Let the point P(α ,β ) be at a unit distance from each of the two lines L_1:3x-4y+12=0, and L_2:8x+6y+11=0. If P lies below L_1 and above L_2, then…

Given, the point $P(\alpha )(,\beta )$P(α)(,β) be at a unit distance from each of the two lines ${L}_{1}:3x-4y+12=0$L1​:3x−4y+12=0, and ${L}_{2}:8x+6y+11=0$L2​:8x+6y+11=0, so on plotting the diagram we get,

Now, ${L}_{1}:3x-4y+12=0$L1​:3x−4y+12=0 and ${L}_{2}:8x+6y+11=0$L2​:8x+6y+11=0

Since {L}_{1}&{L}_{2} are perpendicular so it will form square of unit length, so equation of angle bisector of ${L}_{1}$L1​ and ${L}_{2}$L2​ of angle containing origin and will pass through $(\alpha ,\beta )$(α,β), so

Equation of angle bisector will be, $\frac{(3x-4y+12)}{\sqrt{{3}^{2}+{4}^{2}}}=\frac{8x+6y+11}{\sqrt{{8}^{2}+{6}^{2}}}$32+42​(3x−4y+12)​=82+62​8x+6y+11​

$$\Rightarrow 2(3x-4y+12)=8x+6y+11$$⇒2(3x−4y+12)=8x+6y+11

$$\Rightarrow 2x+14y-13=0$$⇒2x+14y−13=0

$$\Rightarrow 2\alpha +14\beta -13=0\cdots (i)$$⇒2α+14β−13=0⋯(i)

Also given perpendicular distance is one unit so, $\frac{3\alpha -4\beta +12}{5}=1$53α−4β+12​=1

$$\Rightarrow 3\alpha -4\beta +7=0\cdots (\mathrm{ii})$$⇒3α−4β+7=0⋯(ii)

Solving equation (\text{i})&(\text{ii})

$$2\alpha +14\beta -13=0$$2α+14β−13=0

$$3\alpha -4\beta +7=0$$3α−4β+7=0

$\Rightarrow$⇒$P(\alpha )(,\beta ),\alpha =\frac{-23}{25},\beta =\frac{53}{50}$P(α)(,β),α=25−23​,β=5053​

So, $100(\alpha +\beta )=14$100(α+β)=14