Let A(a,b),B(3,4) and (-6,-8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point…

Given: $A(a,b),B(3,4),C(-6,-8)$A(a,b),B(3,4),C(−6,−8) denotes centroid, circumcentre and orthocentre respectively.

We know that, the line joining orthocentre and circumcentre is divided by centroid in the ratio $2:1$2:1, where distance from orthocentre being the larger part.

$$\Rightarrow a=\frac{2\times 3+1(-6)}{2+1},b=\frac{2\times 4+1(-8)}{2+1}$$⇒a=2+12×3+1(−6)​,b=2+12×4+1(−8)​

$$\Rightarrow a=0,b=0$$⇒a=0,b=0

$$\Rightarrow P\equiv (2\times 0+3,7\times 0+5)$$⇒P≡(2×0+3,7×0+5)

$$\Rightarrow P\equiv (3,5)$$⇒P≡(3,5)

Distance from $P$P measured along $x-2y-1=0$x−2y−1=0

$$\Rightarrow x=3+r\mathrm{cos}\theta ,y=5+r\mathrm{sin}\theta$$⇒x=3+rcosθ,y=5+rsinθ

Where, $\mathrm{tan}\theta =\frac{1}{2}$tanθ=21​

So, the distance of point from $2x+3y-4=0$2x+3y−4=0 will be,

$$2(3+r\mathrm{cos}\theta )+3(5+r\mathrm{sin}\theta )-4=0$$2(3+rcosθ)+3(5+rsinθ)−4=0

$$\Rightarrow r(2\mathrm{cos}\theta +3\mathrm{sin}\theta )=-17$$⇒r(2cosθ+3sinθ)=−17

$$\Rightarrow r=|-\frac{17}{(2\cdot \frac{2}{\sqrt{5}}+3\cdot \frac{1}{\sqrt{5}})}|$$⇒r=∣−(2⋅5​2​+3⋅5​1​)17​∣

$$\Rightarrow r=|\frac{-17\sqrt{5}}{7}|$$⇒r=∣7−175​​∣

$$\Rightarrow r=\frac{17\sqrt{5}}{7}$$⇒r=7175​​