f.4math

Two lines are perpendicular
Slope = (-k)(-1/-2) = -1
k=2

Substitute(h,k) into x-2y+3=0,
h - 2k + 3=0
h -2(2) +3 = 0
h = 1

As they are perpendicular,
slope of one line * slope of another line =-1
x-2y+3=0
2y=x+3
y=1/2x+3/2
Slope of x-2y+3 is 1/2
slope of kx+y-4=-1/(1/2)=-2
Slope of a line (Ax+By+C)=0 is -A/B
Therefore, for kx+y-4=0
-k=-2
k=2
The line is 2x+y-4=0
The point of intersection of the lines is
2x+y-4=0 --1)
x-2y+3=0 --2)
2)*2
2x-4y+6=0 --3)
By method of elimination.
1)-3)
5y-10=0
y=2
Put y=2 into 1),
2x+2-4=0
2x=2
x=1
Therefore
h=1
k=2

1find k,the 2 lines are perpendicular to each other,so the product of slopes of
the two lines is -1
(-1/k)*(2/1)= -1
2after you have found the value of k
put the k value into the equation of line
3the last step is simple as you can find the value of x and y by solving
the simutaneous equation