## Integer Type/Fill In The Blanks Type

Q.1₂₀₁₉ Let the point B be the reflection of point A(2, 3) with respect to the line 8x – 6y – 23 = 0. Let S₁ and S₂ be circles of radii 2 and 1 with centers A and B respectively. Let T be a common tangent to the circles S₁ and S₂ such that both the circles are on the same side of T. If C is the point of intersection of T and the line passing through A and B, then the length of the line segment AC is…

Solution:
\begin{aligned}AL=\left| \dfrac{8\left( 2\right) -6\left( 3\right) -23}{\sqrt{8^{2}+6^{2}}}\right| =\left| \dfrac{16-18-23}{10}\right| =\dfrac{5}{2}\\ \Rightarrow \dfrac{CB}{CA}=\dfrac{1}{2}\\ \Rightarrow \dfrac{CA-5}{CA}=\dfrac{1}{2}\Rightarrow CA=10\end{aligned}

Q.2₂₀₀₆ Two rays x + y = |a| and ax – y = 1 in the first quadrant intersect each other in the interval a ∈ (a₀, ∞), then the value of a₀ is..

Solution:
The point of intesection of the given rays can be given as
\begin{aligned}x=\dfrac{\left| a\right| +1}{a+1} >0,y=\dfrac{\left| a\right| -1}{a+1} >0\\ \Rightarrow a+1 >0\Rightarrow a\in \left( 1,\infty \right) \Rightarrow a_{0}=1\end{aligned}
Since x > 0 and |a| + 1 is also greater than zero, hence a + 1 must also be grater than zero.

Q.3₁₉₉₇ Two vertices of an equilateral triangle are (-1, 0) and (1, 0) and its third vertex lies above the x-axis, then the equation of the circumcircle of this triangle will be…

Solution:
Since the triangle is equilateral the midpoint of these two given points will be the origin and the third vertex will be at a distance of √3 unit from the origin on the y-axis, so the third vertex of the equilateral triangle will be (0, √3).
Since the triangle is equilateral, hence circumcentre = centroid = (0, 1/√3).
Radius of the triangle will be = 2/√3. Now the equation of the circle will be given by
(x – 0)2 + (y – 1/√3)2 = (2/√3)2

Q.4₁₉₉₃ The vertices of a triangle are A(-1, -7), B(5, 1), and C(1, 4) the equation of the bisector of angle ABC will be…

Solution:
Equation of the side AB is -4x + 3y + 17 = 0.
Equation of the side BC is 3x + 4y – 19 = 0.
Then the equation of the bisector of the angle ABC can be given by -4x + 3y + 17 = ± (3x + 4y – 19)
Now the bisector of the angle ABC containing origin will be = x – 7y + 2 = 0

Q.5₁₉₉₁ Let the algebraic sum of the perpendicular distance from the points (2, 0), (0, 2) and (1, 1) to a variable straight line be zero, then the straight line passes through a fixed point whose coordinates are…

Solution:
Let the said variable straight line is x cosa + y sina – p = 0
Since the algebraic sum of the perpendicular distance from the points (2, 0), (0, 2) and (1, 1) to a variable straight line x cosa + y sina – p = 0 is zero
Hence (2 cosa – p) + (2 sina – p) + (cosa + sina – p) = 0
⇒ p = sina + cosa
⇒ x cosa + y sina – (sina + cosa) = 0
⇒ (x – 1) + (y – 1)tana = 0
⇒ The variable straight line must pass through the point (1, 1)

Q.6₁₉₈₅ The orthocenter of the triangle formed by the straight lines x + y = 1, 2x + 3y = 6, and 4x – y + 4 = 0 lies in the quadrant number…

Solution:
The points of intersections of x + y = 1, 2x + 3y = 6 is A(-3, 4) and 2x + 3y = 6, and 4x – y + 4 = 0 is B(-3/7, 16/7).
Altitudes through the points A and B is y = -x/4 + 13/4 and y = x/7 + 19/7 respectively.
⇒ The orthocenter of the triangle is given by (3/7, 22/7).
⇒ The orthocenter lies in the First Quadrant.

## True/False Type

Q.1₁₉₈₈ The straight lines 2x + 3y + 19 = 0, and 9x + 6y – 17 = 0 cuts the coordinates in the concyclic points.

Solution:
The straight lines ax + by + c = 0, and lx + my + n = 0 cuts the coordinates in the concyclic points, if al = bm.
Here, 2.9 = 3.6, hence the given statement is TRUE.

Q.2₁₉₈₅ If $\begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{vmatrix}=\begin{vmatrix} a_{1} & b_{1} & 1 \\ a_{2} & b_{2} & 1 \\ a_{3} & b_{3} & 1 \end{vmatrix}$, then the two triangles with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃), (a₁, b₁), (a₂, b₂), and (a₃, b₃) must be congruent.

Solution:
The two Triangles with the same areas need not be congruent, hence the given statement is false.

Q.3₁₉₈₃ The straight line 5x + 4y = 0, passes through the point of intersection of straight lines x + 2y -10 = 0and 2x + y + 5 = 0.

Solution:
The point of intersection of the straight lines x + 2y -10 and 2x + y + 5 = 0 is (-20/3, 25/3), which satisfies the equation of the straight line 5x + 4y = 0.
Hence the given statement is TRUE.

## Multiple Choice Single Correct

Q.1₂₀₁₃ For a > b > c > 0, the distance between the point (1, 1) and the point of intersection of the straight lines ax + by + c = 0 and bx + ay + c = 0 is leass than 2√2, then

(A) a + b – c > 0
(B) a – b + c < 0
(C) a – b + c > 0
(D) a + b – c < 0

Solution:
For the point of intersection of the straight lines ax + by + c = 0 and bx + ay + c = 0, we will have (a – b) x = (a – b) y.
⇒ the point of intersection lies on the line y = x
Let (t, t) be any point on the line y = x
⇒ √(t – 1)2 + (t – 1)2 < 2√2
⇒ |t – 1|< 2 ⇒ -1 < t < 3 ⇒ (-1, -1) lies on the opposite side of the origin for both the straight lines
⇒ – a – b + c < 0 ⇒ a + b - c > 0
Option (A) is correct

Q.2₂₀₁₁ A straight line L through the point (3, -2) is inclined at an angle 60 degree to the line x√3 + y = 1. If straight line L also intersect the x-axis, then the the equation of a straight line L is

(A) x√3 + y + 2 – 3√3 = 0
(B) -x√3 + y + 2 + 3√3 = 0
(C) y√3 – x + 3 + 2√3 = 0
(D) y√3 + x -3 + 2√3 = 0

Solution:
Let the slope of the straight line be ‘m’, then |(m + √3)/(1 – m√3)| = √3, {Since straight line L is inclined at an angle 60 degree to the line x√3 + y = 1}
⇒ (m + √3) = ± (√3 – 3m) ⇒ m = 0 or m = √3
⇒ The equation of straight line L is -x√3 + y + 2 + 3√3 = 0
Option (B) is correct

Q.3₂₀₀₉ The locus of the orthocenter of the triangle formed by the straight lines (1 + p)x – py + p(1 + p) = 0, (1 + q)x – qy + q(1 + q) = 0, and y = 0, where p ≠ q, is
(A) Straight line
(B) Parabola
(C) Ellipse
(D) Hyperbola

Solution:
Point of intersection of (1 + q)x – qy + q(1 + q) = 0 and y = 0 is A(-q, 0)
Point of intersection of (1 + p)x – py + p(1 + p) = 0 and y = 0 is B(-p, 0)
and the Point of intersection of (1 + p)x – py + p(1 + p) = 0 and (1 + q)x – qy + q(1 + q) = 0 is C(pq, (p + 1)(q + 1))
Equation of altitude from C to AB is x = pq
and the equation of altitude from B to AC is (1 + q)y = -q(x + p)
⇒ The point of intersection of the given altitudes is (pq, -pq)
⇒ The locus of the orthocenter is x + y = 0
Option (A) is correct

Q.4₂₀₀₈ Consider the three points P(-Sin(b – a), -Cos(b)), Q(Cos(b – a), Sin(b)), R(Cos(b – a + c), Sin(b – c)), where 0 < a, b, c < π/4, then
(A) P lies on the line segment RQ
(B) Q lies on the line segment PR
(C) R lies on the line segment PQ
(D) P, Q, and R are non-colinear

Solution:
Since the area of the triangle formed by P, Q, and R is not equal to zero, hence P, Q, and R are non-colinear
Option (D) is correct

Q.5₂₀₀₄ Area of the triangle formed by the straight line x + y = 3 and the angle bisectors of the pair of straight lines x2 – y2 + 2y = 1 is
(A) 2
(B) 4
(C) 6
(D) 8

Solution:
The angle bisectors of the pair of straight lines x2 – y2 + 2y = 1 are x = 0, and y – 1 = 0
Thus the area of the triangle formed by the straight line x + y = 3, x = 0, and y – 1 = 0 is (2 x 2)/2 = 2
Option (A) is correct

Q.6₂₀₀₃ The orthocenter of the triangle with vertices O(0, 0), A(3, 4), and B(4, 0) is
(A) (3, 5/4)
(B) (3, 12)
(C) (3, 3/4)
(D) (3, 9)
Solution:
Let the foot of the altitudes from the vertices O and A on the sides AB and OB are U and V respectively
⇒ angle UOB = angle VAB and Let the orthocenter is (3, k)
⇒ tan (angle UOB) = tan (angle VAB) ⇒ k/3 = 1/4 ⇒ k = 3/4
⇒ the orthocentre is (3, 3/4)
Option (C) is correct

Q.7₂₀₀₂ Let P(-1, 0), Q(0, 0), and R(3, 3√3) be three points, then the equation of the bisector of the angle PQR is
(A) (√3/2)x + y = 0
(B) √3y + x = 0
(C) √3x + y = 0
(D) (√3/2)y + x = 0
Solution:
Since the angle PQR = 120ᵒ
⇒ The slope of the required angle bisector is = Tan(120ᵒ), and the bisector passes through the origin
⇒ The equation of the bisector of the angle PQR is √3x + y = 0
Option (C) is correct

Q.8₂₀₀₂ The straight line through the origin O meet the parallel straight lines 4x + 2y = 9 and 2x + y + 6 = 0 at the points P and Q respectively, then the point O divides the segment PQ in the ratio
(A) 1 : 2
(B) 3 : 2
(C) 2 : 1
(D) 4 : 3
Solution:
OP/OQ = 9/(4×3) = 3/4
Option (D) is correct

Q.9₂₀₀₁ The area of the parallelogram formed by the straight lines y = mx, y = mx + 1, y = nx, and y = nx + 1 is
(A) |m + n|/(m – n)
(B) 2/|m + n|
(C) 1/|m + n|
(D) 1/|m – n|
Solution:
The vertices of the parallelogram are O(0, 0), A(0, 1), B(1/(m – n), a), and C(1/(m – n), b), Here a, and b are any arbitrary real numbers
⇒ The area of the parallelogram is = 2 The area of the triangle AOB = 1/|m – n|
Option (D) is correct

Q.10₂₀₀₁ The number of integer value(s) of m for which the x-coordinate of the point of intersection of the straight lines 3x + 4y = 9 and y = mx + 1 is also an integer is/are
(A) 0
(B) 2
(C) 1
(D) 4
Solution:
The x-coordinate of the point of intersection of the straight lines 3x + 4y = 9 and y = mx + 1 is x = 5/(4m + 3)
Since x is an integer ⇒ 4m + 3 = ± 1, ± 5
⇒ m = -1, -2
Option (B) is correct

Q.11₂₀₀₀ The incentre of the triangle with vertices (1, √3), (0, 0), and (2, 0) is
(A) (1, √3/2)
(B) (2/3, 1/√3)
(C) (2/3, √3/2)
(D) (1, 1/√3)
Solution:
Since the vertices of a triangle are given as A(1, √3), B(0, 0), and C(2, 0)
⇒ The Sides of the triangle ABC are AB = BC = CA = 2
⇒ Hence triangle ABC is an equilateral triangle
⇒ Incentre of triangle ABC = Centroid of triangle ABC = (1, 1/√3)
Option (D) is correct

Q.12₂₀₀₀ Let PS be the median of the triangle with vertices P(2, 2), Q(6, -1), and R(7, 3). The equation of the line passing through (1, -1) and parallel to PS is
(A) 2x – 9y – 7 = 0
(B) 2x – 9y – 11 = 0
(C) 2x + 9y – 1 = 0
(D) 2x + 9y + 7 = 0
Solution:
The slope of the line segment PS is = -(2/9)
Now the equation of the line passing through (1, -1) and parallel to PS will be (y + 1) = -(2/9)(x – 1)
⇒ The equation of the line is 2x + 9y + 7 = 0
Option (D) is correct

Q.13₁₉₉₉ If x₁, x₂, x₃ as well as y₁, y₂, y₃ are in G.P. with the same common ratio, then the points (x₁, y₁), (x₂, y₂), and (x₃, y₃)
(A) Lie on the stright line
(B) Lie on the circle
(C) Lie on the ellipse
(D) Are vertices of a triangle
Solution:
Let ‘r’ be the common ratio of the given G.P.
Now the area of the triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) = 0
⇒ The points (x₁, y₁), (x₂, y₂), and (x₃, y₃) lie on the stright line
Option (A) is correct

Q.14₁₉₉₉ let PQR be the right angled isosceles triangle right angled at P if the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is
(A) 3x² – 3y² + 8xy + 20x + 10y + 25 = 0
(B) 3x² – 3y² + 8xy – 20x – 10y + 25 = 0
(C) 3x² – 3y² + 8xy + 10x + 15y + 20 = 0
(D) 3x² – 3y² – 8xy – 20x – 10y – 25 = 0
Solution:
Since the angle PQR = angle PRQ = 45ᵒ
⇒ Tan(45ᵒ) = ± ((m + 2)/(1 – 2m))
⇒ m = 3, -(1/3) {Slopes of lines PQ and PR}
⇒ The joint eqiation of lines PQ and PR is (3x – y – 5)(x + 3y – 5) = 0
⇒ 3x² – 3y² + 8xy – 20x – 10y + 25 = 0
Option (B) is correct

Q.15₁₉₉₈ The diagonals of a parallelogram PQRS are along the lines x + 3y = 4 and 6x – 2 = 7 then PQRS must be a
(A) Rectangle
(B) Square
(D) Rhombus
Solution:
Since the diagonals (x + 3y = 4 and 6x – 2 = 7) of a parallelogram PQRS are perpendicular to each other
⇒ The given parallelogram PQRS is a Rhombus
Option (D) is correct

Q.16₁₉₉₈ P(1, 2), Q(4, 6), R(5, 7), and S(a, b) are the vertices of the parallelogram PQRS, then
(A) a = 2, b = 4
(B) a = 3, b = 4
(C) a = 2, b = 3
(D) a = 3, b = 5
Solution:
Since PQRS is a parallelogram ⇒ mid-point of PR = mid-point of QS
⇒ (6/2, 9/2) = ((a + 4)/2, (b + 6)/2)
⇒ (a + 4)/2 = 6/2 and (b + 6)/2 = 9/2
⇒ a = 2, b = 3
Option (C) is correct

Q.17₁₉₉₂ The sum of the distances of a point from two perpendicular lines in a plane is 1. Then its locus is
(A) Square
(B) Circle
(C) Straight Line
(D) Ellipse
Solution:
Consider the two perpendicular lines as the x-axis and y-axis
⇒ |x| + |y| = 1 ⇒ the locus is Square
Option (A) is correct

Q.18₁₉₉₀ Line ‘L’ has intercepts a and b on the co-ordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same line ‘L’ has intercepts p and q then
(A) a² + b² = p² + q²
(B) 1/a² + 1/b² = 1/p² + 1/q²
(C) a² + p² = b² + q²
(D) 1/a² + 1/p² = 1/b² + 1/q²
Solution:
The length of the intercepts remain same on rotating the axes
⇒ 1/a² + 1/b² = 1/p² + 1/q²
Option (B) is correct

Q.19₁₉₈₈ If P(1, 0), Q(-1, 0), and R(2, 0) are three given points, then the locus of the point S satisfying the relation SQ² + SR² = 2SP² is
(A) A straight line parallel to x-axis
(B) A circle passing through origin
(C) A circle with center at origin
(D) A straight line parallel to y-axis
Solution:
Since SQ² + SR² = 2SP²
⇒ (x + 1)² + y² + (x – 2)² + y² = 2(x – 1)² + 2y²
⇒ x + 1 = 0 ⇒ A straight line parallel to y-axis
Option (D) is correct

## Multiple Choice Multi Correct

Q.1₁₉₉₉ let L₁ be a straight line passing through the origin and L₂ be the straight line x + y = 1. If the intercepts made by the circle x² + y² – x + 3y = 0 on L₁ and L₂ are equal, then which of the following equations can represent L₁
(A) x + y = 0
(B) x – y = 0
(C) x + 7y = 0
(D) x – 7y = 0
Solution:
Let straight line L₁ be y = mx
The x-coordinate of the point of intersection of y = mx and x² + y² – x + 3y = 0 are x = 0, (3m -1 )/(m² + 1)
⇒ the point of intersection of L₁ and the circle are P₁(0, 0) and P₂((3m -1 )/(m² + 1), m(3m -1 )/(m² + 1))
Now the point of intersection of L₂ and the circle are Q₁(1, 0) and Q₂(2, -1)
⇒ |P₁P₂| = |Q₁Q₂| ⇒ m = 1, -(1/7)
⇒ straight line L₁ will be y = x and y = -x/7 ⇒ x + 7y = 0
Options (B) and (C) are correct

Q.2₁₉₈₆ All points inside the triangle formed by the points (1, 3), (5, 0), and (-1, 2) satisfy
(A) 3x + 2y ≥ 0
(B) 2x + y – 13 ≥ 0
(C) 0 ≥ 2x – 3y -12
(D) -2x + y ≥ 0
Solution:
Options A and C are satisfied by the given points (1, 3), (5, 0), and (-1, 2)
Options (A) and (C) are correct

Q.3₁₉₈₅ The straight lines px + qy + r = 0, qx + ry + p = 0, and rx + py + q = 0 are concurrent if
(A) p + q + r = 0
(B) p² + q² + r² = pq + qr + rp
(C) p³ + q³ + r³ = 3pqr
(D) (p – q)² + (q – p)² + (r – p)² = 0
Solution:
Since the straight lines px + qy + r = 0, qx + ry + p = 0, and rx + py + q = 0 are concurrent
⇒ p(rq – p²) – q(q² – pr) + r(pq – r²) = 0
⇒ (p + q + r)((p – q)² + (q – p)² + (r – p)²) = 0
⇒ (p + q + r) = 0 or (p – q)² + (q – p)² + (r – p)² = 0
All options are correct