JEE Main 2022 — Mathematics Coordinate Geometry

Given the equations of the sides $AB,BC$AB,BC and $CA$CA of a triangle $ABC$ABC are $2x+y=0,x+py=15a$2x+y=0,x+py=15a and $x-y=3$x−y=3 respectively.

Now on solving equation $2x+y=0,x+py=15a$2x+y=0,x+py=15a and $x-y=3$x−y=3 we get coordinates of $A(1,-2),B(\frac{15a}{1-2p},\frac{-30a}{1-2p})$A(1,−2),B(1−2p15a​,1−2p−30a​) and $C=(\frac{18p-30}{p+1},\frac{15p-33}{p+1})$C=(p+118p−30​,p+115p−33​)

And let orthocentre be $H(2,a)$H(2,a)

Slope of $AH=\frac{a+2}{1}$AH=1a+2​

Sloe of $BC=-\frac{1}{p}$BC=−p1​

As $AH\perp BC$AH⊥BC, so $p=a+2$p=a+2

Now coordinate of $C=(\frac{18p-30}{p+1},\frac{15p-33}{p+1})$C=(p+118p−30​,p+115p−33​)

So, slope of $HC$HC

$$=\frac{\frac{15p-33}{p+1}-a}{\frac{18p-30}{p+1}-2}=\frac{15p-33-(p-2)(p+1)}{18p-30-2p-2}$$=p+118p−30​−2p+115p−33​−a​=18p−30−2p−215p−33−(p−2)(p+1)​

$$=\frac{16p-{p}^{2}-31}{16p-32}$$=16p−3216p−p2−31​

Now $HC\perp AB$HC⊥AB, so $\text{slope of}HC\times \text{slope of }AB\text{=-1}$slope ofHC×slope of AB=-1

$$\Rightarrow \frac{16p-{p}^{2}-31}{16p-32}\times -2=-1$$⇒16p−3216p−p2−31​×−2=−1

$$\Rightarrow {p}^{2}-8p+15=0$$⇒p2−8p+15=0

$$\Rightarrow p=3\mathrm{or}5$$⇒p=3or5

But if $p=5$p=5 then $a=3$a=3 not acceptable as $p=a+2$p=a+2

$$\therefore p=3$$∴p=3