2) = (y – (-3))/(4 – 1) = (y + 3)/3
ATQ, l1 ⊥ l2
Therefore, m1 x m2 = -1
-9/5 x (y + 3)/3 = -1
[3(y + 3)]/3 – 1
3y + 9 = 5
3y = 5 – 9
y = -4/3
Q.4 Using slopes, determine which of the following sets of three points are collinear:
(i) (−2, 3), (8, −5), (5, 4)
(ii) (5, 1), (1,-1), (11,4)
Ans. (i) A(−2, 3), B(8, −5), C(5, 4)
Using, m = (y2 – y1)/(x2 – x1), we get,
m1 (slope of AB) = (-5 – 3)/(8 – (-2)) = -8/10 = -4/5
m2 (slope of AC) = (4 – 3)/(5 – (-2)) = 1/7
Since m1 ≠ m2
Therefore, A, B & C are not collinear.
(ii)Given : Three points P(5, 1), Q(1,-1), R(11,4)
Using, m = (y2 – y1)/(x2 – x1), we get,
m1 (slope of PQ) = (-1 – 1)/(1 – 5) = -2/-4 = 1/2
m2 (slope of QR) = (4 – (-1))/(11 – 1) = 5/10 = 1/2
Since m1 = m2
Therefore, PQ & QR lie on the same line,
P, Q & R are collinear.
Q.5 Using slopes, find the value of x so that the points (6, -1), (5, 0) and (x, 3) are collinear.
Ans. Three collinear points, A(6, -1), B(5, 0) and C(x, 3)
Using, m = (y2 – y1)/(x2 – x1), we get,
m1 (slope of AB) = (0 – (-1))/(5 – 6) = 1/-1 = -1
m2 (slope of BC) = (3 – 0)/(x – 5) = 3/(x – 5)
ATQ, A, B & C are collinear
Therefore, m1 = m2.
3/(x – 5) = -1
3 = -1 (x – 5)
3 = -x + 5
x = 5 – 3 = 2
So, x = 2
Q.6 If A(-2, 1), B(2, 3) and C(-2,-4) are three points, find the angle between the lines AB and BC.
Ans. Given, three points A(-2, 1), B(2, 3) and C(-2,-4)
Using, m = (y2 – y1)/(x2 – x1), we get,
m1 (slope of AB) = (3 – 1)/(2 – (-2)) = 2/4 = 1/2
m2 (slope of BC) = (-4 – 3)/(-2 – 2) = -7/-4 = 7/4
Using, tan θ = |(m2 – m1)/(1 + m1m2)|, we get,
tan θ = |[(7/4) – (1 x 2/2)]/[1 + (7/4 x 1/2)]|
tan θ = |[(7 – 4)/2]/[1 + (7/8)]|
tan θ = |(5/4)/[(8 + 7)/8]|
tan θ = |(5/4)/(15/8)|
tan θ = |5/4 x 8/15|
tan θ = 2/3
θ = tan ^-1(2/3)
Therefore, θ is an acute angle.
Q.7 Without using Pythagoras theorem, show that the points (-2, 2), (8,-2) and (-4,-3) are the vertices of a right angled triangle.
Ans. Given, A triangle ABC with coordinates A(-2, 2), B(8,-2) C(-4,-3).
To prove : ΔABC is a right angled Δ.
Proof,
Using, m = (y2 – y1)/(x2 – x1), we get,
m1 (slope of AB) = (-2 – 2)/(8 – (-2)) = -4/10 = -2/5
m2 (slope of BC) = (-3 – (-2))/(-4 – 8) = -1/-12 = 1/12
m3 (slope of AC) = (-3 – 2)/(-4 – (-2)) = -5/-2 = 5/2
Now, m1 x m3 = -2/5 x 5/2
m1 x m3 = -1
Therefore, AB ⊥ AC
A is a right angle.
Therefore, ΔABC is a right angled Δ with ∠A being right angle.
Q.8 If A(2,-3) and B (3, 5) are two vertices of a rectangle ABCD, find the slope of
(i) BC
(ii) CD
(iii) DA.
Ans. Given, A rectangle ABCD with coordinates of A(2,-3) and B (3, 5)
Using, m = (y2 – y1)/(x2 – x1), we get,
m1 (slope of AB) = (5 – (-3))/(3 – 2) = 8/1
(i) Since ABCD is a rectangle BC ⊥ AB
Therefore, slope of BC x slope of AB = -1
Slope of BC x 8 = -1
Therefore, slope of BC = -1/8
(ii) Since, ABCD is a rectangle CD || AB (opposite sides)
Therefore, Slope of CD = Slope of AB
Slope of CD = 8
(iii) Since, ABCD is a rectangle DA || BC (opposite sides)
Therefore, Slope of DA = Slope of BC
Slope of DA = -1/8.
Q.9 Using slopes, prove that the points (-4, -1), (-2,-4), (4, 0) and (2, 3), taken in order, are the vertices of a rectangle.
Ans. Given, A quadrilateral PQRS with coordinates P(-4, -1), Q(-2,-4), R(4, 0) and S(2, 3).
To prove : PQRS is a rectangle
Proof :
Using, m = (y2 – y1)/(x2 – x1), we get,
m1 (slope of PQ) = (-4 – (-1))/(-2 – (-4)) = (-4 + 1)/(-2 + 4) = -3/2
m2 (slope of QR) = (0 – (-4))/(4 – (-2)) = (0 + 4)/(4 + 2) = 4/6 = 2/3
m3 (slope of RS) = (3 – 0)/(2 – 4) = 3/-2 = -3/2
m4 (slope of SP) = (-1 – 3)/(-4 – 2) = -4/-6 = 2/3
Now, m1 = m3 & m2 = m4
We know that, when slopes are equal, line segment are parallel.
Therefore, PQ||RS & QR||SP
So, PQRS is a ||gm.
Also, m1 x m2 = -3/2 x 2/3 = -1
Therefore, PQ ⊥ QR.
PQRS is a parallelogram with atleast one right angle.
So. PQRS is a rectangle.
Q.10 The vertices of a quadrilateral are (-4, 2), (2, 6), (8, 5) and (9,-7). Using slopes, show that the mid-points of the sides of the quadrilateral form a parallelogram.
Ans. Coordinates of midpoints :
P(-1, 4), Q(5. 11/2), R(17/2, -1), S(5/2, -5/2)
Using, m = (y2 – y1)/(x2 – x1), we get,
m1 (slope of PQ) = [(11/2) – 4)/(5 – (-1)) = [(11-8)/2]/[(6/1) = 3/2 X 1/6 = 1/4
m2 (slope of QR) = [-1 – (11/2)]/[(17/2) – 5] = [(-2 – 11)/2]/[(17-10)/2] = -13/7
m3 (slope of RS) = [(-5/2)-(-1)]/[(5/2)-(17/2)] = [(-5+2)/2]/[(5-17)/2] = -3/-12 = 1/4
m4 (slope of SP) = [(-5/2) – 4][(5/2) – (-1)] = [(-5-8)/2]/[(5+2)/2] = -13/7
Now, m1 = m3 & m2 = m4
Therefore, PQ||RS & QR||SP
(Line segment with equal slopes are parallel to each others)
Therefore, PQRS is a ||gm.
Q.11 If the points A(x1, y1), B(x2, y2) and P(h, k) are collinear, prove that (h – x1) (y2 – y1) = (k – y1) (x2 – x1).
Ans. Using, m = (y2 – y1)/(x2 – x1), we get,
Slope of AB = (y2 – y1)/(x2 – x1)
Slope of AP = (k – y1)/h – x1)
ATQ, A, B, and P are collinear.
Therefore, Slope of AB = Slope of AP
So, (y2 – y1)/(x2 – x1) = (k – y1)/(h – x1)
(h – x1)(y2 – y1) = (k – y1)(x2 – x1)
Hence proved.
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.
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