MATH SOLVE

3 months ago

Q:
# The equation of line ab is y = 5x + 1. write an equation of a line parallel to line ab in slope-intercept form that contains point (4, 5). y = 5x − 15 y = 5x + 15 y = x + y = x −

Accepted Solution

A:

METHOD 1:

The equation of line parallel to ax+by=c is,

ax+by=k

Similarly,in this case equation of line parallel to the st line y=5x+1 (i.e, 5x-y=-1) is given by,

5x-y=k......(i)

As equation (i) is the line containing (4,5),

k=5×4-5=15

Hence equation (i) becomes,

5x-y=15

y=5x-15➡required equation (ANS)

METHOD 2:

As slope of AB is 5, the slope of line parallel to AB should also be 5.

Thus, the equation of st line parallel to AB (with slope 5) & passing through (4,5) is:

m=(y-y1)/(x-x1)

5=(y-5)/(x-4)

5x-20=y-5

y=5x-15➡ required eqn

The equation of line parallel to ax+by=c is,

ax+by=k

Similarly,in this case equation of line parallel to the st line y=5x+1 (i.e, 5x-y=-1) is given by,

5x-y=k......(i)

As equation (i) is the line containing (4,5),

k=5×4-5=15

Hence equation (i) becomes,

5x-y=15

y=5x-15➡required equation (ANS)

METHOD 2:

As slope of AB is 5, the slope of line parallel to AB should also be 5.

Thus, the equation of st line parallel to AB (with slope 5) & passing through (4,5) is:

m=(y-y1)/(x-x1)

5=(y-5)/(x-4)

5x-20=y-5

y=5x-15➡ required eqn