A triangle with vertices (4, 0), (–1, –1), (3, 5) is
In isosceles triangle side AB = CA For right angled triangle, BC^{2} = AB^{2} + AC^{2}
So, here BC = or BC^{2} = 52
or
Locus of mid point of the portion between the axes of x cosα + y sina = p whre p is constant is [2002]
Equation of AB is x cosα + y sinα = p;
So coordinates of A and B are
So coordinates of midpoint of AB are
⇒ cosα = p/2x_{1} and sinα = p/2y_{1} ;
cos^{2}α + sin^{2}α = 1
Locus of (x_{1}, y_{1}) is
If the pair of lines ax^{2}+2hxy+by^{2}+2gx+2fy+c=0 intersect on the yaxis then [2002]
Put x = 0 in the given equation ⇒ by^{2} + 2 fy + c = 0.
For unique point of intersection f^{2}– bc = 0 ⇒ af^{2} – abc = 0.
Since abc + 2fgh – af^{2} –bg^{2} – ch^{2} = 0 ⇒ 2fgh – bg^{2} – ch^{2} = 0
The pair of lines represented by 3ax^{2} + 5xy + (a^{2} – 2)y^{2} = 0 are perpendicular to each other for [2002]
3a + a^{2} – 2 = 0 ⇒ a^{2} + 3a – 2 = 0.;
A square of side a lies above the xaxis and has one vertex at the origin. The side passing through the origin makes an angle with the positive direction of xaxis. The equation of its diagonal not passing through the origin is
Coordinates of A = (a cosα , a sinα) Equation of OB,
∴ slope of CA =
Equation of CA
y  a sinα = (x  a cosα)
⇒ ( y  a sinα) (1 + tan a) = (a cosα  x )(1  tana)
⇒ (y  a sinα) (cosα+ sinα) = (a cosα x )(cosα sinα)
⇒ y(cos+ sinα)  a sinα cosα  a sin^{2}α = a cos^{2}α  a cosα sinα x (cosα sinα)
⇒ y(cosα + sinα) + x(cosα  sinα)= a
y(sinα + cosα ) + x(cosα  sinα ) = a.
If the pair of straight lines x^{2 } 2 pxy  y^{2} = 0 and x^{2 } 2qxy y^{2}=0 be such that each pair bisects the angle between the other pair, then [2003]
Equation of bisectors of second pair of straight lines
is, qx^{2} + 2 xy  qy^{2}=0 ........(1)
It must be identical to the first pair
x^{2}  2 pxy y^{2}=0 ........(2)
from (1) and (2)
Locus of centroid of the triangle whose vertices are (a cos t, a sin t ), (b sin t,  b cos t ) and (1, 0), where t is a parameter, is [2003]
Squaring & adding,
If x_{1}, x_{2} , x_{3} and y_{1}, y_{2},y_{3} are both in G.P. with the same common ratio, then the points ( x_{1}, y_{1}), (x_{2} , y_{2}) and (x_{3} , y_{3}) [2003]
Taking coordinates as
Then slope of line joining
and slope of line joining (x, y) and (xr, yr)
∴ m_{1} = m_{2}
⇒ Points lie on the straight line.
If the equation of the locus of a point equidistant from the point (a_{1}, b_{1}) and (a_{2}, b_{2}) is (a_{1} b_{2}) x + (a_{1} b_{2}) y + c= 0 , then the value of `c` is [2003]
Let A(2,  3) and B ( 2, 3) be vertices of a triangle ABC. If the centroid of this triangle moves on the line 2x + 3y= 1, then the locus of the vertex C is the line
Let the vertex C be (h, k), then the
centroid of Δ ABC is
orIt lies on 2x + 3y = 1
.
= Locus of C is 2x + 3y = 9
The equation of the straight line passing th rough the point (4, 3) and making intercepts on the coordinate axes whose sum is –1 is [2004]
Let the required line be ......(1)
then a + b = –1 .......(2)
(1) passes through (4, 3) ,
⇒ 4b +3a= ab .................(3)
Eliminating b from (2) and (3), we get a^{2} 4 = 0
⇒ a =±2 ⇒b =3 or1
∴ Equations of straight lines are
If the sum of t he slopes of the lines given by x^{2}  2cxy  7y^{2 }= 0 is four times their product c has the value[2004]
Let the lines be y = m_{1}x and y = m_{2}x then
m_{1} + m_{2} = and 4 m_{1}m_{2} =
Given m_{1} + m_{2}= 4 m_{1}m_{2}
If one of the lines given by 6x^{2}  xy + 4cy^{2 }= 0 is 3x + 4y = 0, then c equals [2004]
3x + 4y=0 is one of the lines of the pair
6 x^{2}  xy + 4cy^{2}=0 , Put
we get
The line parallel to the x axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx – 2ay – 3a = 0, where (a, b) ≠ (0, 0) is [2005]
The line passing through the intersection of lines
ax + 2by = 3b=0 and bx  2ay  3a=0
is ax + 2by + 3b + λ (bx – 2ay – 3a) = 0
⇒ (a + b λ) x + (2b – 2aλ)y + 3b – 3λa = 0
As this line is parallel to xaxis.
∴ a + bλ = 0 ⇒ λ = – a/b
⇒ ax + 2by + 3b – (bx – 2ay – 3a) = 0
⇒ ax + 2by + 3b – ax
So it is 3/2 units below xaxis.
If a vertex of a triangle is (1, 1) and the mid points of two sides through this vertex are (–1, 2) and (3, 2) then the centroid of the triangle is [2005]
Vertex of tringle is (1, 1) and midpoint of sides through this vertex is (– 1, 2) and (3, 2)
⇒ vertex B and C come out to be (– 3, 3) and (5, 3)
∴ Centroid is
A straight line through the point A (3, 4) is such that its intercept between the axes is bisected at A. Its equation is
∵ A is the mid point of PQ , therefore
a = 6, b = 8
∴ Equation of line is or 4x + 3y = 24
If (a,a^{2}) falls inside the angle made by the lines y = x > 0 and y = 3x , x > 0 , then a belong to [2006]
Clearly for point P,
a^{2 } 3a<0 and
Let A (h, k), B(1, 1) and C (2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1square unit, then the set of values which 'k' can take is given by [2007]
Given : The vertices of a right angled triangle A(l, k), B(1, 1) and C(2, 1) and Area of ΔABC = 1 square unit
We know that, area of right angled triangle
⇒ ± (k 1)=2 ⇒ k = – 1, 3
Let P = (–1, 0), Q = (0, 0) and R = (3, ) be three point. The equation of the bisector of the angle PQR is [2007]
Given : The coordinates of points P, Q, R are (–1, 0),
(0, 0), respectively..
Slope of QR =
⇒
Let QM bisects the ∠PQR ,
∴Slope of the line QM =
∴ Equation of line QM is (y – 0) = 
If one of the lines of my^{2} + (1– m^{2}) xy – mx^{2 }= 0 is a bisector of the angle between the lines xy = 0, then m is [2007]
Equation of bisectors of lines, xy = 0 are y = ± x
∴ Put y = ± x in the given equation my^{2} + (1 – m^{2})xy – mx^{2} = 0
∴ mx^{2} + (1 – m^{2})x^{2} – mx^{2} = 0
⇒ 1 – m^{2} = 0 ⇒ m = ± 1
The perpendicular bisector of the line segment joining P (1, 4) and Q(k, 3) has yintercept –4. Then a possible value of k is[2008]
Slope of
∴ Slope of perpendicular bisector of PQ = ( k –1)
Also mid point of
∴ Equation of perpendicular bisector is
⇒ 2y – 7 = 2(k –1) x –(k^{2} –1)
⇒ 2(k – 1)x – 2y + ( 8 – k^{2}) = 0
∴ yintercept
⇒ 8 – k^{2} = –8 or k^{2} = 16 ⇒ k = ± 4
The shortest distance between the line y – x = 1 and the curve x = y^{2} is : [2009]
Let (a^{2}, a) be the point of shortest distance on x = y^{2}
Then distance between (a^{2}, a) and line x – y + 1 = 0 is given by
It is min when
The lines p(p^{2} +1)x – y + q = 0 and (p^{2 }+ 1)2x + (p^{2 }+ 1)y + 2q = 0 are perpendicular to a common line for : [2009]
If the lines p (p^{2} + 1) x – y + q = 0
and (p^{2} + 1)^{2 }x + (p^{2} + 1) y +2q = 0
are perpendicular to a common line then these lines must be parallel to each other,
∴ m_{1} = m_{2}
⇒ (p^{2} + 1) ( p + 1) = 0
⇒ p = – 1
∴ p can have exactly one value.
Three distinct points A, B and C are given in the 2dimensional coordinates plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point (–1, 0) is equal to . Then the circumcentre of the triangle ABC is at the point: [2009]
Given that
P (1, 0), Q (– 1, 0) and
⇒ 3AP = AQ
Let A = (x, y) then 3AP = AQ ⇒ 9 AP^{2}= AQ^{2 }
⇒ 9 (x – 1)^{2} + 9y^{2 }= (x + 1)^{2} + y^{2 }
⇒ 9 x^{2} – 18x + 9 + 9y^{2} = x^{2} +2x +1 + y^{2 }
⇒ 8x^{2} – 20x + 8y^{2 }+ 8 = 0
....(1)
∴ A lies on the circle given by eq (1). As B and C also follow the same condition, they must lie on the same circle.
∴ Centre of circumcircle of Δ ABC
The line L given by passes through the point (13, 32). The line K is parallel to L and has the equation Then the distance between L and K is [2010]
Slope of line L =
Slope of line K =
Line L is parallel to line k.
(13, 32) is a point on L.
Equation of K : y  4x=3 ⇒ 4x y + 3=0
Distance between L and K =
The lines L_{1} : y – x = 0 and L_{2 }: 2x + y = 0 intersect the line L_{3} : y + 2 = 0 at P and Q respectively. The bisector of the acute angle between L_{1} and L_{2} intersects L_{3} at R.
Statement1: The ratio PR : RQ equals
Statement2: In any triangle, bisector of an angle divides the triangle into two similar triangles. [2011]
L_{1} : y – x = 0
L_{2} : 2x + y = 0
L_{3} : y + 2 = 0
On solving the equation of line L_{1} and L_{2} we get their point of intersection (0, 0) i.e., origin O.
On solving the equation of line L_{1} and L_{3}, we get P = (– 2, – 2).
Similarly, we get Q = (– 1, – 2) We know that bisector of an angle of a triangle, divide the opposite side the triangle in the ratio of the sides including the angle [Angle Bisector Theorem of a Triangle]
If the line 2x + y = k passes through the point which divides the line segment joining the points (1,1) and (2,4) in the ratio 3 :2, then k equals : [2012]
Let the joining points be A (1,1) and B (2,4).
Let point C divides line AB in the ratio 3 : 2.
So, by section formula we have
Since Line 2x + y = k passes through
∴ C satisfies the equation 2x + y = k.
A ray of light along gets reflected upon reaching xaxis, the equation of the reflected ray is [JEE M 2013]
Suppose B(0, 1) be an y point on given line and coordinate of A is So, equation of
Reflected Ray is
The xcoordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (0, 1) (1, 1) and (1, 0) is : [JEE M 2013]
From the figure, we have
a = 2, b = , c = 2
x_{1} = 0, x_{2} = 0, x_{3} = 2
Now, xcoordinate of incentre is given as
⇒ xcoordinate of incentre =
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: [JEE M 2014]
Let P, Q, R, be the vertices of ΔPQR
Since PS is the median, S is midpoint of QR
So,
Now, slope of PS
Since, required line is parallel to PS therefore slope of required line = slope of PS Now, eqn of line passing through (1, –1) and having slope is
y(1) =
9y + 9 = –2x + 2 ⇒ 2x + 9y + 7 = 0
Let a, b, c an d d be n on zer o numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 and 5bx + 2by + d =0 lies in the fourth quadrant and is equidistant from the two axes then [JEE M 2014]
Given lines are
4ax + 2ay + c = 0
5bx + 2by + d = 0
The point of intersection will be
∵ Point of intersection is in fourth quadrant so x is positive and y is negative.
Also distance from axes is same So x = – y (∵ distance from xaxis is –y as y is negative)
The number of points, having both coordinates as integers, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0) is : [JEE M 2015]
Total number of integral points inside the square OABC = 40 × 40 = 1600
No. of integral points on AC
= No. of integral points on OB
= 40 [namely (1, 1), (2, 2) ... (40, 40)]
∴ No. of integral points inside the ΔOAC
Two sides of a rhombus are along the lines, x – y + 1 = 0 and 7x – y – 5 = 0. If its diagonals intersect at (–1, –2), then which one of the following is a vertex of this rhombus? [JEE M 2016]
Let other two sides of rhombus are
x – y + λ = 0 and
7x –y + μ = 0
then O is equidistant from AB and DC and from AD and BC
and
∴ Other two sides are x – y – 3 = 0 and 7x – y + 15 = 0 On solving the eqns of sides pairwise, we get
the vertices as
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