Step 1 :

Let the required equation be

$2x-3y+ \lambda (4x-5y-2)=0$----------(1)

It passes through the point (2,1) is

$2(2)-3(1)+ \lambda (4(2)-5(1)-2)=0$------------(1)

$ \Rightarrow 1+ \lambda = 0$

$ \Rightarrow \lambda = -1$

Hence the equation of the line is

$ 2x-3y+(-1) [ 4x-5y-2]=0$

$ \Rightarrow x-y-1=0$

(a) through the point (2,1) , matches (iii) x-y-1=0

Slope of the line (1)

$x(2+4 \lambda )-y (3+5\lambda )-2\lambda=0$

is $m = \large\frac{-(2+4\lambda )}{-(3+5\lambda)}$$ = \large\frac{2+4\lambda}{3+5 \lambda}$

Slope of the line $x+2y+1=0$ is

$ -\large\frac{1}{2}$

$ \therefore \large\frac{2+4\lambda}{3+5\lambda} $$ \times -\large\frac{1}{2} $$ = -1$

$ \Rightarrow (2+4\lambda)=2(3+5\lambda)$

$ 2+4\lambda=6+10 \lambda$

$ \Rightarrow \lambda = -\large\frac{2}{3}$

Hence equation of this required line is

$x \bigg(2+4 \bigg(-\large\frac{2}{3} \bigg)$$ - y \bigg( 3+5 \bigg( -\large\frac{2}{3} \bigg)$$ - 2 \bigg( -\large\frac{2}{3} \bigg) = $$0$

$ x \bigg( -\large\frac{2}{3} \bigg)-y \bigg( -\large\frac{1}{3} \bigg)+ \large\frac{4}{3}=0$

$ \Rightarrow -2x+y+4=0$

$ \therefore 2x-y=4$

(b) perpendicular to the line x+2y+1=0 matches (i) 2x-y=4

Equation of the given line is $3x-4y+5=0$

Slope of this line is $ - \bigg( \large\frac{3}{-4} \bigg)$

$ = \large\frac{3}{4}$

Since the above line is parallel to the line (1) their slopes are equal.

(i.e) $ \large\frac{2+4\lambda}{3+5\lambda}$$ = \large\frac{3}{4}$

$ \Rightarrow 4(2+ 4\lambda )=3(3+5\lambda)$

$8+16\lambda = 9+ 15\lambda$

$ \Rightarrow \lambda = 1$

Hence the equation of the required line is $x (2+4(1))-y(3+5(+1)-2(1)=0$

(i.e) 6x-8y-2=0$

or $3x-4y-1=0$

(c) parallel to the line 3x-4y+5=0 matches (iv) 3x-4y-1=0

When it is equally inclined between the axes then $m = \pm 1$

When m = -1

$ \therefore \large\frac{2+4\lambda}{3+5\lambda}=-1$

$ \Rightarrow 2+4\lambda = -1(3+5\lambda)$

$ \Rightarrow 2+4\lambda = -3-5\lambda$

$ \Rightarrow -9\lambda = 5$

$ \therefore \lambda = -\large\frac{5}{9}$

Hence the equation of the required line is $x \bigg(2+4 \bigg( -\large\frac{5}{9} \bigg)\bigg)$$ - y\bigg( 3+5 \bigg( -\large\frac{5}{9} \bigg) \bigg) $$ - 2 \bigg( -\large\frac{5}{9} \bigg)=$$0$

(i.e) $x\bigg( -\large\frac{2}{9} \bigg) - y \bigg( \large\frac{2}{9} \bigg) + \large\frac{10}{9}=$$0$

$ \Rightarrow 2x+2y-10=0$

or $x+y-5=0$

(d) equally inclined to the axes is matches (ii) x+y-5=0