Question

The equation of a line is 3x - 4y + 12 = 0. It intersects the x axis in point A and Y axis in point B , finds the coordinates of points A and B ,and finds 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

Hint: We will apply the distance formula between two points let us say A and B. This can be found out by the formula $ \text{Distance = }\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} $ . Here A is $ \left( {{x}_{1}},{{y}_{1}} \right) $ and B is $ \left( {{x}_{2}},{{y}_{2}} \right) $ . We will also use the trick here of substituting x = 0 and y = 0 into the equation of line due to the point lying on that axis.

Complete step-by-step answer:
We will first consider the equation that is given to us as 3x – 4y + 12 = 0. Now according to the question we have that this line intersects at x axis in point A. Since, the line intersects at x axis then at that point y will be 0. Therefore, we are going to substitute y = 0 here. Thus, we will get 3x – 4(0) + 12 = 0 or 3x + 12 = 0. Now, we will take the value 12 to the right side of the equation. Thus, we will get 3x = - 12. So, x = - 4. This means that the point A is (- 4, 0).
Similarly the line 3x – 4y + 12 = 0 intersects at Y axis in point B. Since, the line cuts at y axis so at that point x = 0. So, we will substitute this value in the equation of line 3x – 4y + 12 = 0. Thus, we get 3(0) – 4y + 12 = 0 or – 4y + 12 = 0. This results into 4y = 12. That is y = 3 here. Therefore the point B is (0, 3). The required graph of the line 3x – 4y + 12 = 0 is given below.

So, the required coordinates are A (- 4, 0) as $ \left( {{x}_{1}},{{y}_{1}} \right) $ and B (0, 3) as $ \left( {{x}_{2}},{{y}_{2}} \right) $ . Now we will find the length of the line AB. For this we will use distance formula. This is given by $ \begin{align}
  & \text{Distance = }\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} \\
 & \Rightarrow \text{Distance = }\sqrt{{{\left( 0-\left( -4 \right) \right)}^{2}}+{{\left( 3-0 \right)}^{2}}} \\
 & \Rightarrow \text{Distance = }\sqrt{{{\left( 0+4 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\
 & \Rightarrow \text{Distance = }\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\
 & \Rightarrow \text{Distance = }\sqrt{16+9} \\
 & \Rightarrow \text{Distance = }\sqrt{25} \\
 & \Rightarrow \text{Distance = }\pm \text{5} \\
\end{align} $
Since, the distance cannot be negative so we will have the distance as 5 units.
Hence, the correct option is (b).

Note: We can also find out the points at the axis by looking at the graph. After this we have solved as usual by using the distance formula. There is no chance of choosing the distance negative because it is not possible at all. We can also solve this question by using Pythagoras theorem in which we will take the sides as their values in the graph and find the distance AB by considering it as a hypotenuse. This is because the angle made by the line to the two axes is actually $ {{90}^{\text{o}}} $ . One should focus on the question during calculation. If we do wrong calculations then eventually we will get the wrong answer.