Question

Draw the graph of 5x + 6y – 30 = 0 and use it to find the area of the triangle formed
by the line and the coordinate axes.​

Answer 1

Answer:

you can also use shortcut

area of triangle with coordinate axis = c^2/2ab

where c is constant term and 'a' is coefficient of x and 'b' is coefficient of y.

so area of triangle = (30)^2/2×6×5 = 900/60 = 15 square units

Answer 2

Given:

A line of equation, $5x+6y-30=0$

To find:

Area of triangle formed by the line and coordinate axes.

Solution:

On plotting the line $5x+6y-30=0$ in a graph sheet, we get the graph as shown in figure. The line forms a triangle with x-axis as base of triangle and y-axis as the third side of the triangle. The line meets x-axis at $B(6,0)$ and y-axis at $A(0,5)$. The third vertex is given by the origin O(0,0).

Distance BO = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

where $x_{1}=0, x_{2}=0, y_{1}=5,y_{2}=0$

$BO=\sqrt{(0-0)^{2}+(5-0)^{2}}$

$BO=\sqrt{5^{2}+0}=\sqrt{25}=5$

Distance AO = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

where $x_{1}=6, x_{2}=0, y_{1}=0,y_{2}=0$

$AO=\sqrt{(6-0)^{2}+(0-0)^{2}}$

$AO=\sqrt{6^{2}+0}=\sqrt{36}=6$

Area of triangle = $\frac{1}{2}$ × base × height

a(ΔABO) = $\frac{1}{2}$ × BO × AO

a(ΔABO) = $\frac{1}{2}*6*5$

a(ΔABO) = $3*5=15$ unit²

Area of triangle formed by the line $5x+6y-30=0$ and the coordinate axes is 15 unit².