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The points (–a, – b), (0, 0), (a, b) and (a^{2}, ab) are : (1979)
The given points are A (– a, – b), B (0, 0), C (a, b) and D (a^{2}, ab).
Slope of Slope of BC = slope of BD
∴ A, B, C, D are collinear.
Th e poin t (4, 1) under goes th e following t hr ee transformations successively. (1980)
(i) Reflection about the line y = x.
(ii) Translation through a distance 2 units along the positive direction of xaxis.
(iii) Rotation through an angle p/4 about the origin in the counter clockwise direction.
Then the final position of the point is given by the coordinates.
Reflection about the line y = x, changes the point (4, 1) to (1, 4).
On translation of (1, 4) through a distance of 2 units along +ve direction of xaxis the point becomes (1 + 2, 4), i.e., (3, 4).
On rotation about origin through an angle p/4 the point P takes the position P' such that OP = OP'
Also OP = 5 = OP' and
Now, x = OP'
The straight lines x + y = 0, 3x + y – 4 = 0, x + 3y – 4 = 0 form a triangle which is (1983  1 Mark)
Solving the given equations of lines pairwise, we get the vertices of Δ as A (– 2, 2) B (2, – 2), C (1, 1)
Then AB =
BC =
CA = ∴ Δ is isosceles.
If P = (1, 0), Q = (–1, 0) and R = (2, 0) are three given points, then locus of the point S satisfying the relation SQ^{2} + SR^{2 }= 2SP^{2}, is (1988  2 Marks)
We have P = (1, 0), Q = (– 1, 0), R = (2, 0)
Let S = (x, y) ATQ SQ^{2} + SR^{2} = 2SP^{2}
⇒ (x + 1)^{2} + y^{2} + (x – 2)^{2} + y^{2} = 2[(x – 1)^{2} + y^{2}]
⇒ 2x^{2} + 2y^{2} – 2x + 5 = 2x^{2} + 2y^{2 }– 4x + 2
⇒ 2x + 3 = 0 ⇒ x = – 3/2
Which is a straight line parallel to yaxis.
Line L has intercepts a and b on the coordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same line L has intercepts p and q, then (1990  2 Marks)
As L has intercepts a and b on axes, equation of L is
.....… (1)
Let x and y axes be rotated through an angle q in anticlockwise direction.
In new system intercepts are p and q, therefore equation of L becomes
.....… (2)
KEY CONCEPT : As the origin is fixed in rotation, the distance of line from origin in both the cases should be same.
∴ We get d =
∴ (b) is the correct answer.
If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is (1992  2 Marks)
Let the two perpendicular lines be the coordinate axes.
Let (x, y) be the point sum of whose distances from two axes is 1 then we must have  x  +  y  = 1 or ± x ± y = 1
These are the four lines x + y = 1, x – y = 1, – x + y = 1, – x – y = 1
Any two adjacent sides are perpendicular to each other.
Also each line is equidistant from origin. Therefore figure formed is a square.
The locus of a variable point whose distance from (–2, 0) is 2/3 times its distance from the line x =  is (1994)
If variable point is P and S (– 2, 0) then PS = PM where PM is the perpendicular distance of point P from given line x = – 9/2
∴ By definition P describes an ellipse.
The equation s to a pair of opposite sides of parallelogram are x^{2} – 5x + 6 = 0 and y^{2 }– 6y + 5 = 0, the equations to its diagonals are (1994)
The sides of parallelogram are x = 2, x = 3, y = 1, y = 5.
∴ Diagonal AC is or y = 4x – 7
Equation diagonal BD is or 4x + y = 13
The orthocentre of the triangle formed by the lines xy = 0 and x + y = 1 is (1995S)
The lines by which Δ is formed are x = 0, y = 0 and x + y = 1.
Clearly, it is right Δ and we know that in a right Δ orthocentre coincides with the vertex at which right ∠ is formed.
∴ Orthocentre is (0, 0).
Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (1999  2 Marks)
Let m be the slope of PQ then
⇒ m + 2 = 1 – 2m or – 1 + 2m = m + 2
⇒ m = – 1/3 or m = 3
As PR also makes ∠ 45° with RQ.
∴ The above two values of m are for PQ and PR.
∴ Equation of PQ,
⇒ 3y – 3 = – x + 2 ⇒ x + 3y – 5 = 0
and equation of PR is ⇒ 3x – y – 5 = 0
∴ Combined equation of PQ and PR is (x – 3y – 5) (3x – y – 5) = 0
⇒ 3x^{2} – 3y^{2 }+ 8xy – 20x – 10y + 25 = 0
If x_{1}, x_{2}, x_{3} as well as y_{1}, y_{2}, y_{3}, are in G.P. with the same common ratio, then the points (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}). (1999  2 Marks)
x_{2} = x_{1}r, x_{3} = x_{1}r^{2} and so is y_{2 }= y_{1}r, y^{3} = y_{1}r^{2}
Hence the points lie on a line, i.e., they are collinear.
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 (2000S)
S is the midpoint of Q and R
Therefore,
Now slope of PS = m
Now equation of the line passing through (1, – 1) and parallel to PS is
y + 1 = or 2x + 9y + 7 = 0
The incentre of the triangle with vertices (1, √3 ), (0, 0) and (2, 0) is (2000S)
Here AB = BC = CA = 2. So, it is an equilaterial triangle and the incentre coincides with centroid.
Therefore,
Incentre = Centroid =
The number of integer values of m, for which the xcoordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer, is (2001S)
Intersection of 3x + 4y = 9 and y = mx + 1.
For x coordinate 3x + 4 (mx + 1) = 9 ⇒ (3 + 4m) x = 5
For x to be an integer 3 + 4m should be a divisor of 5 i.e., 1, – 1, 5 or – 5. 3 + 4m = 1
⇒ m = – 1/2 (not integer) 3 + 4m = – 1 ⇒ m = – 1 (integer) 3 + 4m = 5
⇒ m = 1/2 (not an integer) 3 + 4m = – 5 ⇒ m = – 2 (integer)
∴ There are 2 integral values of m.
∴ (a) is the correct alternative.
Area of the parallelogram formed by the lines y = mx, y = mx + 1, y = nx and y = nx + 1 equals (2001S)
The vertices,
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 (2002S)
⇒ bisector will have slope tan 120°
⇒ equation of bisector is
Let be fixed angle. If P = (cos q, sin q) and Q = (cos(a  q), sin(a  q)) , then Q is obtained from P by (2002S)
Clearly OP = OQ = 1 and
The bisector of ∠QOP will be a perpendicular to PQ and also bisect it. Hence Q is reflection of P in the line OM which makes an angle ∠MOP + ∠POX with x axis,
So that slope of OM is tan a/2.
A straight line through the origin O meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at points P and Q respectively.Then the point O divides the segemnt PQ in the ratio (2002S)
The given lines are 2x + y = 9/2 .....… (1)
and 2x + y = – 6 .....… (2)
Signs of constants on R.H.S. show that two lines lie on opp. sides of origin. Let any line through origin meets these lines in P and Q respectively then req. ratio is OP : OQ
Now in DOPA and DOQC, Ð POA = Ð QOC (ver. opp. Ð' s) Ð PAO = Ð OCQ (alt. int. Ð' s) ∠
∴DOPA ~ DOQC (by AA similarly)
∴ Req. ratio is 3 : 4.
The number of intergral points (integral point means both the coordinates should be integer) exactly in the interior of the triangle with vertices (0,0), (0,21) and (21,0), is (2003S)
Total no. of points within the square OABC = 20 × 20 = 400
Points on line AB = 20 ((1, 1), (2, 2), ....… (20, 20))
∴ Points within DOBC and DABC = 400 – 20 = 380
By symmetry points within
Orthocentre of triangle with vertices (0, 0), (3, 4) and (4, 0) is (2003S)
We know that or thocen tre is the meeting point of altitudes of a Δ.
Equation of alt. AD
⇒ line parallel to yaxis through (3, 4)
⇒ x = 3 ......... (1)
Similarly eq^{n} of OE ⊥ AB is
⇒ y = x/4 ......... (2)
Solving (1) and (2), we get orthocentre as (3, 3/4)
Area of the triangle formed by th e lin e x + y = 3 an d an gle bisectors of the pair of straight lines x^{2} – y^{2} + 2y = 1 is (2004S)
x^{2 }– y^{2} + 2y = 1 ⇒ x = ± (y – 1)
Bisectors of above lines are x = 0 and y = 1.
So area between x = 0, y = 1 and x + y = 3 is shaded region shown in figure.
Area = sq. units.
Let O(0, 0), P(3, 4), Q(6, 0) be the vertices of the triangles OPQ. The point R inside the triangle OPQ is such that the triangles OPR, PQR, OQR are of equal area. The coordinates of R are (2007 3 marks)
∵ Ar (ΔOPR) = Ar (ΔPQR) = Ar (ΔOQR)
∴ By simply geometry R should be the centroid of ΔPQO
A straight line L through the point (3, –2) is inclined at an angle 60° to the line If L also intersects the xaxis, then the equation of L is (2011)
Let the slope of line L be m.
∵ L intersects xaxis,
∴ Equation of L is y + 2 = (x – 3)
or
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