Maths 20分

The y-intercept of a stright line L is 1, and L is parallel to another line which passas through the points(1,4) and (-5,-2). Find the equation of L ?

slope of the other line,
=(-2-4)/(-5-1)
=(-6)/(-6)
=1
since L is parallel to the above line, so L has the same slope it
then slope of L is also 1
By slope intercept form, equation of L,
y=mx +c
y=(1)x +(1)
y=x +1
x-y+1 = 0

The y-intercept of a straight line L is 1 => L passes through (0,1)

L is parallel to another line which passas through the points(1,4) and (-5,-2)
=> slope of L = slope of another line
= [4 - (-2)] / [1 - (-5)] = 1

Let L be y = mx + c

Then, m = 1 and c = 1 (since L passes through (0,1), 1 = m(0) + c => c = 1).
Therefore, the equation of L is y = x + 1 or x - y + 1 = 0

slope of the other line[L(1)]
=(4-(-5)) / (1-(-2))
=1

beacuse L(1) // L
so slope of L(1) = slope of L = 1

the y-intercept L is 1

By pt.-slope form:
y=(1)x+1
x-y+1=0

slope of the other line:(4-(-2))/(1-(-5))=1

therefore, the slope of L=1
Let L:y=x+Cwhere C is a real constant
sub(0,1)into L
C=1

therefore the equation of L:y=x+1, SO, x-y+1=0