Question

Find the area of triangle formed by the straight line 3x-4y+12=0 with coordinate axes

Answer 1

Eq. of line is 3x + 2y = 6………..(1)

Eq. of x axis is y=0……….(2)

Eq. of y axis is x=0…………(3)

Let point of intersection of eq.(1)and (2) is A, put y=0 in eq.(1)

3x+2×0=6 => x=2, A( 2 , 0)

Point of intersection of eq.(2) &(3) is O(0 , 0 )

Point of intersection of eq.(1) & (3) is B , put x=0 in eq. (1)

3×0+2y=6 => y=3 , B(0 ,3 )

Area of triangle AOB=1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)].

=1/2[2(0–3)+0(3–0)+0(0–0)]sq.unit

=1/2(-6)sq.unit

=-3 sq.unit

=3 sq.unit.Answer.

Answer 2

$\maltese$$\tt{solution:-}\maltese$

the equation of line is 3x+4y-12=0

Dividing the equation both side by 2

then,

$\rightarrow \tt\frac{3x}{12}+\frac{4y}{12}=\frac{12}{12}\\ \rightarrow \tt \frac{x}{4}+\frac{y}{3}=1\\ \tt Area\:of\:the\: triangle=\frac{1}{2}×base×height\\ \rightarrow\tt\frac{1}{2}×4×2\\ \rightarrow \tt{\underline{\fbox{6 sq.unit}}}$