Question

Line l bisects segment BC of triangle ABC, where A is at (3, -3), B is at (1, 4), and C is at (3, -2). If line l also travels through point A, what is its equation?
A. y=-5x-1
B. y=-2x+4
C. y=-4x+9
D. y=4x-1
E. y=5x+4

Answer 1

Line l bisects segment BC of triangle ABC, so the midpoint of BC is also on line l. The midpoint of BC is ((1+3)/2, (4-2)/2) = (2,1). Therefore, the equation of line l can be found using the point-slope form of a line with the point (3,-3) and slope -2/1 (since line l bisects segment BC, it is perpendicular to segment BC, which has slope 1/2):
y - (-3) = -2(x - 3)
y = -2x + 3
Since line l also travels through point A, we can substitute (3,-3) into the equation of line l:
-3 = -2(3) + 3
-3 = -3
The point (3,-3) satisfies the equation of line l, so the answer is: y = -2x + 3. Answer choice B is the correct answer.

Answer 2

Answer:

Option C y = -4x + 9

Step-by-step explanation:

Equation of a line:

            The line l bisects BC. The line l passes through the midpoint of BC.

B(1, 4) ; C(3 , -2)

  $\sf Midpoint \ of \ BC = \left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

                           $\sf = \left(\dfrac{1+3}{2},\dfrac{4-2}2{}\right)\\\\\\=\left(\dfrac{4}{2},\dfrac{2}{2}\right)\\\\\\=(2 , 1)$

Line l passes through (2,1) and A(3 , -3),

    $\sf \boxed{Slope =\dfrac{y_2-y_1}{x_2-x_1}}$

                $\sf =\dfrac{-3-1}{3-2}\\\\\\=\dfrac{-4}{1}\\\\=-4$

m = -4

Equation of line in slope intercept form: y =mx +c

Here, m is the slope and c is the y-intercept.

        y = -4x + c

    As the line l is passing through (2,1), substitute the point (2,1) in the above equation and find c.

        1 = -4*2 + c

        1 = -8 +  c

  1 + 8 = c

        c = 9

Equation of the line l:

       y = -4x + 9