Let the mid points of the sides of a triangle ABC be ((5)/(2),7), ((5)/(2),3) and (4,5).

Question

Let the mid points of the sides of a triangle ABC be $(\frac{5}{2},7)$, $(\frac{5}{2},3)$ and $(4,5)$. If its incentre is $(h,k)$, then $3h+k$ is equal to :

Accepted Answer

Let the midpoints of the sides of triangle $ABC$ be

$D(\frac{5}{2},7),\, E(\frac{5}{2},3),\, F(4,5)$

We have to find the incentre $(h,k)$ of triangle $ABC$.

Use the fact about the medial triangle: The triangle formed by joining the midpoints of the sides of a triangle is called the medial triangle.

If the midpoints are known, then the original triangle can be reconstructed using:

$A=E+F-D,\, B=F+D-E,\, C=D+E-F$

Now calculate each vertex.

Vertex $A$
$A=(\frac{5}{2},3)+(4,5)-(\frac{5}{2},7)$

$A=(4,1)$

Vertex $B$
$B=(4,5)+(\frac{5}{2},7)-(\frac{5}{2},3)$

$B=(4,9)$

Vertex $C$
$C=(\frac{5}{2},7)+(\frac{5}{2},3)-(4,5)$

$C=(1,5)$

So the triangle has vertices

$A(4,1),\, B(4,9),\, C(1,5)$

Find the side lengths: Using distance formula:

Side opposite $A$, i.e. $BC$
$BC=\sqrt{(4-1{)}^{2}+(9-5{)}^{2}}=\sqrt{{3}^{2}+{4}^{2}}=5$

Side opposite $B$, i.e. $CA$
$CA=\sqrt{(4-1{)}^{2}+(1-5{)}^{2}}=\sqrt{{3}^{2}+(-4{)}^{2}}=5$

Side opposite $C$, i.e. $AB$
$AB=\sqrt{(4-4{)}^{2}+(9-1{)}^{2}}=8$

Thus,

$a=5,\, b=5,\, c=8$

Use incentre formula: The incentre of triangle with vertices $A({x}_{1},{y}_{1})$, $B({x}_{2},{y}_{2})$, $C({x}_{3},{y}_{3})$ is

$(\frac{a{x}_{1}+b{x}_{2}+c{x}_{3}}{a+b+c},\frac{a{y}_{1}+b{y}_{2}+c{y}_{3}}{a+b+c})$

where

$a=BC$

$b=CA$

$c=AB$

Substitute the values:

$h=\frac{5\cdot 4+5\cdot 4+8\cdot 1}{5+5+8}=\frac{20+20+8}{18}=\frac{48}{18}=\frac{8}{3}$

$k=\frac{5\cdot 1+5\cdot 9+8\cdot 5}{18}=\frac{5+45+40}{18}=\frac{90}{18}=5$

So the incentre is

$(\frac{8}{3},5)$

Find $3h+k$: $3h+k=3\cdot \frac{8}{3}+5=8+5=13$

Therefore, the answer is

$13$

So, the correct option is Option C.

Solutions · 1