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The coordinates of the vertices P, Q, R & S of square PQRS inscribed in the triangle ABC with vertices A(0, 0), B(3, 0) & C(2, 1) given that two of its vertices P, Q are on the side AB are respectively
Let coordinates of P are (x_{1,}0) and side of square is 'a'
∴ Q(x1+a,0)
S(x1, a)
R(x1+a, a)
Now, m_{AS} = m_{AC}
⇒ a/x1 = 1/2
⇒ x1 = 2a ......(1)
m_{BR} = m_{BC}
⇒ a/(x_{1}+a−3) = −1
⇒ x_{1}+2a−3=0 ......(2)
from (1) & (2) a = 3/4
& x_{1} = 3/2
Hence coordinates of P, Q, R, & S can be determined.
(3/2,0) (9/4,0) (9/4,3/4) (3/2,3/4)
Points A & B are in the first quadrant ; point 'O' is the origin. If the slope of OA is 1, slope of OB is 7 and OA = OB, then the slope of AB is
If origin and (3, 2) are contained in the same angle of the lines 2x + y – a = 0, x – 3y + a = 0, then 'a' must lie in the interval
A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) are three noncollinear points in cartesian plane. Number of parallelograms that can be drawn with these three points as vertices are
Three vertices of triangle ABC are A(1, 11), B(–9, –8) and C(15, 2). The equation of angle bisector of angle A is
If line y – x + 2 = 0 is shifted parallel to itself towards the positive direction of the xaxis by a perpendicular distance of 3√2units, then the equation of the new line is
Given line:− y−x+2=0
line parallel to y−x+2=0 is
y−x+c=0
distance between these two lines=3√2
Distance, d = c2−c1/(√a^{2}+b^{2})
so, 3√2 = c−2/√(−1)^{2} + (1)^{2}
3√2 = c−2/√2
⇒c−2 = 6
so, c−2 = 6 (since line is shifted towards positive x−axis)
c=8
Equation of new line is y−x+8=0
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One of the diameter of the circle circumscribing the rectangle ABCD is 4y = x + 7. If A and B are the points (–3, 4) and (5, 4) respectively then the area of rectangle is equal to
Midpoint of AB=(5−3/2 , 4+4/2) = (1,4)
E≡(1,4)
ycoordinate of A, E, B are same
⇒AB∣∣xaxis
⇒ xcoordinate of =xcoordinate of E centre O
⇒Ox = 1
Now, 4y=x+7 is a diameter hence, it necessarily passes through the centre of the circle O
⇒4y=x+7(x=1)
⇒4y=8
⇒y=2
⇒O≡(1,2)
Now, OE=OF
∵OE=4−2=2
⇒OF=2
⇒EF=2+2=4
AB=5+3=8
EF=BC=4
⇒ Area of rectangle ABCD=L×B=AB×BC=8×4=32
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The pair of straight lines x^{2} – 4xy + y^{2} = 0 together with the line x + y + = 0 form a triangle which is
Let Aº (3, 2) and Bº (5, 1). ABP is an equilateral triangle is constructed on the side of AB remote from the origin then the orthocentre of triangle ABP is
The line PQ whose equation is x – y = 2 cuts the xaxis at P and Q is (4, 2). The line PQ is rotated about P through 45º in the anticlockwise direction. The equation of the line PQ in the new position is
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The point (4, 1) undergoes the following three transformations successively
(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.
The final position of the points is given by the coordinates
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