Answer: perpendicular
Step-by-step explanation: Line 1 has the following points
(-2,5) and (1,4)
We know that the equation of a straight has the form
y=mx+c
m=gradient or the slope
c= where the graph crosses the y axis
let us label our points
let(x1, y1)=(-2, -5)
let (x2, y2)=(1, 4)
Does not matter the labelling of points
we now need to find the gradient or the slope
It is given by the following formula
m=(y2-y1)/(x2-x1)
substitute the values into formula
m=(4--5)/(1--5)=9/3=3
Reminder that two negatives facing each other gives a postive
m=3
We substitute this into our general equation of a straight line to get the following:
y=3x+c
We now to solve for c
and we can use any of the two co-ordinate points
lets use(1,4) this means x=1 and y=4
substitute into the above equations and solve for c
4=3*1+c
4=3+c
c=1
the equation for Line1 is y=3x+1
We now look at line2
(0,6) and (6,4)
let(x1, y1)=(0,6) and (x2, y2)=(6,4)
We following the same procedure as for the first line
we look for the gradient or slope m
m=(y2-y1)/x2-x1)
m=(4-6)/(6-0)=-2/6=-1/3
We then substitute this value into the general equation for a line
y=mx+c
y=-1/3x+c
we now need to solve for c
we can choose any of the co-ordinate points to substitute into the equation
say we choose(0,6) this means x=0 and y=6
y=-1/3x+c
6=-1/3 *0+c
6=0+c
c=6
we then get the equation for line 2 which is
y=-1/3x+6
Line 1 has a gradient of 3
Line2 has a gradient of -1/3
when we multiply the two together we get -1
therefore the two line are perpendicular to each other
m1*m2=-1
gradient of line multiplied by the gradient of line 2 gives -1
perpendicular line