• Slope (gradient) of a line
• Equations: point-slope, slope-intercept, intercept form, two-point form
• Angle between two lines
• Perpendicular distance from a point
• Application in demand curves
2. Circle
• Circle as a locus
• Standard form: (x–h)² + (y–k)² = r²
• Central, diameter & general forms
• Finding centre and radius
• Real-world applications
3. Parabola
• Definition: focus and directrix
• Standard equation: y² = 4ax
• Axis, vertex, latus rectum
• Eccentricity (e = 1)
• Applications: reflectors, arches
The slope of a line making an angle of 45° with the positive direction of the x-axis is:
A0
B1
C√3
DUndefined
✓ Correct Answer: (B) 1
Explanation: Slope m = tan θ = tan 45° = 1. The slope is the tangent of the angle of inclination with the positive x-axis.
Question 2Straight Line
The equation of a line with slope 3 passing through the point (1, –2) is:
Ay = 3x + 5
By = 3x – 5
Cy = –3x + 1
Dy = 3x + 1
✓ Correct Answer: (B) y = 3x – 5
Explanation: Using point-slope form: y – y₁ = m(x – x₁) y – (–2) = 3(x – 1) y + 2 = 3x – 3 y = 3x – 5
Question 3Straight Line
The perpendicular distance of the point (3, 4) from the line 3x – 4y + 5 = 0 is:
A2/5
B2
C1
D5
✓ Correct Answer: (A) 2/5
Explanation: Using the perpendicular distance formula: d = |Ax₁ + By₁ + C| / √(A² + B²) Here A = 3, B = –4, C = 5, and the point is (3, 4). d = |3(3) – 4(4) + 5| / √(3² + (–4)²) = |9 – 16 + 5| / √(9 + 16) = |–2| / √25 = 2 / 5
Question 4Straight Line
Two lines are parallel if their slopes m₁ and m₂ satisfy:
Am₁ × m₂ = –1
Bm₁ = m₂
Cm₁ + m₂ = 0
Dm₁ – m₂ = 1
✓ Correct Answer: (B) m₁ = m₂
Explanation: Two lines are parallel when they have equal slopes (m₁ = m₂) and different y-intercepts. Two lines are perpendicular when m₁ × m₂ = –1.
Question 5Straight Line
The equation of a line with x-intercept 4 and y-intercept –3 (intercept form) is:
Ax/4 + y/3 = 1
Bx/4 – y/3 = 1
C3x + 4y = 12
Dx/3 + y/4 = 1
✓ Correct Answer: (B) x/4 – y/3 = 1
Explanation: Intercept form: x/a + y/b = 1, where a = x-intercept = 4, b = y-intercept = –3. So: x/4 + y/(–3) = 1 → x/4 – y/3 = 1.
Question 6Circle
The centre and radius of the circle (x – 3)² + (y + 2)² = 25 are:
ACentre (3, 2), radius 25
BCentre (–3, 2), radius 5
CCentre (3, –2), radius 5
DCentre (3, –2), radius 25
✓ Correct Answer: (C) Centre (3, –2), radius 5
Explanation: Comparing with central form (x–h)² + (y–k)² = r²: h = 3, k = –2, r² = 25 → r = 5. Centre = (3, –2), radius = 5.
Question 7Circle
The standard equation of a circle with centre at the origin and radius r is:
Ax + y = r
Bx² + y² = r
Cx² + y² = r²
D(x – r)² + (y – r)² = 0
✓ Correct Answer: (C) x² + y² = r²
Explanation: The standard form of a circle centred at origin (0, 0) with radius r is x² + y² = r². This is obtained from (x–0)² + (y–0)² = r².
Question 8Circle
In the general equation of a circle x² + y² + 2gx + 2fy + c = 0, the radius is given by:
A√(g² + f² – c)
B√(g² + f² + c)
Cg² + f² – c
D√(g + f – c)
✓ Correct Answer: (A) √(g² + f² – c)
Explanation: The general form x² + y² + 2gx + 2fy + c = 0 has centre (–g, –f) and radius = √(g² + f² – c). The condition g² + f² – c > 0 ensures a real circle.
Question 9Circle
The equation of a circle with diameter endpoints A(2, 0) and B(–2, 0) is:
Ax² + y² = 2
Bx² + y² = 4
Cx² + y² = 16
D(x–2)(x+2) = 0
✓ Correct Answer: (B) x² + y² = 4
Explanation: Centre = midpoint of AB = (0, 0). Radius = distance from centre to A = 2. Equation: x² + y² = 4.
Question 10Circle
A circle is the locus of a point that moves such that its distance from a fixed point is:
AVariable
BZero
CConstant
DEqual to the x-coordinate
✓ Correct Answer: (C) Constant
Explanation: A circle is defined as the locus of all points that are at a constant distance (called the radius) from a fixed point (called the centre).
Question 11Parabola
The standard equation of a parabola opening to the right is:
Ax² = 4ay
By² = 4ax
Cy² = –4ax
Dx² = –4ay
✓ Correct Answer: (B) y² = 4ax
Explanation: y² = 4ax (a > 0) opens to the right. y² = –4ax opens to the left. x² = 4ay opens upward. x² = –4ay opens downward.
Question 12Parabola
For the parabola y² = 12x, the focus is at:
A(0, 3)
B(3, 0)
C(–3, 0)
D(12, 0)
✓ Correct Answer: (B) (3, 0)
Explanation: Comparing y² = 12x with y² = 4ax: 4a = 12 → a = 3. Focus of y² = 4ax is at (a, 0) = (3, 0).
Question 13Parabola
The eccentricity of a parabola is always:
A0
BLess than 1
CGreater than 1
DEqual to 1
✓ Correct Answer: (D) Equal to 1
Explanation: The eccentricity e = 1 for a parabola. (e < 1 → ellipse/circle; e = 1 → parabola; e > 1 → hyperbola.)
Question 14Parabola
The length of the latus rectum of the parabola y² = 8x is:
A2
B4
C8
D16
✓ Correct Answer: (C) 8
Explanation: Comparing y² = 8x with y² = 4ax: 4a = 8 → a = 2. Length of latus rectum = 4a = 4 × 2 = 8.
Question 15Parabola
A parabola is defined as the locus of a point that is equidistant from:
ATwo fixed points
BA fixed point and a fixed circle
CA fixed point (focus) and a fixed line (directrix)
DA fixed line only
✓ Correct Answer: (C) A fixed point (focus) and a fixed line (directrix)
Explanation: A parabola is the set of all points P such that the distance from P to the focus equals the distance from P to the directrix. This gives eccentricity e = 1.
📋 Assertion-Reason Questions
Statement I is Assertion (A) and Statement II is Reason (R). Choose the correct option:
(a) Both A and R are True and R is the correct explanation of A
(b) Both A and R are True but R is not the correct explanation of A
(c) A is True but R is False
(d) A is False but R is True
Assertion-Reason 1Straight Line
Assertion (A): Two lines with slopes m₁ = 2 and m₂ = –½ are perpendicular to each other. Reason (R): Two lines are perpendicular if and only if the product of their slopes equals –1.
(a) Both A and R are True and R is the correct explanation of A
(b) Both A and R are True but R is not the correct explanation of A
(c) A is True but R is False
(d) A is False but R is True
✓ Correct Answer: (a) Both A and R are True and R is the correct explanation of A
m₁ × m₂ = 2 × (–½) = –1, so the lines are perpendicular — Assertion is true. The condition m₁ × m₂ = –1 is the standard criterion for perpendicularity — Reason is true and directly explains A.
Assertion-Reason 2Circle
Assertion (A): The equation x² + y² – 4x + 6y + 13 = 0 represents a point circle (radius = 0). Reason (R): For the general form x² + y² + 2gx + 2fy + c = 0, if g² + f² – c = 0, the circle degenerates into a single point.
(a) Both A and R are True and R is the correct explanation of A
(b) Both A and R are True but R is not the correct explanation of A
(c) A is True but R is False
(d) A is False but R is True
✓ Correct Answer: (a) Both A and R are True and R is the correct explanation of A
2g = –4 → g = –2; 2f = 6 → f = 3; c = 13. Radius = √(4 + 9 – 13) = √0 = 0, so it is a point circle — Assertion is true. The Reason states the exact condition for this — it is true and is the correct explanation of A.
Assertion-Reason 3Parabola
Assertion (A): The parabola y² = –8x opens towards the left side of the coordinate plane. Reason (R): For the parabola y² = 4ax, if a > 0 it opens to the right; if a < 0 it opens to the left.
(a) Both A and R are True and R is the correct explanation of A
(b) Both A and R are True but R is not the correct explanation of A
(c) A is True but R is False
(d) A is False but R is True
✓ Correct Answer: (a) Both A and R are True and R is the correct explanation of A
4a = –8 → a = –2 (negative). Since a < 0, the parabola opens to the left — Assertion is true. The Reason correctly states the direction rule based on the sign of a — it is true and directly explains A.
Assertion-Reason 4Straight Line
Assertion (A): The slope of a vertical line (parallel to the y-axis) is zero. Reason (R): Slope m = tan θ, where θ is the angle of inclination. For a vertical line θ = 90° and tan 90° is undefined.
(a) Both A and R are True and R is the correct explanation of A
(b) Both A and R are True but R is not the correct explanation of A
(c) A is True but R is False
(d) A is False but R is True
✓ Correct Answer: (d) A is False but R is True
The slope of a vertical line is undefined, not zero (a horizontal line has slope = 0) — Assertion is false. The Reason is true: for a vertical line θ = 90° and tan 90° is undefined, which is exactly why the slope does not exist.
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Find the equation of a straight line passing through (3, –2) and perpendicular to the line 4x – 3y = 10. Also find the perpendicular distance from the origin to this line.
A circular garden has its diameter defined by the points A(–4, 3) and B(4, –3). Find the equation of the circular boundary and verify whether the point P(0, 5) lies inside, on, or outside the garden.
Complete Solution:
Centre = midpoint of AB = ((–4+4)/2, (3–3)/2) = (0, 0)
Radius = distance from centre to A = √((–4–0)² + (3–0)²) = √(16 + 9) = 5
Equation of circle: x² + y² = 25
For P(0, 5): 0² + 5² = 25. Since 25 = r², P lies exactly ON the circle.
Equation: x² + y² = 25 · Point P(0, 5) lies exactly ON the boundary of the garden.
Question 4Parabola
Find the equation of the parabola whose vertex is at the origin, axis is along the x-axis, and it passes through the point (4, 6). Also find its focus and directrix.
Complete Solution:
Since vertex is at origin and axis along x-axis, equation is y² = 4ax
The parabola passes through (4, 6): 6² = 4a(4) → 36 = 16a → a = 9/4
Find the equation of a circle passing through the origin, with centre at (2, –3). Also find the equation in general form.
Complete Solution:
Radius = distance from centre (2, –3) to origin (0, 0) = √(4 + 9) = √13
Equation (central form): (x – 2)² + (y + 3)² = 13
Expanding: x² – 4x + 4 + y² + 6y + 9 = 13
x² + y² – 4x + 6y + 13 – 13 = 0
General form: x² + y² – 4x + 6y = 0
Question 6Straight Line
A demand curve for a product is a straight line. When the price is ₹50, the quantity demanded is 200 units. When price drops to ₹30, quantity demanded rises to 400 units. Find the demand equation and estimate quantity demanded when price is ₹40.
Complete Solution:
Let p = price, q = quantity. Two points on demand curve: (q₁, p₁) = (200, 50) and (q₂, p₂) = (400, 30)
Slope = (30 – 50)/(400 – 200) = –20/200 = –1/10
Using point-slope: p – 50 = (–1/10)(q – 200) p = –q/10 + 20 + 50 = –q/10 + 70
A market research firm studied the demand for a product. The data showed that when price p = ₹100, quantity demanded q = 50 units, and when price p = ₹60, quantity demanded q = 90 units. Assuming the demand curve is a straight line (as in linear demand models used in economics), answer the following questions.
Q1
What is the slope of the demand curve?
A–1
B–2
C1
D2
✓ Correct Answer: (A) –1
Slope (treating p as y-axis, q as x-axis): m = (p₂–p₁)/(q₂–q₁) = (60–100)/(90–50) = –40/40 = –1. Negative slope reflects the law of demand.
Q2
The demand equation is:
Ap = q + 150
Bp = –q + 150
Cp = –q + 50
Dp = 2q – 100
✓ Correct Answer: (B) p = –q + 150
Using point-slope: p – 100 = –1(q – 50) → p = –q + 50 + 100 → p = –q + 150.
Q3
At what price will the quantity demanded be zero? (Price at which no one buys)
A₹50
B₹100
C₹150
D₹200
✓ Correct Answer: (C) ₹150
When q = 0: p = –(0) + 150 = ₹150. This is the y-intercept of the demand curve — the price at which demand drops to zero.
Q4
How many units will be demanded when the price is ₹80?
A50 units
B60 units
C70 units
D80 units
✓ Correct Answer: (C) 70 units
When p = 80: 80 = –q + 150 → q = 70 units.
Case Study 2Circle — Real World Design
Circular Fountain Design
An architect is designing a circular fountain for a city park. The fountain is to be placed such that its centre is at coordinates (5, –3) on the design plan and it must pass through the point (2, 1) on the plan. Another decorative lamp post is to be placed at coordinates (5, 2).
Q1
What is the radius of the fountain?
A3 units
B4 units
C5 units
D√25 = 5 units
✓ Correct Answer: (C) 5 units
r = distance from centre (5,–3) to point (2,1) = √((5–2)² + (–3–1)²) = √(9 + 16) = √25 = 5 units.
Q2
The equation of the fountain's circular boundary is:
A(x – 5)² + (y + 3)² = 5
B(x – 5)² + (y + 3)² = 25
C(x + 5)² + (y – 3)² = 25
Dx² + y² = 25
✓ Correct Answer: (B) (x – 5)² + (y + 3)² = 25
Using central form with h = 5, k = –3, r = 5: (x–5)² + (y–(–3))² = 25 → (x–5)² + (y+3)² = 25.
Q3
Does the lamp post at (5, 2) lie inside, on, or outside the fountain?
AInside the fountain
BOn the boundary
COutside the fountain
DAt the centre
✓ Correct Answer: (B) On the boundary
Substitute (5, 2) into the circle equation (x–5)² + (y+3)² = 25: (5–5)² + (2+3)² = 0 + 25 = 25 = r² Since the value equals r², the point lies exactly ON the boundary of the fountain.
A civil engineer is designing a parabolic arch for a bridge. When placed on a coordinate system with the vertex at the origin, the arch follows the equation x² = –16y (opening downward). Satellite dish antennae and car headlights also use parabolic shapes. Answer the following questions based on the parabola y² = 16x.
Q1
For the parabola y² = 16x, the value of 'a' is:
A2
B4
C8
D16
✓ Correct Answer: (B) 4
Comparing y² = 16x with y² = 4ax: 4a = 16 → a = 4.
Q2
The focus of the parabola y² = 16x is at:
A(0, 4)
B(–4, 0)
C(4, 0)
D(16, 0)
✓ Correct Answer: (C) (4, 0)
Focus of y² = 4ax is at (a, 0) = (4, 0). In satellite dishes, the receiver is placed at the focus to collect all incoming parallel signals that reflect off the parabolic surface.
Q3
The equation of the directrix of y² = 16x is:
Ax = 4
Bx = –4
Cy = 4
Dy = –4
✓ Correct Answer: (B) x = –4
Directrix of y² = 4ax is x = –a = –4. Every point on the parabola is equidistant from focus (4,0) and directrix x = –4.
Q4
The length of the latus rectum and the eccentricity of y² = 16x are respectively:
A16 and e = 1
B4 and e = 0
C8 and e = 1
D16 and e = 2
✓ Correct Answer: (A) 16 and e = 1
Latus rectum = 4a = 4 × 4 = 16. Eccentricity of every parabola is always e = 1 (defining property).
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