Question

If the graph of the equation $4x + 3y = 12$ cuts the coordinate axes at points $A$ and $B$ then hypotenuse of right angle triangle $AOB$ is
A. $4$units
B. $3$units
C. $5$units
D. None of these

Answer 1

Draw the graph of equation and see where it cuts the coordinate axes. Then find the length of the hypotenuse of the triangle by using Pythagoras theorem.
Pythagoras theorem: If $a$and $b$ is the length of base and perpendicular respectively of a right angle triangle. Then the length of hypotenuse is $\sqrt {{a^2} + {b^2}}$.

Complete step-by-step answer:
Step-1 draws the graph of the given equation.

Step-2
From the graph given in step-1 we can see that the equation $4x + 3y = 12$ cuts the coordinate axes at $\left( {3,0} \right)$ and $\left( {0,4} \right)$.
Now see the $\vartriangle AOB$(this is a right angle triangle with right angle at $o$)
In $\vartriangle AOB$, $OA$ is the base of the triangle and $OB$ is the perpendicular of the triangle.
$OA = 3$units
$OB = 4$units
$AB$ is the hypotenuse of triangle $\vartriangle AOB$
Step-3
(Pythagoras theorem: if $a$and $b$ is the length of base and perpendicular respectively of a right angle triangle. Then the length of hypotenuse is $\sqrt {{a^2} + {b^2}}$.)
Apply Pythagoras theorem on $\vartriangle AOB$
$\Rightarrow AB = \sqrt {{{\left( {OA} \right)}^2} + {{\left( {OB} \right)}^2}}$
Substitute the value of $OA$ and $OB$ in the above expression.
$\Rightarrow AB = \sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2}}$
$\Rightarrow AB = \sqrt {9 + 16}$
$\Rightarrow AB = \sqrt {25}$
$\Rightarrow AB = 5$
Therefore the length of hypotenuse of the triangle is $5$ units.
Hence, Option (C) is the correct answer.

Note: We can also find the intercepts as follows:
If the equation of line is in the form $\dfrac{x}{a} + \dfrac{y}{b} = 1$ then x-intercept is $a$ and y-intercept is $b$.
Convert the given equation $4x + 3y = 12$ in the form of $\dfrac{x}{a} + \dfrac{y}{b} = 1$
Divide the given equation by $12$ to get it in the required form.
Equation becomes $\Rightarrow \dfrac{{4x}}{{12}} + \dfrac{{3y}}{{12}} = 1$
$\Rightarrow \dfrac{x}{3} + \dfrac{y}{4} = 1$
On comparing the above equation with $\dfrac{x}{a} + \dfrac{y}{b} = 1$ we get $a = 3$ and $b = 4$
Hence, x-intercept is $3$ and y-intercept is $4$
After finding the intercepts follow the step-3.