Given:
- Point A: (2, -2)
- Equations of two lines: 2x + 3y - 5 = 0 and 7x - 5y - 2 = 0

To Find:
- Equation of the line passing through point A and the point of intersection of the two given lines

Solution:

To find the equation of the line passing through point A and the point of intersection of the given lines, we need to follow these steps:

Step 1: Find the point of intersection of the given lines.
- Solve the system of equations: 2x + 3y - 5 = 0 and 7x - 5y - 2 = 0

Multiplying the first equation by 5 and the second equation by 3, we get:
10x + 15y - 25 = 0
21x - 15y - 6 = 0

Adding the two equations, we eliminate the y variable:
10x + 21x - 25 - 6 = 0
31x - 31 = 0
31x = 31
x = 1

Substituting the value of x into the first equation, we can find the value of y:
2(1) + 3y - 5 = 0
2 + 3y - 5 = 0
3y - 3 = 0
3y = 3
y = 1

Therefore, the point of intersection is (1, 1).

Step 2: Find the slope of the line passing through point A and the point of intersection.
- Use the slope formula: slope = (y2 - y1) / (x2 - x1)

Let's label the point of intersection as B (1, 1).
Using the coordinates of A and B, we can calculate the slope:
slope = (-2 - 1) / (2 - 1)
slope = -3 / 1
slope = -3

Step 3: Use the point-slope form to find the equation of the line.
- The point-slope form is given by: y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.

Using point A (2, -2) and the slope -3, we have:
y - (-2) = -3(x - 2)
y + 2 = -3x + 6

Rearranging the equation, we get:
3x + y = 4

Therefore, the equation of the line passing through point A and the point of intersection of the two given lines is 3x + y = 4.

2x+3y=5 1*5 10x+15y=25
7x-5y=2 2*3 21x-15y=6

31x=31
x=1
x=1 sub in equation 1,
2(1)+3y=5
3y= 3
y=1

(1,1),(2,-2)
y-y1
=y2-y1
x-x1
x2-x1
y-1=-2-1

x-1 2-1
y-1(1)=x-1(-3)
y-1=-3x+3
3x-3+y-1=0
3x+y-4=0