I'm guessing there's a menu of choices to go with this one, and we just need to try (-2,9) with each choice. But it's a much more interesting problem without the menu.
First we verify (-2,9) is on the line y=-7x-5,
-7(-2)-5 = 14-5 = 9, good.
There's a whole family of possibilities for the second equation, basically all the other lines through (-2,9). Believe it or not, that's called a pencil of lines. We need two parameters a and b, whose values only matter in ratio, to represent all these. If we tried to do it with just the slope we'd miss one line, the vertical line x=-2.
We'll do the line through (-2,9) and (a,b). A little thought tells us it's
(a + 2)(y - 9) = (b - 9)(x + 2)
When we plug in (-2,9) we get 0=0, good. When we plug in (a,b) we get
(a + 2)(b - 9) = (b - 9)(a + 2)
which is also true.
(a + 2)(y - 9) = (b - 9)(x + 2)
(b - 9)x - (a+2)y = -9(a + 2) - 2(b-9)
(b - 9)x - (a+2)y = -9a - 2b
We can choose any a and b and we'll get the line through (-2,9) and (a,b), so it will have a solution (-2,9) where it meets y = -7x - 5.
We need to make sure (a,b) isn't already on y = -7x - 5 which is the same line twice.
For arbitrary a,b, b ≠ -7a - 5, the lines y = -7x - 5 and
(b - 9)x - (a+2)y = -9a - 2b
meet at exactly one point, namely (-2,9)
Let's generate a few and plot. We'll do a vertical and horizontal an a couple of others.
a=-2, b=0, (b - 9)x - (a+2)y = -9a - 2b
-9x = 18
x = -2
a=0, b=9, (b - 9)x - (a+2)y = -9a - 2b
-2y = -18
y = 9
a=1, b=1, (b - 9)x - (a+2)y = -9a - 2b
-8x - 3y = -11
a=1, b=3, (b - 9)x - (a+2)y = -9a - 2b
-6x - 3y = -15
2x + y = 5
plot y=-7x+5, x = -2, y = 9, -8x - 3y = -11, 2x - y = 5
