12x²+7xy+ky²+13x-y+3=0
a=12, h=\(\frac { 7 }{ 2 } \), b=k, g=\(\frac { 13 }{ 2 } \), f =\(\frac { 1 }{ 2 } \), c=3
af²+bg²+ch²-abc-2fgh=0
12\(\left( -\frac { 1 }{ 2 } \right) ^{ 2 }+k\left( \frac { 13 }{ 2 } \right) ^{ 2 }+3\left( \frac { 7 }{ 2 } \right) ^{ 2 }\)-12(k)(3)-2\(\left( -\frac { 1 }{ 2 } \right) \left( \frac { 13 }{ 2 } \right) \left( \frac { 7 }{ 2 } \right) \)=0
\(\frac { 12 }{ 4 } +\frac { 169k }{ 4 } +\frac { 147 }{ 4 } -36k+\frac { 91 }{ 4 } \)=0
⇒ 12 + 169k+ 147 -144k+ 91 = 0
25k = -250 ⇒ k = -10
The equation is 12x²+7xy-10y²+13x-y+3=0
To find separate equations: 12x²+7xy-10y=(3x-2y)(4x+5y)
Let 12x²+7xy-10y²+13x-y+3=0 (3x-2y+l)(4x+5y+m)
Equating the coefficient of x ⇒ 4l+ 3m = 13.....(1)
Equating the coefficient of y ⇒ 5l-2m = -1 .....(2)
(1) x 2 ⇒ 8l+6m=26
(2) x 3 ⇒ 15l-6m=-3
23l=23
l=1
4+3=13
3m=9 ⇒ m=3
The separate equations are 3x - 2y + 1 = 0 and 4x + 5y + 3 = 0

12x²-10xy+2y²+14x-5y+c=0
ax² +2hxy+ by² +2gx+2fy+c = 0
\(\left| \begin{matrix} 2h=-10 \\ h=-5 \end{matrix} \right| b=2\left| \begin{matrix} 2g=14 \\ g=7 \end{matrix} \right| \left| \begin{matrix} 2f=-5 \\ f=\frac { -5 }{ 2 } \end{matrix} \right| \)
af² +bg² +ch² -2fgh-abc=0 is the condition
12\(\left( \frac { 25 }{ 4 } \right) \)+2(7)²+c(-5)²-2\(\left( \frac { -5 }{ 2 } \right) \)(7)(-5)-12(2)(c)=0
75+98+25c-175-24c=0, c=2
The equation is 12x²-10xy+2y²+14x-5y+2=0
12x²-10xy+2y=(3x-y)(4x-2y)
Let 12x²-10xy+2y²+14x-5y+2=(3x-y+l)(4x-2y+m)
So that 4l+3m=14, -2l-m=-5
On solving we get t=\(\frac { 1 }{ 2 } \), m=4
∴ The separate equations are 3x-y+\(\frac { 1 }{ 2 } \)=0 ⇒ 6x-2y+1=0 and
4x-2y+4=0 ⇒ 2x-y+2=0
m₁=\(\frac { -6 }{ -2 } \)=3; m₂=\(\frac { -2 }{ -1 } \)=2
∴ tanθ=\(\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| =\left| \frac { 3-2 }{ 1+3.2 } \right| =\frac { 1 }{ 7 } \)
θ=tan^-1\(\left( \frac { 1 }{ 7 } \right) \).

Let a and b be the intercepts on the given line.
\(\therefore \)  Coordinates of A and B are (a, 0) and (0, b) respectively
\(\therefore -5 = \frac{1\times 0+2\times a}{1+2}\Rightarrow 2a=-15\)
\(\Rightarrow a =\frac{-15}{2}\)
\(\therefore A=(\frac{-15}{2},0)\)
and \(4=\frac{l\times b+0\times 2}{1+2}\Rightarrow 4=\frac{b}{3}\Rightarrow b=12\)
\(\therefore \)  b = (0, 12)
So the equation of line AB is y - y₁ = \(\frac{y_2-y_1}{x_2-x_1}(x-x_1)\)
\(y-0=(\frac{12-0}{0+\frac{15}{2}})(x+\frac{15}{2})\Rightarrow y=\frac{12\times 2}{15}(x+\frac{15}{2})\)
\(\Rightarrow y=\frac{8}{5}(x+\frac{15}{2})\)
\(\Rightarrow 5y=8x+60\Rightarrow 8x-5y+60=0\)
Hence, the required equation is 8x - 5y + 60 = 0.

Given that:
2x+y=5.....(i)
x+3y+8=0...(ii)
3x+4y=7 ....(iii)
Equation of any line passing through the point of intersection of equation (i) and (ii) is
(2x+y-5)+λ(x+3y+8)=0 ...(iv)     (λ=constant)
⇒ 2x+y-5+λx+3λy+8λ=0
⇒ (2+λ)x+(1+3λ)y-5+8λ=0
Slope of line m₁ (say) = \(\frac { -(2+\lambda ) }{ 1+3\lambda } \)   \(\left[ \because m=\frac { -a }{ b } \right] \)
Now slope of line 3x + 4y = 7 is
m₂(say) = -\(\frac { 3 }{ 4 } \)
If equation (iii) is parallel to equation (iv) then m₁ = m₂
⇒ \(\frac { -(2+\lambda ) }{ 1+3\lambda } =-\frac { 3 }{ 4 } \)
⇒ \(\frac { 2+\lambda }{ 1+3\lambda } =\frac { 3 }{ 4 } \) ⇒ 8+4λ=3+9λ
⇒ 9λ-4λ=5  ⇒ 5λ=5  ⇒  λ=1
On putting the value of A.in equation (iv) we get
(2x+y-5)+1(x+3y+8)=0
⇒ 2x+y-5+x+3y+8=0  ⇒ 3x+4y+3=0
Hence, the required equation is 3x+4y+3=0

Let (x)be any point lying in the equation x + y = 4
x₁+y₁=4....(i)
Distance of the point (x₁, y₁) from the equation 4x + 3y = 10
\(\frac { 4{ x }_{ 1 }+3{ y }_{ 1 }-10 }{ \sqrt { ({ 4 })^{ 2 }+(3)^{ 2 } } } \)=1
\(\left| \frac { 4{ x }_{ 1 }+3{ y }_{ 1 }-10 }{ 5 } \right| \)=1
4x₁+3y₁-10=±5
Taking 4x₁+3y₁-10=5
⇒ 4x₁+3y₁=15    ........(ii)
From equation (i) we get y₁=4-x₁
Putting the value of y1 in equation (ii) we get
4x₁+3(4-x₁)=15
⇒ 4x₁+12-3x₁=15
⇒ x₁+12=15
x₁=3 and y₁=4-3=1
So, the required point is (3, 1)
Now taking(-) sign, we have
4x₁+3y₁-10=-5
⇒ 4x₁+3y₁=5.....(iii)
From equation (i) we get y1=4-x1
⇒ 4x₁+3(4-x₁)=5
⇒ 4x₁+12-3x₁=5
⇒ x₁=5-12=-7
and y₁=4-(-7)=11
So, the required point is (- 7, 11)
Hence, the required points on the given line are (3, 1) and (-7, 11).