\(y=\frac{5}{3}x,m=\frac{5}{3},c=\frac{11}{3}\)

Let m₁ be the slope of the line joining A(a, 2a) and B(-2,3).
Then, \({ m }_{ 1 }=\frac { 2a-3 }{ a+2 } \)
Let m₂ be the slope of the line 4x + 3y + 5 = 0.
Then, \({ m }_{ 2 }=-\frac { 4 }{ 3 } \)
Since, given lines are perpendicular.
Therefore, m₁m₂ = -1
\(\Rightarrow\) \(\frac { 2a-3 }{ a+2 } \times -\frac { 4 }{ 3 } =-1\)
\(\Rightarrow\) 8a - 12 = 3a + 6 \(\Rightarrow\) \(a=\frac { 18 }{ 5 } \)

Let(h,k) be the coordinates of the foot of the perpendicular from the point(2,3) on the line x+y-11=0
Then, the slope of the perpendicular line is\(\cfrac { k-3 }{ h-2 } \)
Again the slope of the given line x + y - 11= 0 is -1
using the condition of perpendicularity of lines,we have
(\(\cfrac { k-3 }{ h-2 } \))(-1)=-1 0r k-h=1...(i)
since,(h,k) lies on the given line, we have
h + k - 11 = 0 or h + k = 11.....(ii)
On solving eqs. (i) and (ii) we get h=5 and k=6
hence,(5,6) are the required coordinates of the foot of the perpendicular.

\(m_1=\frac{7-y}{2-3}, m_2=\frac{6-4}{0-(-1)}\)
Also, \( m_1=m_2 \Rightarrow \frac{7-y}{-1}=2 \Rightarrow y=9\)

Here A = 3, B = –4, C₁ = 7 and C₂ = 5. Therefore, the required distance is
\(d=\frac{|7-5|}{\sqrt{3^{2}+(-4)^{2}}}=\frac{2}{5}\)

Given line is 3x – 4y –26 = 0 ... (1)
Comparing (1) with general equation of line Ax + By + C = 0, we get
A = 3, B = – 4 and C = – 26.
Given point is (x₁, y₁) = (3, –5). The distance of the given point from given line is
\(d=\frac{\left|\mathrm{A} x_{1}+\mathrm{B} y_{1}+\mathrm{C}\right|}{\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}}=\frac{|3.3+(-4)(-5)-26|}{\sqrt{3^{2}+(-4)^{2}}}=\frac{3}{5} \text { . }\)

Here x₁ = 1, y₁ = – 1, x₂ = 3 and y₂ = 5. Using two-point form (2) above for the equation of the line, we have
\(y-(-1)=\frac{5-(-1)}{3-1}(x-1)\)
or –3x + y + 4 = 0, which is the required equation

Let equation line is \(\frac { x }{ a } +\frac { y }{ b } =1\)
The sum of intercepts is 14, therefore = a + b = 14
\(b=14-a\Longrightarrow \frac { x }{ a } +\frac { y }{ 14-a } =1\)
The line passes through (3,4)
\(\frac { 3 }{ a } +\frac { 4 }{ 14-a } =1\Longrightarrow a=6or7\)
Case I When a = 6, then Eq.(i) becomes \(\frac { x }{ 6 } +\frac { y }{ 14-6 } =1\)
Case II When a=7, then Eq.(i) becomes \(\frac { x }{ 7 } +\frac { y }{ 14-7 } =1\)
4x + 3y - 24 = 0 and x+ y - 7 =0

The given equations are x-7y+5=0 and 3x+y-7 = 0
Equation of any line passing through the point of intersection of the given lines is
(x-7y+5)+k(3x+y-7)=0....(i)
\(\Rightarrow\) x-7y+5+3kx+ky-7k=0  \(\Rightarrow\) (1+3k)x+(k-7)y+5-7k=0
Slope of this equation = \(\frac { -(1+3k) }{ k-7 } ={ m }_{ 1 }\) (Say)
and the slope of the given equation 2x-5y+1=0=\(\frac { -2 }{ -5 } \) i.e.,  \(\frac { 2 }{ 5 } ={ m }_{ 2 }\) (Say)
as the two lines are perpendicular to each other
\(\therefore\)  m₁m₂ = -1
\(\therefore \frac { -(1+3k) }{ k-7 } \times \frac { 2 }{ 5 } =-1\Rightarrow \frac { 1+3k }{ k-7 } =\frac { 5 }{ 2 } \)
\(\Rightarrow\)  5k-35=2+6k  \(\Rightarrow\)  k = -37
Now substituting the value of k in equation (i)
(x-7y5)-37(3x+y-7)=0  \(\Rightarrow\) x-7y+5-11x-37y+259=0
\(\Rightarrow\) - 110x-44y+264 = 0
\(\Rightarrow\) 5x+2y-12 = 0 Required Equation.

13x - y - 9 = 0