Question

The equation of the line which makes right angled triangle with axes whose area is 6 sq. units and whose hypotenuse is of 5 units, is
A. \[\dfrac{x}{4} + \dfrac{y}{3} = \pm 1\]
В. \[\dfrac{x}{4} - \dfrac{y}{3} = \pm 3\]
C. \[\dfrac{x}{6} + \dfrac{y}{1} = \pm 1\]
D. \[\dfrac{x}{1} - \dfrac{y}{6} = \pm 1\]

Answer 1

Hint: As stated in the question, the line forms a triangle with coordinate axes, implying that the triangle will be right-angled. Because the area of the triangle is also given, we must compute the height and base to complete the necessary equation.

Formula Used: Triangle’s area or dimensions can be calculated using the formula
\[\dfrac{1}{2} \times {\rm{height}} \times {\rm{base}}\]

Complete step by step solution: We have been given that the right angled triangle with axes whose area is 6 sq. units and whose hypotenuse is of 5 units.
Let us assume that \[{X^\prime }OX\] and \[YO{Y^\prime }\] be the coordinate axes.
And let us also consider that \[AB\] be the part of the given line intercepted by the axes.
Now, let’s have \[OA = a\] and \[OB = b\]
We already know that the Area of triangle is
\[AOB = \dfrac{1}{2} \times base \times height\]
We can write the above as,
\[ = \dfrac{1}{2}{\rm{ab}}\](As per the given data)
On substituting, we get
\[ = 6\]
Now, we can write that as,
\[ \Rightarrow {\rm{ab}} = 12\]------ (1)
\[{a^2} + {b^2} = 25\]-------- (2) {As per the given data \[AB = \]hypotenuse}

On substituting the value of \[b = \dfrac{{12}}{a}\] in the equation (2), we get
\[{a^2} + \dfrac{{144}}{{{a^2}}} = 25\]
Now, we have to simplify by multiplying the denominator with the terms on the LHS of the equation, we get
\[ \Rightarrow {a^4} - 25{a^2} + 144 = 0\]
We have to assume that \[{a^2} = p\] thus we get
\[ \Rightarrow {p^2} - 25p + 144 = 0\]
Let us expand the above equation in order to factorize, we get
\[ \Rightarrow {p^2} - 16p - 9p + 144 = 0\]
Now, let’s factorize the above equation, we get
\[ \Rightarrow p(p - 16) - 9(p - 16) = 0\]
On simplification, we get
\[ \Rightarrow (p - 16)(p - 9) = 0\]
On substituting the value of\[p = {a^2}\] we get
\[ \Rightarrow \left( {{a^2} - 16} \right)\left( {{a^2} - 9} \right) = 0\]
Now, we have to equate each term to \[0\] we get
\[ \Rightarrow {a^2} = 16\] OR \[{a^2} = 9\]
On squaring the above terms we get
\[ \Rightarrow a = 4\] OR \[a = 3\] (because, length must be always a positive value)
Now, we have ( \[a = 4 \Rightarrow b = 3\]) and ( \[a = 3 \Rightarrow b = 4\])
Now, we have obtained the required equation as
\[\dfrac{x}{4} + \dfrac{y}{3} = 1\] OR \[\dfrac{x}{3} + \dfrac{y}{4} = 1\]
That is, the above can be restructured as,
\[3x + 4y - 12 = 0\] OR \[4x + 3y - 12 = 0\]
Therefore, the equation of the line which makes right angled triangle with axes whose area is 6 sq. units and whose hypotenuse is of \[5\] units is \[\dfrac{x}{4} + \dfrac{y}{3} = \pm 1\].

Option ‘A’ is correct

Note: Three vital facts to recall in this question are: the locations where the line intersects the coordinate axes are called intercepts, the area of a right-angled triangle is \[\dfrac{1}{2} \times {\rm{height}} \times {\rm{base}}\] and the intercept version of the straight line equation is \[\dfrac{{\rm{x}}}{{\rm{a}}} + \dfrac{{\rm{y}}}{{\rm{b}}} = 1\].