Question

In a triangle abc coordinates of a are (1,2) and the equation of the medians through b and c are respectivelyx+y=5 and x=4 then are of triangle abc is

Answer 1

Answer:

9 sq. units.

Step-by-step explanation:

Let us draw a triangle Δabc, where, a(1,2), b(h,k), c(p,q).

We have to find coordinates of b and c to calculate the area of Δabc.

Now assume that d,e, and f are the midpoints of ab, ac and bc.

Given, equation of cd is x=4....... (1)  

and equation of eb is x+y=5 ....... (2)

Again assume that all the three medians be, cd, and af meet at point g.

Now, coordinates of d=($\frac{h+1}{2},\frac{k+2}{2}$) and it satisfy equation (1).

So, (1+h)/2=4 ⇒h=7 and (k+2)/2=0 ⇒k=-2.

So, coordinates of b=(7,-2) ...... (3)

Now, solving (1) and (2) we can get the coordinates of g.

So, 4+y=5 ⇒y=1 and x=4.  

Hence, coordinates of g are (4,1).

Therefore the equation of ag or af is given by

$\frac{y-2}{2-1}=\frac{x-1}{1-4}$

⇒3(y-2)=1-x

⇒x+3y=7 ....... (4)

Now, f is mid point of b and c.

So, coordinates of f=[(7+p)/2 , (-2+q)/2]

This will satisfy equation (4).

Therefore, $\frac{7+p}{2}+3(\frac{-2+q}{2} )=7$

After simplification, we get the equation,

⇒p+3q=13 ........ (5)

Again, e is mid point of a and c.

So, coordinates of e are given by [(1+p)/2, (2+q)/2] and this will satisfy equation (2).

Hence, $\frac{1+p}{2}+\frac{2+q}{2}=5$

⇒p+q=7 ...... (6)

Now solving (5) and (6) we will get

13-3q=7-q, ⇒q=3 and from (6) p=7-3=4.

So, the coordinates of c will be (4,3).

Therefore, the are of Δabc whose vertices are a(1,2), b(7,-2), and c(4,3) will be =$\frac{1}{2}[1(-2-3)+7(3-2)+4(2+2)]$

=1/2[-5+7+16]

=9 sq. units (Answer)