### Video Transcript

Which of the following lines is
perpendicular to the line 19π₯ minus three π¦ equals five?

(a) three π₯ minus 19π¦ equals five,
(b) two minus 19π¦ equals three π₯, (c) three π¦ equals one minus 19π₯, (d) three π¦
equals 19π₯ plus four, (e) three plus 19π¦ equals two π₯.

Before we choose one of these
functions, letβs remember what perpendicular lines are, more specifically what
perpendicular lines have. Perpendicular lines have negative
reciprocal slopes. So first weβll take 19π₯ minus
three π¦ equals five and find its slope. To do that, weβll take the function
given in standard form and convert it into slope-intercept form by isolating π¦.

First subtract 19π₯ from both sides
of the equation. 19π₯ minus 19π₯ cancels out,
leaving us with negative three π¦ equals negative 19π₯ plus five. Remember the goal: isolate π¦. We divide π¦ by negative three and
that means weβll have to divide both terms on the right side by negative three. On the left, negative three divided
by negative three equals one, and one times π¦ equals π¦, equals β our π₯ term has a
negative in the numerator and the denominator. We can simplify that by saying 19
over three π₯ minus five over three. This is the slope-intercept form of
the same equation we started with.

Slope-intercept form is π¦ equals
ππ₯ plus π. The π value is the slope. The slope of the line we were given
is 19 over three. We want the negative reciprocal of
19 over three. The reciprocal of 19 over three is
three over 19, and we need the negative value. Any function with the slope of
negative three over 19 will be perpendicular to our line. Weβll examine all five of these
functions to see which of them has a slope of negative three over 19.

Instead of trying to find the slope
of all five of these lines, letβs see if we can eliminate any of the options. To do that, I want you to notice
the relationship between π₯ and π¦ in the original function. We need the opposite to be
true. We need a constant value of three
associated with the π₯ and 19 associated with the π¦. Itβs also important to note that
the π¦-intercept doesnβt matter. In our new equation, the constant
value, the π¦- intercept, can be anything.

So weβll walk through these five
functions and see which of them do not have a relationship of three π₯ to 19π¦. (a) follows this pattern. (b) follows this pattern. (c) does not, neither does (d) or
(e). Now weβre down to two functions we
have to check. Weβll find the slope of both (a)
and (b). In function (a), we subtract three
π₯ from both sides. Three π₯ minus three π₯ cancels
out. Negative 19π¦ equals negative three
π₯ plus five. To find the slope-intercept form,
we need to isolate π¦. Weβll divide every term by negative
19. π¦ equals three over 19π₯ minus
five over 19. The slope of function (a) is three
over 19.

Three over 19 is the reciprocal of
our first function, but it is not the negative reciprocal of our first function. So now we move to function (b). Our first step: subtract two from
both sides of the equation. Two minus two equals zero. And now we have negative 19π¦
equals three π₯ minus two. We divide every term by negative
19, which simplifies to π¦ equals negative three over 19π₯ plus two over 19. The slope here is negative three
over 19, which is the negative reciprocal slope weβre looking for. The function two minus 19π¦ equals
three π₯ is perpendicular to the function 19π₯ minus three π¦ equals five.