CBSE
Gujarat Board
Haryana Board
Class 10
Class 12
The orthocentre of the triangle formed by the lines x + y = 1 and 2y2 - xy - 6x2 = 0
23, 23
23, - 23
43, - 43
A.
43, 43
We have x + y = 1and 2y2 - xy - 6x2 = 0⇒ 2y2 - 4xy + 3xy - 6x2 = 0⇒ Equation of sides of ∆ABC arex + y = 1, 2y + 3x = 0 and y - 2x = 0
Solving these equations simultaneously, we get A0, 0B13, 23 and C - 2, 3Equation of altitude AD isx - y = 0 ... iEquation of altitude CF isx + 2y = λSince, this passes through - 2, 3∴ - 2 + 6 = λ ⇒ λ = 4On solving eqs. i and ii, we getx = 43, y = 43
The mid-point of the line segment joining the centroid and the orthocentre of the triangle whose vertices are (a, b),(a, c) and (d, c), is
5a + d6, b + 5c6
`a + 5d6, 5b + c6
(a, 0)
(0, 0)
Centroid of ∆ABC = a +a + d3, b + c + c3≡ 2a + d3, 2c + b3Since, ∆ABC is a, cMid point of centroid and orthocentre is= 2a + d3 + a2, 2c + b3 + c2= 5a + d6, 5c + b6
If the pair of straight lines xy - x - y + 1 = 0 and the line ax + 2y - 3a = 0 are concurrent, then a is equal to
0
1
- 1
3
B.
We have,xy - x - y + 1 = 0 ...i ax + 2y - 3a = 0 ...iiOn compairing Eqs. i with,ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we geta = 0, b = 0, h = 12, g = - 12, f = - 12, c=1The point of concurrent is hf - bgab - h2, gh - afab - h2 = 12 - 12 - 00 - 122, - 1212 - 00 - 122 = 1, 1The point lies on equation ii, thena + 2 - 3a = 0 ⇒ 2a = 2 a = 1
The longest distance of the point (a, 0) from the curve 2x2 + y2 = 2x is
1 + a
1 - a
1 - 2a +2a2
1 - 2a + 3a2
C.
Given curve is2x2 + y2 = 2x2x2 - 2x + y2 = 0⇒ 2x - 122 + y2 = 12⇒ x - 122 14 + y212 = 1which represents an ellipseHere, a = 12, b = 12, h = 12, k = 0Consider a point P(h + acosθ, k + bsinθ)= P12 + 12cosθ, 12sinθ on the ellipse from which the distanceof point (a, 0) is maximumlet Q(a, 0)Now,PQ = 12 + 12cosθ - a2 + 12sinθ - 02 = 14 + 14cos2θ + a2 + 12cosθ - acosθ - a + 12sin2θPQ = 12 + a2 - a + 12 - acosθ + 14sin2θLet y = PQ2 = 12 + a2 - a + 12 - acosθ + 14sin2θ
For maxima and minima, put dydθ = 0⇒ - 12 - asinθ + 14 . 2sinθcosθ = 0⇒ sinθ- 12 + a + 12cosθ = 0⇒ sinθ = 0 or - 12 + a + 12cosθ = 0⇒ θ = 0 or cosθ = 1 - 2a⇒ sin2θ = 1 - cos2θ = 1 - 1 - 2a2 = - 4a2 + 4a
Now, d2ydθ2 < 0 for cosθ = 1 - 2aThus, distance PQ is maximum, whencosθ = 1 - 2a and sin2θ = - 4a2 + 4aNow, required longest distance is =12 + a2 - a + 12 - a1 - 2a + 14- 4a2 + 4a= 12 + a2 - a + 12 - a - a + 2a2 - a2 + a= 2a2 - 2a + 1= 1 - 2a + 2a2
The incentre of the triangle formed by the straight lines y = 3x, y = - 3x and y = 3 is and y = 3 is
(0, 2)
(1, 2)
(2, 0)
(2, 1)
a The triangle formed by the given lines is shown in the adjacent figure
Clearly, the tnangle ABC is an isosceles triangle
∴ The incentre he on the median to the base
∵ D is mid-point of BC
∴ OD is median to the base BC
Thus, incentre lie on Y-axis