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




x + 3y = 16




2x + y = 12




-7x − 5y = -48




7x − 5y = 48

Respuesta :

Gradient of AB = [tex] \frac{y_2-y_1}{x_2-x_1} = \frac{4-(-1)}{4-(-3)} = \frac{4+1}{4+3} =5/7[/tex]
Since AB and BC forms a right angle, this means that AB and BC are perpendicular.
For perpendicular lines, m2 = -1/m1; where m1 and m2 are the gradient of the two lines.
i.e. m2 = -1/(5/7) = -7/5
Therefore, the equation of BC is given by
[tex] \frac{y-y_1}{x-x_1} =m_2 \\ \frac{y-4}{x-4} = -\frac{7}{5} \\ 7(x-4)=-5(y-4) \\ 7x-28=-5y+20 \\ -7x-5y=-48 [/tex]

Long story short the answer your looking for is C. -7 - 5y= -48