Question
The line segment joining $A (2, 3)$ and $B (6, -5)$ is intercepted by $x-$ axis at the point $K.$ Write down the ordinate of the point $K.$ Hence, find the ratio inwhich $K$ divides $AB.$ Also find the coordinates of the point $K.$
✓
Answer
Since, point $K$ lies on x-axis, its ordinate is $0.$
Let the point $K (x, 0)$ divides $AB$ in the ratio $k: 1.$
We have,
$y=\frac{k \times(-5)+1 \times 3}{k+1}$
$0=\frac{-5 k+3}{k+1}$
$k=\frac{3}{5}$
Thus, K divides AB in the ratio $3: 5.$
Also, we have:
$x=\frac{k \times 6+1 \times 2}{k+1}$
$x=\frac{\frac{3}{5} \times 6+2}{\frac{3}{5}+1}$
$x=\frac{18+10}{3+5}$
$x=\frac{28}{8}=\frac{7}{2}=3 \frac{1}{2}$
Thus, the co-ordinates of the point $K$ are $\left(3 \frac{1}{2}, 0\right)$
Let the point $K (x, 0)$ divides $AB$ in the ratio $k: 1.$
We have,
$y=\frac{k \times(-5)+1 \times 3}{k+1}$
$0=\frac{-5 k+3}{k+1}$
$k=\frac{3}{5}$
Thus, K divides AB in the ratio $3: 5.$
Also, we have:
$x=\frac{k \times 6+1 \times 2}{k+1}$
$x=\frac{\frac{3}{5} \times 6+2}{\frac{3}{5}+1}$
$x=\frac{18+10}{3+5}$
$x=\frac{28}{8}=\frac{7}{2}=3 \frac{1}{2}$
Thus, the co-ordinates of the point $K$ are $\left(3 \frac{1}{2}, 0\right)$
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