Explanation
Distance between two points $$= \sqrt { \left( { x }_{ 2 }-{x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$Distance between the points $$A (3,1) $$ and $$B (-3,2) = \sqrt { \left( -3-3 \right) ^{ 2 }+\left( 2 - 1 \right) ^{ 2 } } = \sqrt { 36 + 1 } = \sqrt { 37 }$$
Distance between the points $$B(-3,2) $$ and $$C (0,2-\sqrt {3}) = \sqrt { \left( 0 + 3 \right) ^{ 2 }+\left(2 - \sqrt {3} - 2\right) ^{ 2 } } = \sqrt { 9 + 3 } = \sqrt { 12 } $$
Distance between the points $$A(3,1) $$ and $$ C(0,2-\sqrt {3})$$
$$ = \sqrt { \left( 0-3\right) ^{ 2 }+\left( 2-\sqrt {3} - 1\right) ^{ 2 } } = \sqrt { 9 + 1 + 3 -2\sqrt {3} } = \sqrt { 13 - 2\sqrt {3} } $$
Since the length of the sides between all vertices are different, they are the vertices of a scalene triangle.
Area of a triangle $$= \left| \cfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$Area of $$\triangle ABC= \left| \cfrac { (3)(2-2+\sqrt {3})+(-3)(2-\sqrt {3} - 1)+0(1-2) }{ 2 } \right| $$
$$ = \left| \cfrac { 3\sqrt {3} -3 + 3\sqrt {3} }{ 2 } \right| $$
$$ = \cfrac{-3 + 6 \sqrt {3}}{2}$$ sq. units
Let the point on the x-axis be $$ (x,0) $$.Distance between $$ (x,0) $$ and $$ (7,6) = \sqrt { \left( 7-x \right) ^{ 2}+\left( 6 - 0 \right) ^{ 2 } } = \sqrt { { 7}^{ 2 }+ { x }^{ 2 } - 14x + 36 }= \sqrt { { x }^{ 2 } - 14x + 85 } $$Distance between $$ (x,0) $$ and $$ (-3,4) = \sqrt { \left( -3-x \right) ^{ 2}+\left( 4 - 0 \right) ^{ 2 } } = \sqrt { 9 + { x }^{ 2 } + 6x + 16 } =\sqrt { { x }^{ 2 } + 6x + 25 } $$As the point $$ (x,0) $$ is equidistant from the two points, both the distances calculated are equal. $$ \sqrt { { x }^{ 2 } - 14x + 85 } = \sqrt { { x }^{ 2 } + 6x + 25} $$$$ \Rightarrow { x }^{ 2 } - 14x + 85 = { x }^{ 2 } + 6x + 25 $$$$ \Rightarrow 85 - 25 = 6x + 14x $$$$\Rightarrow 60 = 20x $$
$$\Rightarrow x = 3 $$Thus, the point is $$ (3,0) $$.
or, $$ {PA}^{2} = {PB}^{2} $$$$ => {(x-3)}^{2} + {(y-4)}^{2} = {(x-5)}^{2} + {(y+2)}^{2} $$$$ {x} ^{2} -6x + 9 + {y} ^{2} -8y + 16 = {x}^{2} -10x + 25 + {y}^{2} + 4y + 4 $$$$ 4x - 12y = 4 $$$$ x - 3y = 1 $$ .......(i)Also Area of $$ PAB=10 $$$$ | \dfrac {(x (4+2)+3(-2-y)+5(y-4)}{ 2 } =10 $$$$ \left| \dfrac { 6x -6-2y +5y -20 }{ 2 } \right| = 10 $$$$ \dfrac {6x+3y -26}{2} = \pm 10 $$$$ 6x+2y26=\pm 20 $$$$ 6x+2y =46 $$ ........(ii)or $$ 6x + 2y = 6 $$ .......(iii)Solving\ equation\ (i)\ and\ (ii) $$ x = 7, y = 2 $$ and solving equation (i) and (iii) $$ x = 1, y = 0 $$
So, the co-ordinates of P are $$ (7, 2) or (1, 0) $$
$$ => \sqrt { \left( -3-x \right) ^{ 2 }+\left( 0 - y\right) ^{ 2 } } = 4 $$
But $$ x = 0 $$$$=> \sqrt { 9 + { y }^{ 2 } } = 4 $$$$ => 9 + {y}^{2} = 16 $$
$$ {y}^{2} = 7 $$
$$ => y = \sqrt {7} $$ or $$ - \sqrt {7} $$
Using the section formula, if a point $$(x,y)$$ divides theline joining the points $$({ x }_{ 1 },{ y }_{ 1 })$$ and $$({ x }_{ 2 },{ y}_{ 2 })$$externally in the ratio $$ m:n$$, then $$(x,y) = \left( \dfrac { m{ x }_{ 2 }-n{ x }_{ 1 } }{ m-n } ,\dfrac {m{ y }_{ 2 }-n{ y }_{ 1 } }{ m-n } \right) $$Since, $$AC = 2BC => \dfrac { AC }{ BC } =\dfrac { 2 }{ 1 } $$Substituting $$({ x }_{ 1 },{ y }_{ 1 }) = (-3,4) $$ and $$({x }_{ 2 },{ y }_{2 }) = (2,1) $$ and $$ m = 2, n = 1 $$ in the section formula, we get $$ C = \left( \dfrac { 2(2)-1(-3) }{ 2-1 } ,\dfrac { 2(1)-1(4) }{ 2-1 } \right) =\left( 7,-2 \right) $$
Distance between the points $$ (0,3) $$ and $$ (5,0) =\sqrt { \left( 5-0\right) ^{ 2 }+\left( 0-3\right) ^{ 2 } } = \sqrt { 25 + 9 } = \sqrt { 34 } $$
Let the point on the x-axis be $$ (x,0) $$ Distance between $$ (x,0) $$ and $$ (5,4) = \sqrt { \left( 5-x \right) ^{ 2}+\left( 4 - 0 \right) ^{ 2 } } = \sqrt { { 5}^{ 2 }+ { x }^{ 2 } - 10x + 16 }= \sqrt { { x }^{ 2 } - 10x + 41 } $$Distance between $$ (x,0) $$ and $$ (-2,3) = \sqrt { \left( -2-x \right) ^{ 2}+\left( 3 - 0 \right) ^{ 2 } } = \sqrt { { 2}^{ 2 }+ { x }^{ 2 } + 4x + 9 } =\sqrt { { x }^{ 2 } + 4x + 13 } $$
As the point $$ (x,0) $$ is equidistant from the two points, both the distances calculated are equal.
$$ \sqrt { { x }^{ 2 } - 10x + 41 } = \sqrt { { x }^{ 2 } + 4x + 13 } $$
$$ => { x }^{ 2 } - 10x + 41 = { x }^{ 2 } + 4x + 13 $$
$$ 41 - 13 = 10x + 4x $$
$$ 28 = 14x $$$$ x = 2 $$. Thus, the point is $$ (2,0) $$.
Using this formula, mid point of AB $$ = D = \left( \dfrac { 2-4 }{ 2 } ,\dfrac { 2-4 }{ 2 } \right) \quad =\quad (-1,-1) $$
Distance between two points $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 }\right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$Distance between the points C$$ (5,-8) $$ and D $$ (-1,-1) = \sqrt { \left( -1-5\right) ^{ 2 }+\left( -1 +8 \right) ^{ 2 } } = \sqrt { 36 + 49} = \sqrt {85 }$$
Distance between two points $$ \left( { x }_{ 1 },{ y }_{ 1} \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculatedusing the formula $$ \sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2}+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$Distance between the points $$ (x+y,x-y) $$ and $$ (x-y,x+y) = \sqrt { \left(x-y-x-y\right) ^{ 2 }+\left( x+y-x+y \right) ^{ 2 } } = \sqrt { { (2y) }^{ 2 }+{ (2y) }^{ 2 }} = 2\sqrt {2} y $$
Since, $$AD = 3AB => \dfrac { AD }{ BD} =\dfrac { 3 }{2 } $$
Using the section formula, if a point $$(x,y)$$ divides theline joining the points $$({ x }_{ 1 },{ y }_{ 1 })$$ and $$({ x }_{ 2 },{ y}_{ 2 })$$externally in the ratio $$ m:n$$, then $$(x,y) = \left( \dfrac { m{ x }_{ 2 }-n{ x }_{ 1 } }{ m-n } ,\dfrac {m{ y }_{ 2 }-n{ y }_{ 1 } }{ m-n } \right) $$Substituting $$({ x }_{ 1 },{ y }_{ 1 }) = (1,1) $$ and $$({x }_{ 2 },{ y }_{2 }) = (2,-3) $$ and $$ m = 3, n = 2 $$ in the section formula, we get $$ C = \left( \dfrac { 3(2)-2(1) }{ 3-2 } ,\dfrac { 3(-3)-2(1) }{ 3-2 } \right) =\left( 4,-11 \right) $$
Distance between two points $$ \left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and$$ \left( { x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated using theformula $$ \sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{2 }-{ y }_{ 1 } \right) ^{ 2 } } $$Distance between the points $$ (3,a) $$ and $$ (4,1) = \sqrt {10} $$
$$ => \sqrt { \left( 4-3 \right) ^{ 2 }+\left( 1 - a \right) ^{ 2 } } = \sqrt {10} $$
$$ \sqrt { 1 + { (1 - a) }^{ 2 } } = \sqrt {10} $$
$$ 1 + { (1 - a) }^{ 2 } = 10 $$
$$ { (1-a) }^{ 2 } = 9 $$
$$ 1-a = 3 $$ or $$ - 3 $$
$$ => a = -2 $$ or $$ 4 $$
Let $$ A (2,5), B (3,-4) $$ and $$ C (7,10) $$Distance between two points $$ \left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$\left( { x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y}_{ 1 } \right) ^{ 2 } } $$
Distance between the points A and B $$ AB = \sqrt { \left( 3+2 \right) ^{ 2 }+\left( -4-5\right) ^{ 2 } } = \sqrt { 25 + 81 } = \sqrt { 106 } $$
Distance between the points B and C $$ BC = \sqrt { \left(7-3 \right) ^{ 2 }+\left( 10+4\right) ^{ 2 } } = \sqrt { 16 + 196 } = \sqrt { 212 } $$
Distance between the points A and C $$AC = \sqrt { \left(7+2 \right) ^{ 2 }+\left(10-5\right) ^{ 2 } } = \sqrt { 81 + 25 } = \sqrt { 106 } $$
Since, $$ {(\sqrt { 212}) }^{2} = {(\sqrt { 106 }) }^{2} + {(\sqrt { 106 })}^{2} $$, the triangle is a right angled triangle.
Distancebetween two points $$ \left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$ \left( {x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated using the formula $$ \sqrt {\left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 }\right) ^{ 2 } } $$Distance between $$ (a,2) $$ and $$ (3,4) = 8$$ $$ \sqrt { \left( 3-a \right) ^{ 2 }+\left( 4-2 \right) ^{ 2 } } = 8 $$$$ \sqrt { 9 + { a }^{ 2 } - 6a + 4 } = 8 $$$$ \sqrt { { a }^{ 2 } - 6a + 13 } = 8 $$$$ { a }^{ 2 }- 6a + 13 = 64 $$$$ { a }^{ 2 } - 6a - 51 = 0 $$
$$ a= \dfrac { -(-6)\pm \sqrt { \left( { -6 }^{ 2 } \right) -4(1)(-51) } }{ 2(1) } $$
Consider given, distance between the points $$(3,k)$$ and $$(4,1)$$ is $$\sqrt{10}$$.
Now,
$$ \sqrt{{{\left( 3-4 \right)}^{2}}+{{\left( k-1 \right)}^{2}}}=\sqrt{10} $$
$$ {{\left( -1 \right)}^{2}}+{{\left( k-1 \right)}^{2}}=10 $$
$$ {{\left( k-1 \right)}^{2}}=9 $$
$$ k-1=\pm 3 $$
If,
$$k-1=-3$$
$$k=-2$$
if,
$$k-1=+3$$
$$ k=4$$
Hence, this is the answer.
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