NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1 – Straight Lines
NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1 introduces the foundational concepts of straight lines in coordinate geometry. This exercise focuses on finding slopes of lines, understanding the conditions for parallel and perpendicular lines, collinearity of points, and the angle between two lines. These concepts form the basis for all further topics in Chapter 9 and are highly important for board examinations and competitive exams like JEE.
Question 1. Draw a quadrilateral in the Cartesian plane, whose vertices are (−4, 5), (0, 7), (5, −5) and (−4, −2). Also, find its area.
Let the vertices of the quadrilateral be A(−4, 5), B(0, 7), C(5, −5) and D(−4, −2).
NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1 – Straight Lines
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By plotting these points and joining them in order, we get the quadrilateral ABCD.
To find the area, we divide the quadrilateral into two triangles: △ABC and △ACD.
Area of △ABC with A(−4, 5), B(0, 7), C(5, −5): Area = (1/2)|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| = (1/2)|−4(7 − (−5)) + 0((−5) − 5) + 5(5 − 7)| = (1/2)|−4(12) + 0 + 5(−2)| = (1/2)|−48 − 10| = (1/2)(58) = 29 sq. units
Area of △ACD with A(−4, 5), C(5, −5), D(−4, −2): Area = (1/2)|−4((−5) − (−2)) + 5((−2) − 5) + (−4)(5 − (−5))| = (1/2)|−4(−3) + 5(−7) + (−4)(10)| = (1/2)|12 − 35 − 40| = (1/2)|−63| = 63/2 sq. units
Area of quadrilateral ABCD = 29 + 63/2 = 58/2 + 63/2 = 121/2 sq. units
Question 2. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle.
Since the base lies along the y-axis and its midpoint is at the origin, the two base vertices are A(0, a) and B(0, −a).
For an equilateral triangle with side 2a, the third vertex C lies on the x-axis.
Using the distance formula, distance from C(x, 0) to A(0, a): √(x² + a²) = 2a x² + a² = 4a² x² = 3a² x = ±√3a
Therefore, the vertices of the triangle are (0, a), (0, −a) and (√3a, 0) or (0, a), (0, −a) and (−√3a, 0).
Question 3. Find the distance between P(x₁, y₁) and Q(x₂, y₂) when: (i) PQ is parallel to the y-axis (ii) PQ is parallel to the x-axis.
i. When PQ is parallel to the y-axis:
When PQ is parallel to the y-axis, the x-coordinates of P and Q are equal, i.e., x₁ = x₂.
Distance PQ = √((x₂ − x₁)² + (y₂ − y₁)²) = √(0 + (y₂ − y₁)²)
Therefore, PQ = |y₂ − y₁|.
ii. When PQ is parallel to the x-axis:
When PQ is parallel to the x-axis, the y-coordinates of P and Q are equal, i.e., y₁ = y₂.
Distance PQ = √((x₂ − x₁)² + (y₂ − y₁)²) = √((x₂ − x₁)² + 0)
Therefore, PQ = |x₂ − x₁|.
Question 4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Let the point on the x-axis be P(a, 0).
Given that P is equidistant from A(7, 6) and B(3, 4): PA² = PB² (a − 7)² + (0 − 6)² = (a − 3)² + (0 − 4)² a² − 14a + 49 + 36 = a² − 6a + 9 + 16 −14a + 85 = −6a + 25 −8a = −60 a = 15/2
Therefore, the required point on the x-axis is (15/2, 0).
Question 5. Find the slope of a line which passes through the origin, and the mid-point of the line segment joining the points P(0, −4) and B(8, 0).
The mid-point M of the line segment joining P(0, −4) and B(8, 0) is: M = ((0 + 8)/2, (−4 + 0)/2) = (4, −2)
Slope of the line passing through origin O(0, 0) and M(4, −2): m = (−2 − 0)/(4 − 0) = −2/4
Therefore, the slope of the line is −1/2.
Question 6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (−1, −1) are the vertices of a right-angled triangle.
Let A(4, 4), B(3, 5) and C(−1, −1).
Slope of AB = (5 − 4)/(3 − 4) = 1/(−1) = −1 Slope of BC = (−1 − 5)/(−1 − 3) = −6/(−4) = 3/2 Slope of CA = (4 − (−1))/(4 − (−1)) = 5/5 = 1
Slope of AB × Slope of CA = −1 × 1 = −1
Since the product of slopes of AB and CA is −1, lines AB and CA are perpendicular to each other.
Therefore, the triangle formed by these points is a right-angled triangle, with the right angle at A.
Question 7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
If a line makes an angle of 30° with the positive y-axis, then it makes an angle of (90° + 30°) = 120° with the positive x-axis.
Slope = tan 120° = tan(180° − 60°) = −tan 60° = −√3
Therefore, the slope of the line is −√3.
Question 8. Find the value of x for which the points (x, −1), (2, 1) and (4, 5) are collinear.
Slope of line through (x, −1) and (2, 1): m₁ = (1 − (−1))/(2 − x) = 2/(2 − x)
Slope of line through (2, 1) and (4, 5): m₂ = (5 − 1)/(4 − 2) = 4/2 = 2
For collinearity, m₁ = m₂: 2/(2 − x) = 2 2 = 4 − 2x 2x = 2 x = 1
Therefore, the value of x is 1.
Question 9. Without using distance formula, show that the points (−2, −1), (4, 0), (3, 3) and (−3, 2) are the vertices of a parallelogram.
Let A(−2, −1), B(4, 0), C(3, 3) and D(−3, 2).
Slope of AB = (0 − (−1))/(4 − (−2)) = 1/6 Slope of DC = (2 − 3)/(−3 − 3) = −1/(−6) = 1/6
Since slope of AB = slope of DC, AB ∥ DC.
Slope of BC = (3 − 0)/(3 − 4) = 3/(−1) = −3 Slope of AD = (2 − (−1))/(−3 − (−2)) = 3/(−1) = −3
Since slope of BC = slope of AD, BC ∥ AD.
Since both pairs of opposite sides are parallel, ABCD is a parallelogram.
Hence proved that the given points are the vertices of a parallelogram.
Question 10. Find the angle between the x-axis and the line joining the points (3, −1) and (4, −2).
Slope of the line joining (3, −1) and (4, −2): m = (−2 − (−1))/(4 − 3) = −1/1 = −1
If θ is the angle that the line makes with the positive x-axis: tan θ = m = −1 θ = 135°
Therefore, the angle between the x-axis and the given line is 135°.
Question 11. The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.
Let the slope of one line be m, and the slope of the other line be 2m.
tan θ = |(m − 2m)/(1 + m·2m)| = |−m/(1 + 2m²)| = 1/3
Case 1: −m/(1 + 2m²) = 1/3 ⟹ 2m² + 3m + 1 = 0 ⟹ (2m + 1)(m + 1) = 0 ⟹ m = −1/2 or m = −1
If m = −1, slopes are −1 and −2 If m = −1/2, slopes are −1/2 and −1
Case 2: −m/(1 + 2m²) = −1/3 ⟹ 2m² − 3m + 1 = 0 ⟹ (2m − 1)(m − 1) = 0 ⟹ m = 1/2 or m = 1
If m = 1, slopes are 1 and 2 If m = 1/2, slopes are 1/2 and 1
Question 12. A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that k − y₁ = m(h − x₁).
The slope of the line passing through (x₁, y₁) and (h, k) is: m = (k − y₁)/(h − x₁)
Multiplying both sides by (h − x₁):
k − y₁ = m(h − x₁) (Hence proved)
Question 13. If three points (h, 0), (a, b) and (0, k) lie on a line, show that a/h + b/k = 1.
For collinearity, slope of (h, 0) to (0, k) = slope of (h, 0) to (a, b).
Slope of (h, 0) to (0, k) = (k − 0)/(0 − h) = −k/h
Slope of (h, 0) to (a, b) = b/(a − h)
Setting equal: b/(a − h) = −k/h bh = −k(a − h) = −ka + kh bh − kh = −ka
Dividing by kh: b/k − 1 = −a/h
Therefore, a/h + b/k = 1. (Hence proved)
Question 14. From the graph, line AB passes through A(1985, 92) and B(1995, 97). Find slope and population in 2010.
Slope of AB: m = (97 − 92)/(1995 − 1985) = 5/10 = 1/2
Equation of line AB using point A(1985, 92): y − 92 = (1/2)(x − 1985)
For the year 2010 (x = 2010): y − 92 = (1/2)(2010 − 1985) y − 92 = (1/2)(25) = 12.5 y = 104.5
Therefore, the slope of AB is 1/2 and the population in 2010 will be 104.5 crores.
Frequently Asked Questions (FAQs)
Q1. What is the formula for the slope of a line passing through two points? The slope of a line passing through two points (x₁, y₁) and (x₂, y₂) is given by m = (y₂ − y₁)/(x₂ − x₁). It represents the steepness and direction of the line.
Q2. What is the condition for two lines to be parallel? Two lines are parallel if and only if their slopes are equal, i.e., m₁ = m₂. Parallel lines never intersect each other.
Q3. What is the condition for two lines to be perpendicular? Two lines are perpendicular if and only if the product of their slopes is −1, i.e., m₁ × m₂ = −1. This means their slopes are negative reciprocals of each other.
Q4. How do you check if three points are collinear using slopes? Three points A, B, and C are collinear if the slope of AB equals the slope of BC (or AC). If all three slopes are equal, the points lie on the same straight line.
Q5. How many questions are there in NCERT Class 11 Maths Exercise 9.1? Exercise 9.1 contains 14 questions that cover concepts such as slope of a line, angle between two lines, collinearity of points, conditions for parallel and perpendicular lines, and application-based problems involving coordinates.