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Find the value of k for which the points A (3, 2), B (4, k) and C (5, 3) are collinear.
The points A (3, 2), B (4, K), and C (5, 3) are collinear then area (ΔABC) = 0
⇒ 1/2 [3(k − 3)+ 4(3 − 2) + 5(2 − k)] = 0
⇒ 3k  9 + 4 + 10  5k = 0
⇒ 2k + 5 = 0 ⇒ k = 5/2
Find ratio in which the line 2x + y  4 = 0 divides the line segment joining A(2, 2) and B(3, 7).
Let the point P (x, y) divide the line AB in the ratio k : 1
This point P(x, y) lies on the line
2x + y  4 = 0
⇒ 6k + 4+ 7k  2  4k  4 = 0
⇒ 9k  2 = 0 ⇒ 9k = 2
⇒ k = 2/9 = 2 : 9
The coordinates of ends of a diameter of a circle are (4, 1) and (2, 5). Find the centre of the circle.
Ends of diameter of circle are (4, 1) and (2, 5)
centre = Midpoint of (4, 1) and (2, 5)
The area of a triangle with vertices (a, b + c) and (b, c + a) and (c, b + a) is
Area of triangle
= 1/2 [a(c + a  a  b) + b (a + b  b  c) + c (b + c  c  a)]
= 1/2 [a(c − b)+ b(a − c) + c(b − a)]
= 1/2 [ac − ab + ab − bc + bc − ac]
= 1/2 × 0 = 0
Find the coordinates of the points which trisects the line joining (3, 5) and (6, 7).
Let P(x, y) be the point
The ends of a diagonal of a square have the coordinates (a, 1) and (1, a), find a if the area of the square is 50 square units.
Let side of the square be x.
x^{2} + x^{2} = (1  a)^{2} + (a  1)^{2}
⇒ 2x^{2} = 1 + a^{2} + 2a + a^{2} + 1  2a
⇒ x^{2} = a^{2} + 1
⇒ 50 = a^{2} + 1 ⇒ a^{2} = 49
⇒ a = ±7
Two vertices of a triangle are (2, 4) and (1, 3). If the origin is the centroid of the triangle then what is the third vertex?
third vertex = (3, 1)
What is the locus of a point equidistant from the point (2, 4) and yaxis?
Let, the point be P(x, y).
Given AP = BP
⇒ AP^{2} = BP^{2}
⇒ (x  2)^{2} + (y  4)^{2} = (x  0)^{2} + (y  y)^{2}
⇒ x^{2} + 4  4x + y^{2} + 16  8y = x^{2}
⇒ y^{2}  4x  8y + 20 = 0
If the coordinates of the midpoint of the sides of a triangle are (1, 1) (2, 3), and (3, 4) what is the centroid?
⇒ x_{1 }+ x_{2} = 6
y_{1} + y_{2} = 8
x_{2} + x_{3} = 2
y_{2} + y_{3} = 2
x_{1} + x_{3} = 4
y_{1} + y_{3} = 6
x_{1} + x_{2} + x_{3} = 12/2 = 6
x_{1} = 6  2 = 4; x_{2} = 2, x_{3} = 0
y_{1} + y_{2} + y_{3} = 2
y_{1} = 0; y_{2} = 8; y_{3} = 6
Centroid
What is the value of k, so that the points A(8, 1), B(3, 4), and C(2, K) are collinear?
Given points are A(8, 1), B(3, −4) and C(2, k)
It is also said that they are collinear and hence the area enclosed by them should be 0
Area of the triangle having vertices (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3})
= 1/2 x_{1}(y_{2 } y_{3}) + x_{2}(y_{3 } y_{1}) + x_{3}(y_{1 } y_{2})
Given that area of ∆ABC = 0
∴ 0 = 1/2 8(4 – k) + 3(k – 1) + 2(1 – (4))
∴ 0 = 1/2 32 – 8k + 3k  3 + 10
∴ 5k + 25 = 0
∴ k = 5
Hence, the value of k is 5.
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