Question

4. Line J has a y-intercept of 15 and passes through the ordered pair(6, -8). Line K is perpendicular to line J and passes through the orderedpair (-7,-). Which equation represents line K?AC.-y=Zx15-16fxx +B.y||||D.y563

Answer 1

Answer:

B. -6/7 x - 16/3

Explanation:

The equation of a line in slope-intercept form is given by

$y=mx+b$

where m = slope and b = y-intercept.

First, let us find the equation for line J.

We are told that the y-intercept of line J is -15, meaning J passes through (0, -15). Furthermore, we are told that J also passes through (6, -8). Therefore, the slope of the line is given by

$m=\frac{-8-(-15)}{6-0}=\frac{7}{6}$

Therefore, the equation for J is

$y=\frac{7}{6}x+b$

Now we know that the y-intercept for J is -15. Therefore, b = -15 and so the above equation gives

$y=\frac{7}{6}x-15$

Now, what is a line K that is perpendicular to the above line?

Here we remind ourselves that if we have an equation of the form y = mx + b, then the equation for a perpendicular line is given by

$y=-\frac{1}{m}x+c$

where c is a y-intercept that can be different from b.

Now in our case m = 7/6; therefore, the equation for the perpendicular line

$y=-\frac{1}{7/6}x+c$$y=-\frac{6}{7}x+c$

Now, what is the value of c, the y-intercept?

We are told that the line K passes through the point (-7, 2/3); therefore, putting x = -7 and y = 2/3 into the above equation gives

$\frac{2}{3}=-\frac{6}{7}(-7)+c$

which simplifies to give

$\frac{2}{3}=6+c$

subtracting 6 from both sides gives

$\frac{2}{3}-6=c$$c=-\frac{16}{3}$

with the value of c in hand, we can now write the equation for the line K.

$\boxed{y=-\frac{6}{7}x-\frac{16}{3}.}$

which is our answer!

Looking at the answer choices we see that choice D gives the equation we found above.

Therefore, choice D is the correct answer.