Explanation
$$x+2y-9=0$$ ---(1)
$$3x+5y-5=0$$ ---(2)
$$ax+by-1=0$$ ---(3)
Solving (1) and (2) simultaneously we get
$$y=22, x=-35$$
Now, equation (1), (2) and (3) will be concurrent, that is they will pass through one point if $$y=22, x=-35$$ satisfy the third equation $$ax+by-1=0$$.
Substituting the values of 'x' and 'y' in this equation we get $$-35a+22b-1=0$$ ---(4)
And another equation given is $$22x-35y-1=0$$ ---(5)
$$x=b$$ & $$y=a$$
That is $$22x-35y-1=0$$ passes through $$(b,a)$$
We know that concurrent lines have same intersection points.Then
$$\therefore 3x+2y-5=0$$ --(1)
$$2x-5y+3=0$$ ---(2)
$$5x+by+c=0$$ ---(3)
From 1 & 2,
$$6x+4y-10=0$$
$$6x-15y+9=0$$
$$\Rightarrow 19y-19=0$$
$$y=1$$ & $$x=1$$
So, $$substituting\ \ x=1$$ & $$y=1$$, in equation (3),
$$5x+by+c=0$$
$$5+b+c=0$$
$$\therefore b+c=-5$$
$$ {\textbf{Step -1: Find equation}} $$
$$ {\text{Since A,O,C are linear they will lie on same line}}{\text{.}} $$
$$ {\text{Given points are A}}(1,2),{\text{O}}(0,0)\;{\text{and C}}(a,b) $$
$$ {\text{Therefore,}} $$
$$ {\text{area }}\Delta {\text{AOC = 0}} $$
$$ \therefore \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 0 $$
$$ {\text{Here, }}{x_1} = 1,{x_2} = 0\;{\text{and }}{x_3} = a $$
$$ {\text{and }}{y_1} = 2,\;{y_2} = 0\;{\text{and }}{y_3} = b $$
$$ {\textbf{Step -2: Find the relation}} $$
$$ \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 0 $$
$$ \Rightarrow \dfrac{1}{2}\left[ {1\left( {0 - b} \right) + 0\left( {b - 2} \right) + {\text{a}}\left( {2 - 0} \right)} \right] = 0 $$
$$ \Rightarrow - {\text{b + 2a}} = 0 $$
$$ \Rightarrow {\text{2a = b}} $$
$$ {\textbf{Hence, the correct answer is option C}}{\text{.}} $$
For the points $$AB =(x, y), BC =(1,2)$$ and $$ CD=(7,0)$$ to be collinear they must have area $$=0$$.
Thus, area of a triangle ABC = $$ \frac { 1 }{ 2 } \left( x_{ 1 }\left( y_{ 2 }-y_{ 3 }\right) +x_{ 2 }\left( y_{ 3 }-y_{ 1 } \right) +x_{ 3 }\left( y_{ 1 }-y_{ 2 }\right) \right)= 0 $$
$$=\dfrac{1}{2}(2x\times 2+6y-14)=0$$
$$=(2x+6y-14)=0$$
Thus, we have the relation $$x+3y=7$$ for the points to be collinear.
Given that PQR are three collinear points andPQ=2.5. Let us now find the distance of PR
Recall Distance Formula is =$$ \sqrt { { \left( { x}_{ 2 }{ { x }_{ 1 } } \right) }^{ 2}+{ \left( { y }_{ 2 }{ { y }_{ 1 } } \right) }^{ 2 } } \\$$
Distance PR=$$\sqrt { (113)^{ 2 })+( 104) ^{ 2 } }\\=\sqrt{100}=10$$
Let us now consider S is the midpoint of PR,then we have the co-ordinates of S as $$\dfrac{11+3}{2},\dfrac{10+4}{2}=(7, 7)$$.
$$ \therefore$$ it is clear that Q is the
midpoint of PS as we have know PQ=2.5 units.
Thus, to get the co-ordinates of Q takemidpoint of PS which is $$\dfrac{3+7}{2},\dfrac{4+7}{2}=(5,11/2)$$.
$$ {\textbf{Step - 1: Find equation}} $$
$$ {\text{Let A, B, C are the three points}} $$
$$ {\text{Since A, B, C are linear they will lie on same line}}{\text{.}} $$
$$ {\text{Given points are A}}(0,0),B(1,2)\;{\text{and C}}(x, y) $$
$$ {\text{area }}\Delta {\text{ABC = 0}} $$
$$ \Rightarrow \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 0 $$
$$ {\text{Here, }}{x_1} = 0,{x_2} = 1\;{\text{and }}{x_3} = x $$
$$ {\text{and }}{y_1} = 0,\;{y_2} = 2\;{\text{and }}{y_3} = y $$
$$ {\textbf{Step - 2: Find the relation}} $$
$$ \Rightarrow \dfrac{1}{2}\left[ {0\left( {2 - y} \right) + 1\left( {y - 0} \right) + x\left( {0 - 2} \right)} \right] = 0 $$
$$ \Rightarrow y - 2x = 0 $$
$$ \Rightarrow {\text{2x = y}} $$
$$ {\textbf{Hence, the correct answer is option B}}{\text{.}} $$
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