Question

A line passes through points $\left(-2,1\right)$ and $\left(4,3\right)$ where does the line intersect the $x$ axis and the $y$ axis.

Answer 1

Step-1: Find the equation of a line:

Given points, $\left(-2,1\right)$,$\left(4,3\right)$

Two point slope form $\left(m\right)=\frac{y_2-y_1}{x_2-x_1}$ (we know that)

$$\begin{gathered} \therefore m=\frac{3-1}{4-\left(-2\right)} \\ =\frac{2}{6} \\ =\frac{1}{3} \end{gathered}$$

Equation of a line through a point, $\left(y-y_2\right)=m\left(x-x_2\right)$ (we know that)

Substituting the values,

$$\begin{gathered} \Rightarrow \left(y-3\right)=\frac{1}{3}\left(x-4\right) \\ \Rightarrow 3y-9=x-4 \\ \Rightarrow x-3y+5=0\dots \dots \left(1\right) \end{gathered}$$

Step-2: Finding where the line intersects $x$-axis and $y$-axis :

Thus, the equation of the line passing through both points is $x-3y+5=0$.

$\therefore$ If a line is intersecting $x$-axis , i.e. $y=0$ . Putting $y=0$ in equation $\left(1\right)$.

$$\begin{gathered} \Rightarrow x-3\times 0+5=0 \\ \Rightarrow x=-5 \end{gathered}$$

Therefore, the line intersects the $x$-axis at point $\left(-5,0\right)$.

$\therefore$ If a line is intersecting $y$ - axis , i.e. $x=0$ . Putting $x=0$ in equation $\left(1\right)$.

$$\begin{gathered} \Rightarrow 0-3\times y+5=0 \\ \Rightarrow y=\frac{5}{3} \end{gathered}$$

Therefore, the line intersects the $y$-axis at point $\left(0,\frac{5}{3}\right)$.

Hence, line intersects $x$-axis at $\left(-5,0\right)$ and $y$-axis at $\left(0,\frac{5}{3}\right)$.