Question

If the points $(10,5)$ $(8,4)$ and $(6,6)$ are the midpoints of the sides of a triangle, find its vertices.

• A. $(8,7)$
• B. $(12,3)$
• C. $(4,5)$
• D. All of the above

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $(8,7)$
B $(4,5)$
C $(12,3)$
D All of the above

Given-

In $\Delta ABC$, the mid point of BC is $D(8,4),$

The mid point of AC is $E(10,5),$ and the mid point of AB is $F(6,6).$

To find out -

The coordinates of $A,B$ & $C.$

Solution-

Let the vertices of $\Delta ABC$ be $A(x_1,y_1),B(x_2,y_2)$ & $C(x_3,y_3).$

Then, by applying mid point formula,

$\frac{x_2+x_3}{2}=8$

$⟹x_2+x_3=16$ and

$\frac{y_2+y_3}{2}=4$

$⟹y_2+y_3=8$ ........(i)

$\frac{x_1+x_3}{2}=10$

$⟹x_1+x_3=20$ and

$\frac{y_1+y_3}{2}=5$

$⟹y_1+y_3=10$ ........(ii),

$\frac{x_1+x_2}{2}=6$

$⟹x_1+x_2=12$ and

$\frac{y_1+y_2}{2}=6$

$⟹y_1+y_2=12$ ........(iii).

Adding (i), (ii) & (iii),

${2(x}_1+x_2+x_3)=48$ and

${2(y}_1+y_2+y_3)=30$

$⟹x_1+x_2+x_3=24$ and

$y_1+y_2+y_3=15$ ...........(iv).

Subtracting (i) from (iv)

$x_2=24-20=4$ and $y_2=15-10=5,$

Subtracting (ii) from (iv)

$x_1=24-16=8$ and $y_1=15-8=7,$

Subtracting (iii) from (iv)

$x_3=24-12=12$ and $y_2=15-12=3.$

$\therefore$ The coordinates of the vertices are $A(8,7),B(4,5)$ & $C(12,3)$