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Find the third vertex of a ΔABC if two of its vertices are B (3, 1), C (0, 2), and its centroid is at the origin.
B (3, 1) C (0, 2)
Let the third vertex be A (x, y)
G = (0, 0)
= 0 ⇒ x = 3 and
⇒ y  1 = 0 ⇒ y = 1
A (3, 1)
The circumcentre of a triangle is (3, 3). If its two vertices are (4, 6) and (0, 4) find the third vertex of the triangle.
Let P (3, 3) be the circumcentre of ΔABC, C(x, y) be the third vertex
PA = PB = PC
PA^{2 }= PB^{2 }= PC^{2}
PA^{2} = PC^{2}
⇒ (4  3)^{2} + (6  3)^{2}
= (x  3)^{2} + (y  3)^{2}
⇒ 1 + 9 = x^{2 }+ 9  6x + y^{2} + 9  6y
⇒ x^{2} + y^{2}  6x  6y + 8 = 0 ...(1)
and PB^{2} = PC^{2}
⇒ (0  3)^{2} + (4, 3)^{2} = (x  3)^{2} + (y  3)^{2}
⇒ 9 + 1 = x^{2} + 9  6x + y^{2}  6y + 9
Equations (1) and (2) are identical
(x  3)^{2} + (y  3 )^{2} = 10
So, (6, 2) is the required point.
In which ratio does the point P (1, 2) divides the join of A (2, 1) and B (7, 4)?
Let the ratio be k : 1
⇒ 6k = 3
⇒ k = 3/6 = 1/2
If (2, 2), (2, 1) and (5, 2) are vertices of a right angled triangle, then the area of triangle is
Area of triangle
= 1/2 [x_{1} (y_{2}  y_{3}) + x_{2} (y_{3} y_{1}) + x_{3} (y_{1}  y_{2})
= 1/2 [2 (1 − 2) + (−2) (2 + 2) + 5 (−2 −1)]
= 1/2 [−2−8 −15] = 25/2 = 12.5 sq. units
The centre of a circle is (4, 5) A(8, 10) is a point on the circumference. Find the other end of diameter of the circle through A.
Let the other end be (x, y)
The ends of a diameter of a circle have the coordinates (4, 3) and (4, 3), PQ is another diameter where P has coordinate
find the coordinates of Q.
Midpoint of (4, 3) and (4, 3)
The vertices of a triangle are A(1, 1), B(2, 5) and C(2, 2) find the length of the medium through C.
What is the circumradius of the triangle whose vertices are (2, 2), (8, 6), and (8, 2)?
Let P(x, y) be the circumcentre
PA = PB = PC
⇒ PA^{2} = PB^{2}
⇒ (x  2)^{2} + (y + 2)^{2}
= (x  8)^{2} + (y  6)^{2}
⇒ x^{2} + 4  4x + y^{2} + 4  4y
= x^{2}+ 64  16x + y^{2} + 36  12y
⇒  4x + 16x + 4y + 12y = 100  8
⇒ 12x + 16y = 92 ⇒ 3x+ 4y = 23 ...(1)
and PB^{2} = PC^{2}
⇒ (x  8)^{2} + (y  6)^{2} = (x  8)^{2} + (y + 2)^{2}
⇒ y^{2} + 36  12y = y^{2} + 4 + 4y
⇒ 16y = 32 ⇒ y = 2
3x = 23  4 ⇒ 2 = 15 ⇒ x = 5
P (5, 2)
∴ PA
The sum of the square of the distance of a moving point from two fixed points (a, 0) and (a, 0) is equal to the constant quantity 2c^{2}. Find the equation of its locus.
Here (x a)^{2} + (y  0)^{2} = 2c^{2}
x^{2} + a^{2}  2ax + y^{2} = 2c^{2}
and (x + a)^{2} + (y  0)^{2} = 2c^{2}
⇒ x^{2} + a^{2} + 2ax + y^{2} + 2c^{2}
Now x^{2} + a^{2}  2ax + y^{2} = x^{2} + a^{2} + 2ax + y^{2}
⇒ 4ax = 0 ⇒ x = 0 or a = 0
(x  0)^{2} + y^{2} = 2c^{2}
⇒ x^{0 }+ y^{2} = 2c^{2}
What is the distance of the point (4, 7) from the yaxis?
Distance of the point (4, 7) from (0, y)
= x coordinate
= 4
If the distance between the points (3, 0) and (0, y) is 5 units. y is positive then what is value of y?
(3  0)^{2} + (0  y)^{2 }= 52
9 + y^{2} = 25 ⇒ y^{2} = 16 ⇒ y = ± 4
y is positive, ∴ y = 4
If the centroid of the triangle is formed by points P(a, b), Q(b, c) and R(c, a) is at the origin. What is the value of a + b +c?
If the point P(1, 2) divides externally the line segment joining A(2, 5) and B in the ratio 3 : 4. What is the coordinate of point B?
The points A(3, 1), B(0, 4), C(3, 1), D(0, 2) are vertices of a
The given points are A (0,2), B(3,1), C(0,4) and D(3,1)
Thus, diagonal PR = diagonal QS
Therefore, the given points from a square.
If the point (x, 4) lies on a circle whose center is at the origin and radius is 5, What is the value of x?
OP = 5
⇒ OP^{2} = 25
⇒ (x  0)^{2} + (4  0)^{2} = 25
⇒ x^{2} + 16 = 25
⇒ x^{2} = 9 ⇒ x = ± 3
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