Let A be the point of intersection of the lines 3x+2y=14,5x-y=6 and B be the point of intersection of the lines 4x+3y=8,6x+y=5. The distance of the…

Solving lines ${L}_{1}(3x+2y=14)$L1​(3x+2y=14) and ${L}_{2}(5x-y=6)$L2​(5x−y=6).

$$\Rightarrow x=\frac{14-2y}{3}=\frac{6+y}{5}$$⇒x=314−2y​=56+y​

$$\Rightarrow 70-10y=18+3y$$⇒70−10y=18+3y

$$\Rightarrow 52=13y$$⇒52=13y

$$\Rightarrow y=4$$⇒y=4

$$\Rightarrow x=\frac{14-8}{3}=2$$⇒x=314−8​=2

$$\Rightarrow A\equiv (2,4)$$⇒A≡(2,4)

Now, solving lines ${L}_{3}(4x+3y=8)$L3​(4x+3y=8) and ${L}_{4}(6x+y=5)$L4​(6x+y=5)

$$\Rightarrow x=\frac{8-3y}{4}=\frac{5-y}{6}$$⇒x=48−3y​=65−y​

$$\Rightarrow 48-18y=20-4y$$⇒48−18y=20−4y

$$\Rightarrow 28=14y$$⇒28=14y

$$\Rightarrow y=2$$⇒y=2

$$\Rightarrow x=\frac{8-6}{4}=\frac{1}{2}$$⇒x=48−6​=21​

$$\Rightarrow B\equiv (\frac{1}{2},2)$$⇒B≡(21​,2)

So, equation of $AB$AB is given by,

$$AB:(y-4)=(\frac{2-4}{\frac{1}{2}-2})(x-2)$$AB:(y−4)=(21​−22−4​)(x−2)

$$\Rightarrow AB:(y-4)=(\frac{4}{3})(x-2)$$⇒AB:(y−4)=(34​)(x−2)

$$\Rightarrow AB:(y-4)=(\frac{4}{3})(x-2)$$⇒AB:(y−4)=(34​)(x−2)

$$\Rightarrow 3y-12=4x-8$$⇒3y−12=4x−8

$$\Rightarrow 4x-3y+4=0$$⇒4x−3y+4=0

The distance of $P(5,-2)$P(5,−2) from the line $AB$AB is given by,

$$D=|\frac{4(5)-3(-2)+4}{5}|$$D=∣54(5)−3(−2)+4​∣

$\Rightarrow D=6$⇒D=6 units.