The area (in square units) of the triangle fo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given lines are $x=0$ (the $y$-axis),$y=0$ (the $x$-axis),and $3x+4y=12$.To find the intercepts of the line $3x+4y=12$,we rewrite it in the in

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The given lines are $x=0$ (the $y$-axis),$y=0$ (the $x$-axis),and $3x+4y=12$.To find the intercepts of the line $3x+4y=12$,we rewrite it in the intercept form $\frac{x}{a} + \frac{y}{b} = 1$:$\frac{3x}{12} + \frac{4y}{12} = \frac{12}{12} \implies \frac{x}{4} + \frac{y}{3} = 1$.This line intersects the $x$-axis at $A(4, 0)$ and the $y$-axis at $B(0, 3)$.The triangle formed by the lines $x=0, y=0$,and $3x+4y=12$ is a right-angled triangle with vertices at $O(0, 0)$,$A(4, 0)$,and $B(0, 3)$.The base of the triangle is $OA = 4$ units and the height is $OB = 3$ units.Area of $	riangle OAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4 	imes 3 = 6$ square units.

The area (in square units) of the triangle formed by the lines $x=0, y=0$ and $3x+4y=12$ is:

Similar Questions

Find the area of the triangle formed by the lines $y-x=0$,$x+y=0$,and $x-k=0$.

$A$ square is formed by the lines $x=0, y=0, x=1, y=1$. Then,the equations of its diagonals will be

The circumcentre of the triangle formed by the lines $x = -2$,$y = -2$ and $x + y + 2 = 0$ is

If the orthocenter of the triangle formed by the lines $2x + 3y - 1 = 0$,$x + 2y + 1 = 0$,and $ax + by - 1 = 0$ lies at the origin,then $\frac{1}{a} + \frac{1}{b} =$

If the coordinates of points $A, B, C$ are $(-1, 5), (0, 0)$ and $(2, 2)$ respectively,and $D$ is the midpoint of $BC$,then the equation of the perpendicular drawn from $B$ to the line $AD$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module