Question

find the length of the longest side of the triangle formed by the line 3x+4y=12 with the coordinate axes​

Answer 1

Given line is 3x+4y=12

First we find the points where the line meets coordinate axes

put y=0, we get

3x=12

x=4

put x=0, we get

4y=12

y=3

$\therefore$The given line meets

the coordinate axes at (4,0) and (0,3)

The length of the longest side of the triangle formed by the line 3x+4y=12

= The distance betweeen (4,0) and (0,3)

$=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$=\sqrt{(4-0)^2+(0-3)^2}$

$=\sqrt{16+9}$

$=\sqrt{25}$

$=5\,units$

Answer 2

The longest side has length 5 units.

Step-by-step explanation:

• This question can be solved in the easiest manner, by using the equation of intercept form of line.

We know that:

• Equation for intercept form of line : $\dfrac{x}{a} + \dfrac{y}{b} = 1$

     Where;         'a' is the x - intercept.

                            'b' is the y - intercept.

We are given the equation of line as;

•                                                3x + 4y = 12

We try to convert it in the intercept form. We divide both the sides, by 12.

• $\dfrac{3x}{12} + \dfrac{4y}{12} = \dfrac{12}{12}$

       $\dfrac{3x}{12} + \dfrac{4y}{12} = \dfrac{12}{12}$

         $\dfrac{x}{4} + \dfrac{y}{3} = 1$

Now, x - intercept is 4 and y - intercept is 3.

• Since, the triangle $\Delta AOB$is the right angle triangle.

• We can apply Pythagoras theorem.

        $AB^2 = AO^2 + OB^2$

             $AB^2 = 3^2 + 4^2$

             $AB^2 = 9 + 16 = 25$

               $AB = \sqrt{25}$

                  $AB = \pm5$, since it is length. So it can't be negative.

Therefore;     AB = 5 units

This is the longest side of the triangle.

Thus, the longest side has length 5 units.