Question

Let P(11,7)Q(13.5,4)and R(9.5,4) be the mid point of the sides AB,BC and AC respectively of triangle ABC.Find the coordinates of the vertices A,B and C.Hence find the area of triangle ABC and compare this with area of triangle PQR​

Answer 1

Answer:

Area of ΔABC =24 Sq. Units.

Area of ΔABC=4×(Area of ΔPQR)

Step-by-step explanation:

Let us assume that the coordinates of vertices of ΔABC are A(x₁,y₁), B(x₂,y₂), and C(x₃,y₃).

Now given that, P(11,7), Q(13.5,4), and R(9.5,4) are the midpoints of sides AB, BC, and CA respectively.

So, we can write,

$\frac{x_{1}+x_{2}}{2} =11$, ⇒$x_{1}+x_{2}=22$ ........ (1)

$\frac{x_{2}+x_{3}}{2}=13.5$, ⇒$x_{2}+x_{3}=27$ ........ (2)

$\frac{x_{3}+x_{1}}{2}=9.5$, ⇒$x_{3}+x_{1}=19$ ........ (3)

By adding equations (1), (2), and (3) we get,

2(x₁+x₂+x₃)=68, ⇒ x₁+x₂+x₃ =34 ....... (4)

By {(4)-(2)}, we get, x₁=34-27 =7

By {(4)-(3)}, we get, x₂= 34-19 =15

and, By {(4)-(1)}, we get, x₃= 34-22 =12

Again, we can write,

$\frac{y_{1}+y_{2}}{2} =7$, ⇒$y_{1}+y_{2}=14$ ........ (5)

$\frac{y_{2}+y_{3}}{2}=4$, ⇒$y_{2}+y_{3}=8$ ........ (6)

$\frac{y_{3}+y_{1}}{2}=4$, ⇒$y_{3}+y_{1}=8$ ........ (7)

By adding equations (5), (6), and (7) we get,

2(y₁+y₂+y₃)=30, ⇒ y₁+y₂+y₃ =15 ....... (8)

By {(8)-(6)}, we get, y₁=15-8 =7

By {(8)-(7)}, we get, y₂= 15-8 =7

and, By {(8)-(5)}, we get, y₃= 15-14 =1

Hence, the coordinates of A, B, and C are (7,7), (15,7), and (12,1).

So, the area ofΔABC will be

$\frac{1}{2}[7(7-1)+15(1-7)+12(7-7)]$

=$\frac{1}{2}[42-90]$

=$\frac{1}{2}[-48]$

=24 Sq. Units. {Since area can not be negative. Hence we have taken the positive value} (Answer)

Now, area of ΔPQR will be

$\frac{1}{2}[11(4-4)+13.5(4-7)+9.5(7-4)]$

=$\frac{1}{2}[-12]$

=6 Sq. Units. {Since area can not be negative. Hence we have taken the positive value}

Therefore, the area of ΔABC=4×(Area of ΔPQR) (Answer)