Question

The equation of a line is $3x-4y+12=0$. It intersects X-axis in point A and Y-axis in point B, find the co-ordinates of points A and B, find the length of AB.

• A. $A=(0,3),B=(-4,0),AB=5$
• B. $A=(-4,0),B=(0,3),AB=5$
• C. $A=(4,0),B=(0,-3),AB=5$
• D. $A=(0,-3),B=(-4,0),AB=5$

Answer 1

The correct option is B $A=(-4,0),B=(0,3),AB=5$

As the point A lies on the $x$ - axis, its $y$ coordinate $=0$.
To find the co-ordinates of the point A, substitute $y=0$ in the equation of the line $3x-4y+12=0$.
Hence, $3x-4\left(0\right)+12=0$.
$=>3x=-12$
$=>x=-4$
So, point A $=(-4,0)$
Similarly, as the point B lies on the $y$ - axis, its $x$ coordinate $=0$.
To find the co-ordinates of the point B, substitute $x=0$ in the equation of the line $3x-4y+12=0$.
Hence, $3\left(0\right)-4y+12=0$.
$=>4y=12$
$=>y=3$
So, point B $=(0,3)$
Formula to calculate distance between two points $(x_1,y_1)$ and $(x_2,y_2)=$ $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
So, length of $AB=\sqrt{{(0-(-4\left)\right)}^2+{(3-0)}^2}=\sqrt{16+9}=\sqrt{25}=5$