
The area of a triangle formed by the lines $x = 0,y = 0$ and $3x + 4y = 12$ is
A. $12$
B. $6$
C. $2$
D. None of these
Answer
198.3k+ views
Hint: Find all the vertices of the triangle. There are three vertices, one is on the $x$-axis, another one is on the $y$-axis and the third one is the origin. Find the length and height of the triangle is right-angled. So, simply obtain the area using the formula of finding the area of a triangle using its base and height.
Formula Used:
Area of a triangle is obtained by the formula $\dfrac{1}{2} \times base \times height$.
Complete step by step solution:
The given straight lines are $x = 0,y = 0$ and $3x + 4y = 12$
The straight line $x = 0$ represents $y$-axis.
The straight line $y = 0$ represents $x$-axis
So, one vertex of the triangle is at origin.
Find the other two vertices
Coordinate of any point on $x$-axis is of the form $\left( {a,0} \right)$.
So, putting $x = a$ and $y = 0$ in the equation $3x + 4y = 12$, we get $3a + 4 \times 0 = 12$
Find the value of $a$
$3a + 0 = 12\\ \Rightarrow 3a = 12\\ \Rightarrow a = \dfrac{{12}}{3}\\ \Rightarrow a = 4$
So, the line $3x + 4y = 12$ meets $x$-axis at the point $A\left( {4,0} \right)$
Coordinate of any point on $y$-axis is of the form $\left( {0,b} \right)$.
So, putting $x = 0$ and $y = b$ in the equation $3x + 4y = 12$, we get $3 \times 0 + 4b = 12$
Find the value of $b$
$ \Rightarrow 0 + 4b = 12\\ \Rightarrow 4b = 12\\ \Rightarrow b = \dfrac{{12}}{4}\\ \Rightarrow b = 3$
So, the line $3x + 4y = 12$ meets $y$-axis at the point $B\left( {0,3} \right)$
$\therefore $ All three vertices of the triangle are $O\left( {0,0} \right),A\left( {4,0} \right)$ and $B\left( {0,3} \right)$.
The length of base of the triangle is $OA = 4$ units
and the length of height of the triangle is $OB = 3$ units.
So, the area of the triangle is $\dfrac{1}{2} \times OA \times OB = \dfrac{1}{2} \times 4 \times 3 = 6$ square units.
Option ‘B’ is correct
Note: The set of axes meet at origin. So, clearly, origin is one vertex of the given triangle. The ordinate i.e. the $y$-value of each point on $x$-axis is equal to zero and the abscissa i.e. the $x$-value of each point on $y$-axis is equal to zero.
Formula Used:
Area of a triangle is obtained by the formula $\dfrac{1}{2} \times base \times height$.
Complete step by step solution:
The given straight lines are $x = 0,y = 0$ and $3x + 4y = 12$
The straight line $x = 0$ represents $y$-axis.
The straight line $y = 0$ represents $x$-axis
So, one vertex of the triangle is at origin.
Find the other two vertices
Coordinate of any point on $x$-axis is of the form $\left( {a,0} \right)$.
So, putting $x = a$ and $y = 0$ in the equation $3x + 4y = 12$, we get $3a + 4 \times 0 = 12$
Find the value of $a$
$3a + 0 = 12\\ \Rightarrow 3a = 12\\ \Rightarrow a = \dfrac{{12}}{3}\\ \Rightarrow a = 4$
So, the line $3x + 4y = 12$ meets $x$-axis at the point $A\left( {4,0} \right)$
Coordinate of any point on $y$-axis is of the form $\left( {0,b} \right)$.
So, putting $x = 0$ and $y = b$ in the equation $3x + 4y = 12$, we get $3 \times 0 + 4b = 12$
Find the value of $b$
$ \Rightarrow 0 + 4b = 12\\ \Rightarrow 4b = 12\\ \Rightarrow b = \dfrac{{12}}{4}\\ \Rightarrow b = 3$
So, the line $3x + 4y = 12$ meets $y$-axis at the point $B\left( {0,3} \right)$
$\therefore $ All three vertices of the triangle are $O\left( {0,0} \right),A\left( {4,0} \right)$ and $B\left( {0,3} \right)$.
The length of base of the triangle is $OA = 4$ units
and the length of height of the triangle is $OB = 3$ units.
So, the area of the triangle is $\dfrac{1}{2} \times OA \times OB = \dfrac{1}{2} \times 4 \times 3 = 6$ square units.
Option ‘B’ is correct
Note: The set of axes meet at origin. So, clearly, origin is one vertex of the given triangle. The ordinate i.e. the $y$-value of each point on $x$-axis is equal to zero and the abscissa i.e. the $x$-value of each point on $y$-axis is equal to zero.
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