All questions
- Prove that: 1/(1 – sinθ) + 1/(1 + sinθ) = 2sec²θ Answer: LHS simplifies to RHS, as shown below. LHS=(1+sinθ)+(1-sinθ)/(1-sinθ)(1+sinθ) =2/1-sin²θ=2/cos²θ =2sec²θ=RHS 1
- If A(1, -1), B(0, 4), C(-5, 3) are vertices of a triangle then find the slope of each side. Answer: Slope of AB = -5, slope of BC = 1/5, slope of AC = -2/3 Slope =y₂-y₁/x₂-x₁ . AB: 4-(-1)/0-1=5/-1=-5 BC: 3-4/-5-0=-1/-5=1/5 AC: 3-(-1... 1
- Show that A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are the vertices of a parallelogram. Answer: AB ∥ CD and BC ∥ DA (equal slopes), so ABCD is a parallelogram. Slope =y₂-y₁/x₂-x₁ . Slope of AB =2+7/-1+4=3 , slope of CD =-4-5/5-8... 1
- Find k, if R(1, -1), S(-2, k) and slope of line RS is -2. Answer: k = 5 k-(-1)/-2-1=-2\ ⇒\ k+1/-3=-2 k+1=6\ ⇒\ k=5 1
- Find k, if B(k, -5), C(1, 2) and slope of the line is 7. Answer: k = 0 2-(-5)/1-k=7\ ⇒\ 7/1-k=7 1-k=1\ ⇒\ k=0 1
- Find k, if PQ ∥ RS and P(2, 4), Q(3, 6), R(3, 1), S(5, k). Answer: k = 5 Parallel lines have equal slopes. Slope of PQ =6-4/3-2=2 . Slope of RS =k-1/5-3=k-1/2 . k-1/2=2\ ⇒\ k=5 1
- Seg AB is parallel to Y-axis and coordinates of point A are (1, 3) then co-ordinates of point B can be ........ (A) (3, 1) (B) (5, 3) (C) (3, 0) (D) (1, -3) Answer: (D) (1, -3) All points on a segment parallel to the Y-axis have the same x-coordinate. A has x = 1, so B must also have x = 1. Only... 1
- Out of the following, point ........ lies to the right of the origin on X-axis. (A) (-2, 0) (B) (0, 2) (C) (2, 3) (D) (2, 0) Answer: (D) (2, 0) A point on the X-axis has y = 0, and to the right of the origin it has x > 0. (-2, 0) is to the left, (0, 2) is on the Y-... 1
- Distance of point (-3, 4) from the origin is ........ (A) 7 (B) 1 (C) 5 (D) -5 Answer: (C) 5 d=√((-3)²+4²)=√(25)=5 A distance can never be negative, so (D) is impossible. 1
- A line makes an angle of 30° with the positive direction of X-axis. So the slope of the line is ........ (A) 1/2 (B) √3/2 (C) 1/√3 (D) √3 Answer: (C) 1/sqrt3 Slope =tan30^°=1/sqrt3 . ( 1/2 is sin30^° , sqrt3/2 is cos30^° , and sqrt3 is tan60^° .) 1
- Determine whether the given points are collinear: A(0, 2), B(1, -0.5), C(2, -3). Answer: The points are collinear. Slope =y₂-y₁/x₂-x₁ . Slope of AB =-0.5-2/1-0=-2.5 ; slope of BC =-3+0.5/2-1=-2.5 . The slopes are equal an... 1
- Determine whether the given points are collinear: P(1, 2), Q(2, 8/5), R(3, 6/5). Answer: The points are collinear. Slope =y₂-y₁/x₂-x₁ . Slope of PQ =8/5-2/2-1=-2/5 ; slope of QR =6/5-8/5/3-2=-2/5 . The slopes are equal an... 1
- Determine whether the given points are collinear: L(1, 2), M(5, 3), N(8, 6). Answer: The points are not collinear. Slope =y₂-y₁/x₂-x₁ . Slope of LM =3-2/5-1=1/4 ; slope of MN =6-3/8-5=1 . The slopes are different, so... 1
- Find the coordinates of the midpoint of the line segment joining P(0, 6) and Q(12, 20). Answer: (6, 13) Midpoint =(0+12/2,6+20/2)=(6,13) . 1
- Find the ratio in which the line segment joining the points A(3, 8) and B(-9, 3) is divided by the Y-axis. Answer: AP : PB = 1 : 3 (the dividing point is P(0, 27/4)) A point on the Y-axis has x = 0. Let P(0, y) divide AB in the ratio m:n . 0=m(-9)... 1
- Find the point on X-axis which is equidistant from P(2, -5) and Q(-2, 9). Answer: (-7, 0) Let the point be A(x, 0). AP²=AQ² : (x-2)²+25=(x+2)²+81 x²-4x+29=x²+4x+85 -8x=56\ ⇒\ x=-7 1
- Find the distance between the points A(a, 0), B(0, a). Answer: AB=asqrt2 Distance formula: d=√((x₂-x₁)²+(y₂-y₁)²) . AB=√((0-a)²+(a-0)²)=√(2a²)=asqrt2 (taking a>0 ; in general the distance is |a|s... 1
- Find the distance between the points P(-6, -3), Q(-1, 9). Answer: PQ = 13 Distance formula: d=√((x₂-x₁)²+(y₂-y₁)²) . PQ=√((-1+6)²+(9+3)²)=√(25+144)=√(169)=13 1
- Find the distance between the points R(-3a, a), S(a, -2a). Answer: RS = 5a Distance formula: d=√((x₂-x₁)²+(y₂-y₁)²) . RS=√((a+3a)²+(-2a-a)²) =√(16a²+9a²)=√(25a²)=5a (taking a>0 ; in general 5|a| ). 1
- Find the coordinates of the circumcentre of a triangle whose vertices are (-3, 1), (0, -2) and (1, 3). Answer: (-1/3, 2/3) The circumcentre P(x, y) is equidistant from A(-3, 1), B(0, -2), C(1, 3). PA²=PB² : (x+3)²+(y-1)²=x²+(y+2)² 6x+10-2y=4y+... 1
- Can the segments joining the points L(6, 4), M(-5, -3), N(-6, 8) form a triangle? If triangle is formed, state the type of the triangle considering sides of the triangle. Answer: Yes, a triangle is formed; it is a scalene triangle. Distance formula: d=√((x₂-x₁)²+(y₂-y₁)²) . LM=√(11²+7²)=√(170)≈13.04 MN=√(1²+11... 1
- Can the segments joining the points P(-2, -6), Q(-4, -2), R(-5, 0) form a triangle? If triangle is formed, state the type of the triangle considering sides of the triangle. Answer: No, the points are collinear, so no triangle is formed. Distance formula: d=√((x₂-x₁)²+(y₂-y₁)²) . PQ=√((-2)²+4²)=√(20)=2sqrt5 QR=√(... 1
- Can the segments joining the points A(√2, √2), B(-√2, -√2), C(-√6, √6) form a triangle? If triangle is formed, state the type of the triangle considering sides of the triangle. Answer: Yes; it is an equilateral triangle (each side 4). AB²=(2sqrt2)²+(2sqrt2)²=8+8=16 BC²=(sqrt6-sqrt2)²+(sqrt6+sqrt2)² =2(6+2)=16 AC²=(s... 1
- Find k if the line passing through points P(-12, -3) and Q(4, k) has slope 1/2. Answer: k = 5 k-(-3)/4-(-12)=1/2\ ⇒\ k+3/16=1/2 k+3=8\ ⇒\ k=5 1