Question

AB and BC form a right angle at point B. If A = (-3, -1) and B = (4, 4), what is the equation of BC?

A. x + 3y = 16
B. 2x + y = 12
C. -7x − 5y = -48
D. 7x − 5y = 48

Answer 1

-5y-7y=-48

Explanation:

distance between BC (4,4) C(x,y)

BC is perpendicular on AB (perpendicular line has opposite reciprocal slope)

slope of AB=y2-y1/x2-x1=4-(-1)/4-(-3)=5/7

slope of BC=-7/5

now input the value of B (4.4) and C(x,y)

y=mx+b

4=-7/5(4)+b

b=4+28/5

b=48/5

y=-7/5x+48/5

5y=-7x+48

5y+7x=48 multipy by -1

-5y-7y=-48

Answer 2

Answer: Choice C. -7x - 5y = -48

===============================================

First we need the equation of line AB. Compute the slope through (-3,-1) and (4,4)

m = (y2-y1)/(x2-x1)

m = (4-(-1))/(4-(-3))

m = (4+1)/(4+3)

m = 5/7

This is the slope through points A and B, ie the slope of line AB.

To get the slope of line BC, we flip the fraction and change the sign.

Doing so has us go from 5/7 to -7/5. Note how 5/7 and -7/5 multiply to -1.

The slope of line BC is -7/5. Let m = -7/5.

--------

Use (x,y) = (4,4) along with m =-7/5 to find the equation of line BC

y = mx+b

4 = (-7/5)(4) + b

4 = -28/5 + b

20 = -28 + 5b ... multiply every term by 5 to clear out the fraction

20+28 = 5b

48 = 5b

5b = 48

b = 48/5

With m = -7/5 and b = 48/5, we go from y = mx+b to y = (-7/5)x+48/5

The slope intercept form for line BC is y = (-7/5)x+48/5

--------

Let's get this into standard form Ax+By = C

y = (-7/5)x+48/5

5y = -7x + 48 .... multiply everything by 5

7x+5y = 48 .... is one way to represent the equation in standard form

-7x - 5y = -48 ... your teacher has decided (for some reason) to multiply both sides by -1