Straight Lines Test -8

consider the points (a,b),(0,0),(-a,-b)

It is clear the (0,0) is the midpoint of (a,b) and (-a,-b)

This implies that these three points are collinear

Now consider the points,

(a,b),(a²,ab) and (-a,-b)

If three points are collinear then the area if the triangle formed by these points is zero

That is 1/2[x₁(y₂-y₃) +x₂(y₃-y₁) + x₃(y₁-y₂) = 0

Substituting the values

1/2[a(a +b)+a²(-b-b) +(-a)(b-ab)]

on expanding and simplifying we get the value to be zero

Hence these points are collinear.

The abscissa is equal to the ordinate implies x = y

Hence the locus is x-y=0

Given xy=0 and x+y=1, indicates that the straight line passes through (1,0) and (0,1). Hence it is a right angled triangle whose vertex is origin. Therefore the orthocenter is also at the origin (0,0). this is one of the properties of right angled triangle whose vertex is origin.
 

Let F be the midpoint on AC

Hence the coordiate of F is (x₁+1)/2 and (5-x₁+2)/2

This point lies on x=4

Therefore (x₁+1)/2 = 4

Therefore x₁ = 7 

Therefore y = 5-7 = -2

Hence the coordiates of B is (7,-2)

Let (0,y) be the point on Y axis which is equidistant from the points (-1,2) and (3,4)

By applying the distance formula,

(0+1)² +(y-2)² = (3-0)² +(4-y)²

on simplifying we get

4y = 20

Therefore y = 5

Hence the point on the y axis is (0,5)

Equation of the line through the given points is $$y-3 = x-2 \Rightarrow x-y+1=0$$. 

For a rectangle, the opposite sides need to be equal and diagonals also have to be equal.

Slope of a line $$=m=\tan { \theta  } $$

Area of a triangle ABC = $$ \frac { 1 }{ 2 } \left|[ x_{ 1 }\left( y_{ 2 }-y_{ 3 }

\right) +x_{ 2 }\left( y_{ 3 }-y_{ 1 } \right) +x_{ 3 }\left( y_{ 1 }-y_{ 2 }

\right)  \right]|$$

Substituting the given coordinates, we have

=$$ \frac { 1 }{ 2 } \left[ 5 \left(7+4) \right) -\left( 4+2\right) +7\left( 2-7) \right)  \right]$$

$$ =\frac { 1 }{ 2 } \left( 5 ( 11) +4\times( -6)
+7\times( -5)\right)$$

$$ =\frac { 1 }{ 2 } \left[(55) - ( 24) - ( 35)  \right]$$

$$ =\frac { 1 }{ 2 } \times |-4| = 2$$ square units.

Using distance formula,
$${L}_{1}=\sqrt{({x}_{2}-{x}_{1})^2+({y}_{2}-{y}_{1})^2}$$

The slope of a line passing through the point $$( { x }_{ 1 },{ y }_{ 1 })$$ and $$( { x }_{ 2 },{ y }_{ 2 })$$ is given by 

$$(A)=\ |y_{1}|\rightarrow 2 $$

The distance between two points can be found by the equation

$$To\quad identify-\\ whether\quad ABCD\quad is\quad a\quad (i)\quad parallelogram,\quad (ii)\quad rectangle,\quad (iii)\quad square,\quad \\ (iv)\quad scalene\quad one.\\ Solution-\\ (i)\quad We\quad know\quad that\quad in\quad a\quad parallelogrm\quad the\quad diagonals\quad bisect\quad each\quad \\ other\quad i.e\quad the\quad mid\quad points\quad of\quad AC\quad \& \quad BD\quad are\quad same.\\ Let\quad the\quad mid\quad point\quad of\quad AC\quad be\quad P(x,y).\\ Applying\quad mid\quad point\quad theorem,\\ P(x,y)=\left( \dfrac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } ,\dfrac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 }  \right) =\left( \dfrac { -2+4 }{ 2 } ,\dfrac { -1+3 }{ 2 }  \right) =(1,1).\\ Again\quad \\ let\quad the\quad mid\quad point\quad of\quad BD\quad be\quad Q(p,q).\\ Applying\quad mid\quad point\quad theorem,\\ Q(p,q)=\left( \dfrac { { x }_{ 2 }+{ x }_{ 4 } }{ 2 } ,\dfrac { { y }_{ 2 }+{ y }_{ 4 } }{ 2 }  \right) =\left( \dfrac { 1+1 }{ 2 } ,\dfrac { 0+1 }{ 2 }  \right) =(1,1).\\ \therefore \quad P(x,y)=Q(p,q)\\ \Longrightarrow the\quad mid\quad points\quad of\quad AC\quad \& \quad BD\quad are\quad same.\\ So\quad ABCD\quad is\quad a\quad parallelogram.\\ (ii)\quad If\quad the\quad diagonals\quad of\quad the\quad parallelogram\quad ABCD\quad are\quad equal\quad \\ then\quad it\quad is\quad a\quad rectangle.\\ i.e\quad AC=BD.\\ Applying\quad distance\quad formula,\\ AC=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 3 } \right)  }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 3 } \right)  }^{ 2 } } =\sqrt { { \left( -2-4 \right)  }^{ 2 }+{ \left( -1-3 \right)  }^{ 2 } } units=2\sqrt { 13 } units.\\ BD=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 4 } \right)  }^{ 2 }+{ \left( { y }_{ 2 }-y_{ 4 } \right)  }^{ 2 } } =\sqrt { { \left( 1-1 \right)  }^{ 2 }+{ \left( 0-2 \right)  }^{ 2 } } units=2units.\\ \therefore \quad AC\neq BD.\quad \\ So\quad ABCD\quad is\quad not\quad a\quad rectagle.\\ Consequently,\quad it\quad is\quad not\quad a\quad square.\\ Ans-\quad Option\quad A.\\  $$ 

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y

}_{ 2 })$$  and $$({ x }_{ 3 },{ y }_{ 3 })$$  is $$ \left| \frac { {

x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{

3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 }  \right| $$
Hence, area of the triangle with given vertices is $$ \left| \frac { { a }(c-c+a)+a(c-a-c-a)-a(c+a-c) }{ 2 }  \right|  = \left| \frac { { a }^{ 2 }-2{ a }^{ 2 }-{ a }^{ 2 } }{ 2 }  \right|  = { a }^{ 2 } $$