Question

The area of a triangle whose coordinates of the mid-points of the sides are p(4, 5), q(1,2) and r(1,4) is equal to

Answer 1

Answer: The required area is $\dfrac{3}{4} sq. units$

Step-by-step explanation:

P (4,5) , Q(1,2) ,R(1,4)

Let A is the midpoint of PQ . B is the midpoint of QR and C is the midpoint of PR.

By Mid point formula

$(x,y) = (\dfrac{x_1+x_2}{2} ,\dfrac{y_1+y_2}{2} )$

Coordinates of A is

$(x,y) = (\dfrac{4+1}{2} ,\dfrac{5+2}{2} )= (\dfrac{5}{2} ,\dfrac{7}{2} )=(x_1,y_1)$

Coordinates of B is

$(x,y) = (\dfrac{1+1}{2} ,\dfrac{4+2}{2} )= (1,3)=(x_2,y_2)$

Coordinates of C

$(x,y) = (\dfrac{4+1}{2} ,\dfrac{5+4}{2} )= (\dfrac{5}{2} ,\dfrac{9}{2} )=(x_3,y_3)$

Now area of triangle

$area = \dfrac{1}{2}(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))\\\\=\dfrac{1}{2} (\dfrac{5}{2} (3-\dfrac{9}{2})+1( \dfrac{9}{2}-\dfrac{7}{2} )+\dfrac{5}{2} (\dfrac{7}{2} -3)) \\\\= -\dfrac{3}{4} \\\\ \text { Area cannot be negative } \\\\Area = \dfrac{3}{4} sq. units$

Hence, the required area is $\dfrac{3}{4} sq. units$

Answer 2

Answer:

$\red {Area \:of \: \triangle ABC}\green{= 12 \:square \: units}$

Step-by-step explanation:

$In \: \triangle ABC, \: P, Q \:and\: R \: are \: mid-points \: of \: BC , AC \:and \:AB \: respectively$

$Let \: P(4,5) = \left( x_{1}, y_{1}\right) ,\\</p><p>Q(1,2) = \left( x_{2}, y_{2}\right) ,\\R(1,4) = \left( x_{3}, y_{3}\right)$

$Area \:of \: \triangle PQR \\= \frac{1}{2}|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{2}-y_{1})|$

$= \frac{1}{2}| 4(2-4)+1(4-5)+1(5-2)|\\= \frac{1}{2}|4(-2)+1(-1)+1(3)|\\= \frac{1}{2}| -8-1+3|\\= \frac{6}{2}\\=3\:square \: units$

$Now, \:Area \:of \: \triangle ABC \\= 4 \times Area \:of \: \triangle PQR\\= 4\times 3 \\= 12 \:square \: units$

Therefore.,

$\red {Area \:of \: \triangle ABC}\green{= 12 \:square \: units }$

•••♪