Q32. - Geometry

Question

A line passes through points (-2, -1) and (4, 3). Where does the line intersect the x-axis? the y-axis?

Step-by-Step Solution

Answer

The required point of intersection of line with $x-\text{axis\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}\hspace{0.17em}}(-\frac{1}{2},0)\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}y-\text{axis\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}\hspace{0.17em}}(0,\frac{1}{3})$

Step-1 – Given

Given that a line passes through points$(-2,-1)$ and$(4,3)$ .

Step-2 – To determine

We have to find the point of intersection of line with x-axis and y-axis.

Step-3 – Calculation

The two-point formula using these two points $(x_1,y_1)\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}(x_2,y_2)$ is:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

The given points are $(-2,-1)$ and$(4,3)$ . 

$\begin{array}{l}y-(-1)=\frac{3-(-1)}{4-(-2)}(x-(-2))\\ y+1=\frac{2}{3}(x+2)\\ 3y+3=2x+4\\ 3y-2x=1\end{array}$

The equation of the x-axis is y = 0, we solve it with$3y-2x=1$ .

$\begin{array}{l}3y-2x=1\\ 3\left(0\right)-2x=1\\ -2x=1\\ x=-\frac{1}{2}\\ \end{array}$

Equation of the y-axis is x = 0, solving it with$3y-2x=1$ .

$\begin{array}{l}3y-2x=1\\ 3y-2\left(0\right)=1\\ 3y=1\\ y=\frac{1}{3}\end{array}$

So, the required point of intersection of line with$x-\text{axis\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}\hspace{0.17em}}(-\frac{1}{2},0)$ .

$\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}y-\text{axis\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}\hspace{0.17em}}(0,\frac{1}{3})$