Question 985338
Hi there,
I take it that line 1 is AB.
gradient of AB = y2-y1/x2-x1
= 9-5/10-2 = 4/8 = 1/2
Using the line equation for AB
Using m = 1/2 and point A (2,5)
y - 5 = 1/2(x - 2)
y = 1/2x - 1 + 5
y = 1/2x + 4
This is the equation for line AB
Line C has a gradient of -2
Lines that are perpendicular have 
gradients that multiply together to give -1
m1 x m2 = -1
1/2 x m2 = -1
m2 = -2
...........
Using the line equation:
y - b = m(x - a)
Using m = -2 and pt C (6,2)
y - 2 = -2(x - 6)
y = -2x + 12 + 2
y = -2x + 14
This the equation of line C
Setting up simultaneous equations:
y - 1/2x = 4 ....(1)
y + 2x = 14 .....(2)
Multiply (1) by 4
4y - 2x = 16 .....(1)
y + 2x = 14  .....(2)
Add (1) + (2)
5y = 30
y = 6
Substitute y = 6 into:
y + 2x = 14
6 + 2x = 14
2x = 14 - 6
2x = 8
x = 4
{4,6} the point of intersection
of line AB and C
Hope this helps:-)