PA: x -y+1 = 0 ->Equation A
PB: 2x-3y+5 = 0 ->Equation B
The common point for both these lines is P. Therefore by solving these two equation we can find the coordinates of P.
2*A - B = 2x-2y+2 - (2x-3y+5) = 0
2x-2y+1-2x+3y-5 = 0
y=5 - 1= 4.
Substituting in Equation A,
x - 4+1 = 0
therefore, x = 4-1 = 3.
The coordinates of P = (3,4)
There are two lines that passes through P at a distance 7/5 from (3,2). This lines can be found by deriving the equations of to tangents drawn to a circle with centre at (3,2) and radius of 7/5.
First we have to find the equation of the circle.
Equation for a circle, with centre (a,b) and radius of r is;
(x-a)^2 + (x-b)^2 = a^2
for this circle,
(x-3)^2+(y-2)^2 = (7/5)^2 -> Equation C
we will denote the line as equation D and let its gradient be m,
(y-4)/(x-3) = m
y = mx+4 - 3m -> Equation D
the radius drawn at the point of interesection with circle C, is perpendicular to equation D. Therefore its gradient is (-1/m).
We will denote that line as equation E,
(y -2)/(x-3) = (-1/m)
y = (-x/m)+(3/m) +2 -> Equation E
By solving equations D & e, we can get an expression for the coordinates of the point.
mx+4 - 3m = (-x/m)+(3/m) +2
(m+1/m)x = 3(m+1/m) - 2
x = 3 - 2/(m+1/m)
x = 3 - 2m/(m^2+1)
By substituting in equation D, find y,
y = m[3 - 2m/(m^2+1)]+4 - 3m
y = -2m^2/(m^2+1) + 4
By subsituting in Equation C, we can solve for m.
(x-3)^2+(y-2)^2 = (7/5)^2
[3 - 2m/(m^2+1) - 3]^2 + [-2m^2/(m^2+1) + 4 - 2] = 49/25
[-2m/(m^2+1)]^2 + [-2m^2/(m^2+1) + 2]^2 = 49/25
By solving you get,
m = 1.0202 or -1.0202
therefore the two lines can be written as,
y = mx+4 - 3m = 1.0202x+4 - 3*1.0202
y = 1.0202x+ 0.9394
y = mx+4 -3m = -1.0202x+4 + 3*1.0202
y = -1.0202x+ 7.0606