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QUESTION: 1

**Distance of a point from the origin:**

The distance of a point A(x, y) from the origin O(0, 0) is given by **OA = √x ^{2} + y^{2}**

Find the distance of the point A(4, -2) from the origin.

Solution:

OA = √4 - 0^{2}+(-2 - 0)^{2} = √16+4 = √20 = radic;4*5 = 2√5 units

QUESTION: 2

**Distance between two points :**

If (x_{1}, y_{1}) and B(x_{2}, y_{2}) be two points, then ** AB = √(x _{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}**

Find the distance between the points A(-4, 7) and B(2, -5).

Solution:

AB = √(2+4)^{2} + (-5-7)^{2}

= √6^{2} + (-12)^{2}

= √36+144 = √180

=√36*5 = 6√5 units.

QUESTION: 3

The distance between the points A(b, 0) and B(0, a) is.

Solution:

AB = √(b-0)^{2}-(0-a)^{2}

= √b^{2}+a^{2}

= √a^{2}+b^{2}.

QUESTION: 4

If the distance of the point P(x, y) from A(a, 0) is a + x, then y^{2} = ?

Solution:

√(x-a)^{2}+(y-0)^{2} = a + x

= (x-a)^{2}+y^{2}

= (a+x)^{2} => y^{2} = (x-a)^{2}-(x-a)^{2}-4ax => y^{2} = 4ax

QUESTION: 5

The distance between the points A(5, -7) and B(2, 3) is:

Solution:

AB^{2} = (2 - 5)^{2} + (3 + 7)^{2}

=> (-3)^{2} + (10)^{2}

=> 9 + 100 => √109

QUESTION: 6

**Area of a triangle :**

If A(x_{1},y_{1}), B(x_{2},y_{2} and C(x_{3}, y_{3}) be three vertices of a ΔABC, then its area is given by:

**Δ = 1/2 [x _{1}(y_{2} - y_{3} + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})]**

Find the area of ΔABC whose vertices are A(9, -5), B(3, 7) and (-2, 4).

Solution:

Here, x_{1} = 9, x_{2} = 3, x_{3} = -2 and y_{1} = -5, y_{2} = 7, y_{3} = 4

= 1/2 [9(7-4) + 3(4+5) + (-2)(-5-7)]

= 1/2 [9(3) + 3(9) - 2(-12)]

= 1/2 [27 + 27 + 24]

= 1/2 [78]

= 39 sq.units

QUESTION: 7

Find the area of ΔABC whose vertices are A(2, -5), B(4, 9) and (6, -1).

Solution:

Here, x_{1} = 2, x_{2} = 4, x_{3} = 6 and y_{1} = -5, y_{2} = 9, y_{3} = -1

= 1/2 [2(9+1) + 4(-1+5) + 6(-5-9)]

= 1/2 [2(10) + 4(4) + 6(-14)]

= 1/2 [20 + 16 - 84]

= 1/2 [-48]

= 24 sq.units

QUESTION: 8

The points A(0, 6), B(-5, 3) and C(3, 1) are the vertices of a triangle which is ?

Solution:

AB^{2}= (-5 - 0)^{2} + (-3 - 0)^{2} = 16 + 9 = 25

BC^{2} = (3 + 5)^{2} + (1-3)^{2} = 8^{2} + (-2)^{2} = 64 + 4 = 68

AC^{2} = (3 - 0)^{2} + (1 - 6)^{2} = 9 + 25 = 34.

AB = AC. ==> ΔABC is isosceles.

QUESTION: 9

The points A(-4, 0), B(1, -4), and C(5, 1) are the vertices of

Solution:

AB^{2} = (1 + 4)^{2} + (-4 - 0)^{2}

= 25 + 16 = 41,

BC^{2} = (5 - 1)^{2} + (1 + 4)^{2} = 4^{2} + 5^{2}

= 16 + 25 = 41

AC^{2} = (5 + 4)^{2} + (1 - 0)^{2}

= 81 + 1 = 82

AB = BC and AB^{2} = BC^{2} = AC^{2}

ΔABC is an isosceles right angled triangle

QUESTION: 10

Find the coordinates of the point equidistant from the points A(1, 2), B (3, –4) and C(5, –6).

Solution:

**Given** three points A(1,2) B(3,-4) and C(5,-6)

we have **to find the coordinates of the point equidistant from the points.**

The point that is equidistant from three points is called** circumcenter which can be evaluated to find the perpendicular bisectors.**

**To find the perpendicular bisectors of AB:
**(11,2)

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