Question

A triangle of area 24 sq. units is formed by a straight line and the coordinate axes in the first quadrant. Find the equation of the straight line, if it passes through 3,4.

Answer 1

Hint: Express the equation of the line in the intercept form then pass it through the given point $\left( {3,4} \right)$, after that use the formula for the area of the triangle to get the relation between the base and height of the triangle and then use these relations to get the desired result.

Complete step-by-step answer:
It is given that the area of a triangle is $24$ square meters.
We know that the equation of the line in intercept form is given as:
$\dfrac{x}{a} + \dfrac{y}{b} = 1$, where $a$ is the $x - $ intercept and $b$ is the $y - $ intercept.
If this line passes through the point$\left( {3,4} \right)$ then the equation of a line is given as:
$\dfrac{3}{a} + \dfrac{4}{b} = 1$
Solving the above equation:
$\dfrac{{3b + 4a}}{{ab}} = 1$
Cross multiply the above expression:
$4a + 3b = ab$
We know that the area of the triangle has the formula:
Area of a triangle$ = \dfrac{1}{2}ab$
Substitute $24$as the area of the triangle in the equation:
$24 = \dfrac{1}{2}ab$
$ab = 24 \times 2$
$ab = 48$
$b = \dfrac{{48}}{a}$
The relation between $a$and $b$is given as$b = \dfrac{{48}}{a}$.
We already have the equation:
$4a + 3b = ab$
Substituting the value$b = \dfrac{{48}}{a}$ and $ab = 48$ in the above expression:
$4a + 3\left( {\dfrac{{48}}{a}} \right) = 48$
Solve to equation ofr the value of $a$ :
$4{a^2} + 144 = 48a$
$ \Rightarrow 4{a^2} - 48a + 144 = 0$
$ \Rightarrow 4\left( {{a^2} - 12a + 36} \right) = 0$
$ \Rightarrow {a^2} - 12a + 36 = 0$
Factorize the above equation by breaking the middle term:
$ \Rightarrow {a^2} - 6a - 6a + 36 = 0$
$ \Rightarrow a\left( {a - 6} \right) - 6\left( {a - 6} \right) = 0$
$ \Rightarrow \left( {a - 6} \right)\left( {a - 6} \right) = 0$
$ \Rightarrow {\left( {a - 6} \right)^2} = 0$
So, the value of$a$ is $6$. Now, find the value of $b$ using the relation:
$b = \dfrac{{48}}{a}$
Substitute the value $a = 6$in the expression:
$
  b = \dfrac{{48}}{6} \\
  b = 8 \\
$
Now, we have the values $a = 6$ and $b = 8$.
Equation of a line in intercept form is given as:
$\dfrac{x}{a} + \dfrac{y}{b} = 1$
Substitute the values of $a$ and$b$ in the equation:
$\dfrac{x}{6} + \dfrac{y}{8} = 1$

Therefore, the required equation of the line that passes through the point$\left( {3,4} \right)$ is
$\dfrac{x}{6} + \dfrac{y}{8} = 1$

Note: If the triangle has the height $b$ and the base $a$, then the area of the triangle is given as:
${\text{Area}} = \dfrac{1}{2}\left( {{\text{base}}} \right)\left( {{\text{height}}} \right)$
Area=$\dfrac{1}{2}\left( a \right)\left( b \right)$