how do u do this math problem?

Question

im in Advance Math 11 and i dont get this question:

Two lines intersect at (50 , 500). Their slopes differ by 6. Prove that their y-intercepts differ by 300.

Answer 1

Call the slopes m and n. Then

m-60=n

Write two equations with y-intercepts a and b. Then

y=mx+a
y=nx+b

Substitute,

y=mx+a
y=(m-60)x+b

Now plug in (50,500)

500 = 50m + a
500 = 50m-50(60)+b

Equate the equations since they both equal 100.

50m+a = 50m -50(60)+b
Cancel out 50m, bring b over to the left side and expand -50(60)

a-b = -300, which is a difference of 300.

Answer 2

you know that both lines crosses (50,500) and the difference between both lines's slope is 6

let's do it from the basic equatio of a line y = mx + c

let us say that the first line's equation will be

y1 = m1( x1) + c1

so, for the second line, it will be

y2 = m1( x2) + c2

since we know that the difference of m1 and m2 is 6, we can rewrite y2 as:

y2 = (6 + m1)(x2) + c2

at x = 50, y = 500. So, substotuting y1, y2, x1 and x2 gives

500 = 50 (m1) + c1
500 = 50 (6 + m1) + c2

Both equations are equal, so

50 (m1) + c1 = 50 (6 + m1) + c2

50 (m1) + c1 = 300 + 50m1 + c2

c1 = 300 + c2

This proves that c1 is actually c2 + 300, and since c corresponds to y-intercept, you have your answer from the last equation

Hope that helps

Answer 3

make the two equations

(1) y=mx + b
500=m(50) +b


(2) y=mx +b
500 = (m+6)(50) +b
200 = 50m +b

take into consideration both the equations.... the y-int (b) for both will differ by 300 (500-200)

Answer 4

y-500=m(x-50) (1)
y-500=(m+6)(x-50) (2)
Put x=0
y intercept (1) y1= 500-50m
y intercept(2) y2= 500-50m-300

substract y1-y2= 300.

Answer 5

if two lines intersect then at the point of intersection , they shoul both satisfy the line equation.(y=mx+c)

let at the point of intersection have coordinates:
for line1 =(x1, y1)
for line 2 =(x2, y2)

slope m1= (y1-500)/(x1-50)
sope m2=(y2-500)/(x2-50)
but we know that their slope differ by 6
therefore if slope 1= m1 then slope2 = m1+6

rearanging equation1:
y1-500=m1(x1-50)
y1=m1x1-50m1+500
by comparing with (y=mx+c) we can see that c=(500-50m1)

rearanging equation2:
y2-500=(m1+6)(x1-50)
y2-500=m1x1-50m1+6x1-300
y2=m1x1-50m1+6x-300+500
y2=m1x1-50m1+6x+200
y2=(m1x1+6x)+200-50
by comparing with(y=mx+c) we can see that c=(200-50m1)

therefore: c1-c2 ( intersepts)
=500-50m1
-200-50m1
______________
300+0

therefore c1-c2=300 (proven)

Hope this is helpful

Answer 6

y=mx+b
y=(m+6)x+b

500=m(50)+b
500=50m+b
500-50m=b


500=(m+6)50+b
500=50m+300+b
500-50m-300=b
therefore second problem has a y intercept with a difference of 300