JEE Main 2022 — Mathematics Coordinate Geometry

On plotting the graph we get,

Now by diagram ${m}_{AC}\longrightarrow \infty$mAC​⟶∞ and ${m}_{PD}=0$mPD​=0 as $PD$PD is perpendicular bisector, so $D(\frac{a+a}{2},\frac{b+3}{2})$D(2a+a​,2b+3​)

$$\equiv D(a,\frac{b+3}{2})$$≡D(a,2b+3​)

Now since ${m}_{PD}=0$mPD​=0, so $\frac{b+3}{2}-1=0$2b+3​−1=0

$\Rightarrow b+3-2=0$⇒b+3−2=0$\Rightarrow b=-1$⇒b=−1

Now similarly finding the coordinates of $E(\frac{b+a}{2},\frac{5+b}{2})=(\frac{a-1}{2},2){\text{as}b=-1}$E(2b+a​,25+b​)=(2a−1​,2)asb=−1

Now we know that product of perpendicular slopes is ${m}_{CB}\cdot {m}_{EP}=-1$mCB​⋅mEP​=−1

$$\Rightarrow (\frac{5-b}{b-a})=(\frac{2-1}{\frac{a-1}{2}-1})=-1$$⇒(b−a5−b​)=(2a−1​−12−1​)=−1

$$\Rightarrow (\frac{6}{-1-a})=(\frac{2}{a-3})=-1$$⇒(−1−a6​)=(a−32​)=−1

$$\Rightarrow 12=(1+a)(a-3)$$⇒12=(1+a)(a−3)

$$\Rightarrow 12={a}^{2}-3a+a-3$$⇒12=a2−3a+a−3

$$\Rightarrow {a}^{2}-2a-15=0$$⇒a2−2a−15=0

$$\Rightarrow (a-5)(a+3)=0$$⇒(a−5)(a+3)=0

So, $a=5$a=5 or $a=-3$a=−3

Now given $ab>0$ab>0

So, $a(-1)>0$a(−1)>0$\Rightarrow a<0$⇒a<0

So, $a=-3$a=−3

Now equation of line $AP$AP by two point form will be, {where A(-3,3)&P(1,1)}

$$y-1=(\frac{3-1}{-3-1})(x-1)$$y−1=(−3−13−1​)(x−1)

$$\Rightarrow -2y+2=x-1$$⇒−2y+2=x−1

$\Rightarrow x+2y=3$⇒x+2y=3 $\cdots (1)$⋯(1)

Now equation of line $BC$BC {where $B(-1,5)$B(−1,5) and $C(-3,-1)$C(−3,−1)} will be,

$$(y-5)=\frac{6}{2}(x+1)$$(y−5)=26​(x+1)

$$\Rightarrow y-5=3x+3$$⇒y−5=3x+3

$$\Rightarrow y=3x+8\cdots (2)$$⇒y=3x+8⋯(2)

Now solving $(1)$(1) and $(2)$(2) we get,

$$x+2(3x+8)=3$$x+2(3x+8)=3

$\Rightarrow 7x+16=3$⇒7x+16=3$\Rightarrow 7x=-13$⇒7x=−13

$$\Rightarrow x=-\frac{13}{7}$$⇒x=−713​

And $y=3(-\frac{13}{7})+8$y=3(−713​)+8$=\frac{-39+56}{7}$=7−39+56​

So, $y=\frac{17}{7}$y=717​

So the value of $({k}_{1},{k}_{2})$(k1​,k2​) will be $x+y=\frac{-13+17}{7}=\frac{4}{7}$x+y=7−13+17​=74​ {\text{where}{k}_{1}\text{is}x&{k}_{2}\text{is }y}