Question

Line a and b are perpendicular lines.If line a is represented by y=s/t x+10 and line b is represented by y=q/r -4 where s,t,q and r are integers and t and r are not equal to zero which statement is always true

Answer 1

The options are not shown, so i will answer in a general way.

First, for a line of the form:

y = a*x + b

All the perpendicular lines will have a slope of -(1/a), then we can write a general perpendicular line as:

y = -(1/a)*x + c.

In this problem we have the perpendicular lines:

y=(s/t)*x+10

y = (q/r)*x - 4

Because those lines are perpendicular, we must have that:

(q/r) = -(1/(s/t)) = -(t/s)

q/r = -t/s

and we knew that s, t, q and r are integers (where t and r can not be equal to zero), and because of this relation, where we can see that in the denominator at the right we have the integer s, we can conclude that s can not be zero (we can not divide by zero).

And we know that t and s are both different than zero, then:

-t/s ≠ 0.

This means that:

q/r ≠ 0

And from this, we can conclude that q is different than zero.