Question

A parking lot in an it company is triangular shaped with two of its vertices at B(-2,0) and C(1,12). The third vertex A is at the midpoint of line joining the points (1,1) and (3,11)
(a) find the coordinates of A

(b) find the equation of the line that pases through the points B(2,0) and C(1,12)

(c) find the equation of the line parallel to BC and passing through vertex A

(d) find the equation of a line perpendicular to BC and passing through the vertex A​

Answer 1

Solution :-

given that,

• The third vertex A is at the midpoint of line joining the points (1,1) and (3,11) .

so, A coordinates (x, y) will be :-

→ x = (1 + 3)/2 = 4/2 = 2 .

→ y = (1 + 11)/2 = 12/2 = 6 .

therefore, coordinates of A will be (2,6)..

now , The equation of the line that pases through the points B(2,0) and C(1,12) :-

(since you have written B coordinates as (2,0) , i m taking them same.)

→ slope of line = (y2 - y1)/(x2 - x1) = (12 - 0) / (1 - 2) = 12/(-1) = (-12)

then,

→ Equation of line = (x - x1) = m(y - y1)

taking (x1,y1) as (2,0) we get,

→ (x - 2) = (-12)(y - 0)

→ x - 2 = (-12) * y

→ x - 2 = (-12y)

→ x + 12y - 2 = 0 (Ans.)

now,

→ slope of BC = (y2 - y1)/(x2 - x1) = (12 - 0) / (1 + 2) = 12/3 = 4

so,

→ Slope of line parallel to BC = Slope of BC = 4 .

then, Equation of the line parallel to BC and passing through vertex A(2,6) :-

→ (x - 2) = 4(y - 6)

→ x - 2 = 4y - 24

→ x - 4y - 2 + 24 = 0

→ x - 4y + 22 = 0 (Ans.)

also,

→ Slope of line perpendicular to BC * Slope of BC = (-1)

→ Slope of line perpendicular to BC = (-1/4)

therefore, Equation of a line perpendicular to BC and passing through the vertex A(2,6) :-

→ (x - 2) = (-1/4)(y - 6)

→ 4(x - 2) = (-1)(y - 6)

→ 4x - 8 = - y + 6

→ 4x + y - 8 - 6 = 0

→ 4x + y - 14 = 0 (Ans.)

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Answer 2

Answer:

find the equation of the line that passer through the points B (-2,0)and C(1,12)