You need to find the coordinates of the point P to write the equation of the line passing through this point.
The point P represents the points of intersection of the lines x-y+1=0 and 2x-3y+5=0, hence you need to find the solution of the system of these two equations.
Multiply the first equation by 2, then subtract then subtract the second equation from this new equation.
Eliminating the variable x yields:
y - 7 = 0 => y = 7
Plugging y = 7 in the first equation yields:
x - 7+ 1 = 0 => x = 6
The coordinates of the point P are (6;7).
The line passing through P is 7 = 6m + b.
The point (3,2) is at the distance d = `7/5` from the line 7 = 6m + b , hence you may replace these values in the following formula:
`7/5 = |m*3+b*2 - 7|/sqrt(m^2 + b^2)`
`5|3m + 2b - 7| = 7sqrt(m^2 + b^2)`
`25(9m^2 + 4b^2 + 49 - 21m - 14b + 6bm) = 49m^2 + 49b^2`
`225m^2 + 100b^2 - 49m^2- 49b^2 - 525m - 350b + 150bm + 49 = 0`
`` `176m^2 + 51b^2 - 525m - 350b + 150bm + 49 = 0`
The equation of the line passing through P and found at a distance of `7/5` from the point (3;2) is 6m + b - 7 = 0, m and b being in the following relation `176m^2 + 51b^2 - 525m - 350b + 150bm + 49 = 0.`
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