Question

What is the x-intercept of the line passing through $\left( {1,4} \right)$ and $\left( {4,1} \right)$ ?
A.4.5
B.5
C.4
D.6
E.None of these

Answer 1

As we have given the two points that lie on the line whose equation is to be found, we will use the two-point form of the equation of line i.e. the equation of a line passing through any two points lets say $\left( {a,b} \right)$ and $\left( {c,d} \right)$ is given by
$\Rightarrow \left( {y - d} \right) = \dfrac{{\left( {d - b} \right)}}{{\left( {c - a} \right)}}\left( {x - c} \right)$
After finding the equation of the line for the x-intercept we will substitute $y = 0$ , as it will be that point that lies on the x-axis and the ordinate of all the points that lie on the x-axis is zero(0).

Complete step-by-step answer:
Given data: line passing through $\left( {1,4} \right)$ and $\left( {4,1} \right)$

We know that the equation of a line passing through any two points lets say $\left( {a,b} \right)$ and $\left( {c,d} \right)$ is given by
$\Rightarrow \left( {y - d} \right) = \dfrac{{\left( {d - b} \right)}}{{\left( {c - a} \right)}}\left( {x - c} \right)$
Therefore, the equation of the line that passes through points $\left( {1,4} \right)$ and $\left( {4,1} \right)$ will be
$\Rightarrow \left( {y - 1} \right) = \dfrac{{\left( {1 - 4} \right)}}{{\left( {4 - 1} \right)}}\left( {x - 4} \right)$
On simplifying the brackets
$\Rightarrow \left( {y - 1} \right) = - \dfrac{3}{3}\left( {x - 4} \right)$
$\Rightarrow \left( {y - 1} \right) = 4 - x$
Therefore the equation of the line that passes through $\left( {1,4} \right)$ and $\left( {4,1} \right)$ is
$\Rightarrow y + x = 5$
Now for the x-intercept, we will substitute $y = 0$ , as it will be that point that lies on the x-axis and the ordinate of all the points that lie on the x-axis is zero(0).
$\therefore x = 5$
Therefore the x-intercept of the line is 5.
Hence, Option (B) is correct.

Note: we can find the equation of the line that passes through $\left( {1,4} \right)$ and $\left( {4,1} \right)$ with an alternative method i.e.
We know that the slope(m) of the line joining any two points lets say $\left( {a,b} \right)$ and $\left( {c,d} \right)$ is given by
$\Rightarrow m = \dfrac{{d - b}}{{c - a}}$
Therefore the slope of the line required i.e. that passes through $\left( {1,4} \right)$ and $\left( {4,1} \right)$
$\Rightarrow m = \dfrac{{4 - 1}}{{1 - 4}}$
$= - \dfrac{3}{3}$
$= - 1$
Now using the one-point form of the line i.e. equation of the line that passes through $\left( {c,d} \right)$ and has a slope equal to (m) is given by
$\Rightarrow \left( {y - d} \right) = m\left( {x - c} \right)$
Therefore the equation of the required line is
$\Rightarrow \left( {y - 1} \right) = - 1\left( {x - 4} \right)$
On multiplication and simplification
$\Rightarrow y - 1 = 4 - x$
$\Rightarrow y + x = 5$, i.e. the same line that is the result of the above solution.