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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