Question

Find the equation of the straight lines passing through $(-2,-7)$ & havig and intercept of length $3$ between the straight lines $4x+3y=12,4x+3y=3$,

Answer 1

$4x+3y=124x+3y=3\frac{x}{3}+\frac{y}{4}=\frac{12}{12}\frac{4}{3}x+y=1\frac{x}{3}+\frac{y}{7}=1\frac{x}{3/4}+\frac{y}{1}=1$

Given are two parallel line perpendicular distance between two parallel lines

$d=\sqrt{\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}}=\frac{|12-3|}{\sqrt{4^2+3^2}}=\frac{9}{5}$

Let $y=mx+c$

$4x+3(mx+c)=12(4+3m)x+3c=12x=\frac{12-3c}{4+3m}+c$

$y=\frac{m(12-3c)}{4+3x}+c=\frac{12m-3cm+4c+3xc}{4+3m}=\frac{12m+4c}{4+3m}$

For $4x+3y=3$

$(4+3x)x+3c=3x=\frac{3-3c}{4+3x}y=\frac{3m+4c}{4+3m}$

Given

$\sqrt{(\frac{3m+4c}{4+3x}{)}^2-\frac{12m+4c}{4+3m}+(\frac{3-3c}{4+3m}-\frac{12-3c}{4+3m}{)}^2}=3(\frac{3m+4c-12m-4c}{4+3m}{)}^2+(\frac{3-3c-12+3c}{4+3m}{)}^2=9(\frac{-9}{4+3m}{)}^2+-(\frac{-9}{4+3m}{)}^2=9(-9m{)}^2+(-9{)}^2=9(4+3m)\Rightarrow 81m^2+81=9(4+3m)\Rightarrow 9m^2+9-4-3m=0\Rightarrow 9m^2-3m+5=0$

Given imaginary numbers

Byt if $\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}=\frac{|12-3|}{\sqrt{4^2+3^2}}=\frac{9}{5}$

Given $y=mx+c$

$m_1{m}_2=-1$

$m\left(\frac{-4}{3}\right)=-1m=\frac{4}{3}$

As $y=\frac{4}{3}x+c$ passes through $(-2,-7)$

$-7=\frac{4}{3}(-2)+cc=\frac{8}{3}-7=\frac{-13}{3}y=\frac{4}{3}x-\frac{13}{3}4x-3y=13$

Equation of straight line

$=4x-3y=13=4x-3y-13=0$