JEE Main 2024 — Mathematics Coordinate Geometry

Given: $A(a,-2),B(a,6)\text{and}C(\frac{a}{4},-2)$A(a,−2),B(a,6)andC(4a​,−2)

$$\Rightarrow A{B}^{2}={(a-a)}^{2}+{(6+2)}^{2}$$⇒AB2=(a−a)2+(6+2)2

$$\Rightarrow A{B}^{2}=64...(i)$$⇒AB2=64...(i)

$$\Rightarrow B{C}^{2}={(a-\frac{a}{4})}^{2}+{(6+2)}^{2}$$⇒BC2=(a−4a​)2+(6+2)2

$$\Rightarrow B{C}^{2}=\frac{9{a}^{2}}{16}+64...(ii)$$⇒BC2=169a2​+64...(ii)

$$\Rightarrow A{C}^{2}={(a-\frac{a}{4})}^{2}+{(-2+2)}^{2}$$⇒AC2=(a−4a​)2+(−2+2)2

$$\Rightarrow A{C}^{2}=\frac{9{a}^{2}}{16}...(iii)$$⇒AC2=169a2​...(iii)

Using $(i),(ii)\text{and}(iii)$(i),(ii)and(iii),

$$\Rightarrow A{C}^{2}+A{B}^{2}=B{C}^{2}$$⇒AC2+AB2=BC2

So, $\triangle ABC$△ABC is right angled at $A$A.

So, circumcentre will be mid-point of $BC$BC.

$$\Rightarrow S\equiv (\frac{a+\frac{a}{4}}{2},\frac{6-2}{2})$$⇒S≡(2a+4a​​,26−2​)

$$\Rightarrow S\equiv (\frac{5a}{8},2)=(5,\frac{a}{4})$$⇒S≡(85a​,2)=(5,4a​)

$$\Rightarrow a=8$$⇒a=8

$$\Rightarrow BC=\sqrt{\frac{9\times 64}{16}+64}$$⇒BC=169×64​+64​

$$\Rightarrow BC=\sqrt{100}=10$$⇒BC=100​=10

$$\Rightarrow AB=8$$⇒AB=8

$$\Rightarrow AC=\sqrt{\frac{9\times 64}{16}}$$⇒AC=169×64​​

$$\Rightarrow AC=6$$⇒AC=6

Now, circumradius is given by,

$$R=\frac{BC}{2}=5\Rightarrow \alpha =5$$R=2BC​=5⇒α=5

Now, area of triangle will be $\beta =\frac{1}{2}\times 6\times 8=24$β=21​×6×8=24 and perimeter will be $\gamma =6+8+10=24$γ=6+8+10=24

$$\Rightarrow \alpha =5,\beta =24,\gamma =24$$⇒α=5,β=24,γ=24

$$\Rightarrow \alpha +\beta +\gamma =53$$⇒α+β+γ=53