Question

A $\Delta ABC$ with AB = 13, BC = 5 and AC = 12 slides on the coordinate axis with A and B are on the positive x – axis and positive y – axis respectively, the locus of vertex C is a line 12x – ky = 0 then the value of k is:
(a). 1
(b). 4
(c). 3
(d). 5

Answer 1

Hint: If you can see the values of AB, BC and AC, addition of square of AC and square of BC gives square of AB. This shows that $\Delta ABC$ is a right angled triangle at C. The line 12x – ky = 0 is a straight line passing through origin so the slope of this line has the value of k. And slope is tan θ so by geometry we can find k.

Complete step-by-step solution -
For a better understanding of the question, we have drawn a figure according to the conditions given in question:

The length of AB, BC and AC are 13, 5 and 12 respectively. If we carefully look to these lengths we will find that they are the right angled triangle triads.
$(13)^2 = (12)^2 + (5)^2$
$\Rightarrow $ 169 = 144 + 25
$\Rightarrow $169 = 169
$\Rightarrow (AB)^2 = (AC)^2 + (BC)^2 $
So, we can see that $\Delta ABC$ is right angled at C.
As it is given that point A slides on the x – axis and point B slides on the y – axis and we now that angle made by the intersection of x and y axis is 90° so $\angle BOA={{90}^{0}}$ and $\angle BCA={{90}^{0}}$($\Delta ABC$ is right angled at C). So, the sum of $\angle BOA={{90}^{0}}$ and $\angle BCA={{90}^{0}}$ is 180°.
As from the above discussion that sum of the opposite angles of the quadrilateral OABC are 180° ($\angle BOA+\angle BCA={{180}^{0}}$) so we can say that the quadrilateral AOBC is a cyclic quadrilateral. As concyclic angles are equal in a cyclic quadrilateral so $\angle BOC=\angle BAC$.
Now, let us assume that $\angle COA=\theta $ then $\angle BOC={{90}^{0}}-\theta $. And as we have proved above that $\angle BOC=\angle BAC$ so $\angle BAC={{90}^{0}}-\theta $
Now, in right $\Delta ABC$:
$\tan \left( {{90}^{0}}-\theta \right)=\dfrac{\text{BC}}{AC}$
Substituting the values of BC and AC and the value of tan (90 – θ) as cot θ in the above equation we get,
$\cot \theta =\dfrac{5}{12}$
As tan θ is the reciprocal of cot θ so the value of tan θ is $\dfrac{12}{5}$.
Now, we have to find the value of k in 12x – ky = 0.
We know that slope of the line is:
$\text{slope} =-\dfrac{\left( \text{coefficient of x} \right)}{\left( \text{coefficient of y} \right)}$
So, the slope of the line 12x – ky = 0 is equal to $\dfrac{12}{k}$
And slope = tan θ. So, we are going to equate the above slope with the value of tan θ that we have solved above.
$\begin{align}
  & \dfrac{12}{5}=\dfrac{12}{k} \\
 & \Rightarrow k=5 \\
\end{align}$
Hence, the value of k is equal to 5.
Hence, the correct option is (d).

Note: If a straight line is of the form of ax + by = 0 then without solving we can say that the line is passing through origin because (0, 0) satisfies the straight line equation .And if the slope is positive then we can say that the line is making an acute angle with the positive axis. Having this knowledge of straight lines will reduce the time of solving.