Question

In a triangle $ABC,$ one of the vertex $A$ is $(3,4)$ and vertices $B(x_1,y_1),C(x_2,y_2)$ lie on the line $3x+4y=5$. The area of the triangle $ABC$ is $20$ sq. units. If $x_1+x_2=10$ and $\frac{x_1+y_1}{x_2+y_2}=\frac{3}{k},$ then the smaller value of $\left[k\right]$, (where [.] is greatest integer function) is

Answer 1

In triangle $ABC,AD\perp BC$
$\text{Height}AD=\left|\frac{3\cdot 3+4\cdot 4-5}{\sqrt{3^2+4^2}}\right|=4$
Area of triangle $\Delta ABC=\frac{1}{2}\times AD\times BC=20$
$\therefore BC=10=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
$\Rightarrow (x_2-x_1{)}^2+(y_2-y_1{)}^2=100....\left(i\right)$

We know that $B,C$ lies on the line $3x+4y=5$
$\Rightarrow x_1=\frac{5-4y_1}{3}\text{and}{x}_2=\frac{5-4y_2}{3}{x}_2-x_1=\frac{4(y_1-y_2)}{3}....\left(ii\right)$
From equation $\left(i\right)\text{and}\left(ii\right),$ we get
$y_1-y_2=\pm 6\Rightarrow x_2-x_1=\pm 8$
Also, $x_1+x_2=10$
So from above condition
$x_1=1,9x_2=9,1$ and $y_1=\frac{1}{2},\frac{-11}{2}{y}_2=\frac{-11}{2},\frac{1}{2}$

$\frac{x_1+y_1}{x_2+y_2}=\frac{1+\frac{1}{2}}{9-\frac{11}{2}}=\frac{3}{k}\Rightarrow \left[k\right]=7$
or
$\frac{x_1+y_1}{x_2+y_2}=\frac{9-5.5}{1+.5}=\frac{7}{3}=\frac{3}{k}\Rightarrow \left[k\right]=1$
Hence the smaller value of $\left[k\right]$ is $1$