Question

Find the area of the triangle ABC with A (1,-4) and mid point of the sides through A being (2,-1) n (0,-1)

Answer 1

Let(x2, y2) and (x3, y3) be the co-ordinates of B and C respectively. Since, the co-ordinates of A are (1, -4), therefore

1+x22=21+x22=2
⇒⇒ x2 = 3

−4+y22=−1−4+y22=−1
⇒⇒ y2 = 2

1+x32=01+x32=0
⇒⇒ x3 = -1

−4+y32=−1−4+y32=−1
⇒⇒ y3 = 2

Let A (x1, y1) = A (1, -4), B (x2, y2) = B (3, 2) and C (x3, y3) = C (-1, 2). Now

Area (ΔABCΔABC = 12{x1(y2−y3)+x2(y3−y1)+x3(y1−y2)}12{x1(y2−y3)+x2(y3−y1)+x3(y1−y2)}
= 12∗{1(2–2)+3(2+4)−1(−4–2)}12∗{1(2–2)+3(2+4)−1(−4–2)}
= 12∗{1(0)+3(6)−1(−6)}12∗{1(0)+3(6)−1(−6)}
= 12{0+18+6}12{0+18+6}
= 12{24}12{24}
= 12 sq.unit

Thus, Area of triangle ABC = 12 sq.units

Answer 2

Answer:

Let(x2, y2) and (x3, y3) be the co-ordinates of B and C respectively. Since, the co-ordinates of A are (1, -4), therefore

1+x22=21+x22=2

⇒⇒ x2 = 3

−4+y22=−1−4+y22=−1

⇒⇒ y2 = 2

1+x32=01+x32=0

⇒⇒ x3 = -1

−4+y32=−1−4+y32=−1

⇒⇒ y3 = 2

Let A (x1, y1) = A (1, -4), B (x2, y2) = B (3, 2) and C (x3, y3) = C (-1, 2). Now

Area (ΔABCΔABC = 12{x1(y2−y3)+x2(y3−y1)+x3(y1−y2)}12{x1(y2−y3)+x2(y3−y1)+x3(y1−y2)}

= 12∗{1(2–2)+3(2+4)−1(−4–2)}12∗{1(2–2)+3(2+4)−1(−4–2)}

= 12∗{1(0)+3(6)−1(−6)}12∗{1(0)+3(6)−1(−6)}

= 12{0+18+6}12{0+18+6}

= 12{24}12{24}

= 12 sq.unit

Thus, Area of triangle ABC = 12 sq.units

Step-by-step explanation: