Question

The point where the line $x=my+c$ cuts ‘x’ and ‘y’ axes are ……
A. $\left( c,0 \right),\left( 0,-\dfrac{c}{m} \right)$
B. $\left( 0,c \right),\left( -\dfrac{c}{m},0 \right)$
C. $\left( -\dfrac{c}{m},0 \right),\left( 0,c \right)$
D. $\left( c,\dfrac{-c}{m} \right),\left( 0,0 \right)$

Answer 1

Hint: To answer this question, we specify the general equation of a line in the intercept form. We do this since we are required to find the intersection point of the line and the coordinate axes which are nothing but the intercepts of the line. We then convert the given equation in question to intercept form and compare the two equations and obtain the answer.

Complete step by step solution:
We first define the general form of a line. The two basic equations of a straight line in slope intercept form are as given below:
$\Rightarrow y=mx+c\ldots \ldots \left( 1 \right)$
For the above equation, m is the slope and c is the y-intercept. Hence the equation is called slope intercept form.
$\Rightarrow x=my+c\ldots \ldots \left( 2 \right)$
For this above equation, m is the slope and c is the x-intercept. Hence the equation is called slope intercept form.
Now in order to calculate the two points, where the given line $x=my+c$ cuts the x and y axes, we need to convert the given equation to intercept form. The given equation $x=my+c$ is in slope intercept form and we need to convert this to the intercept form. The intercept form for a straight line is given as:
$\Rightarrow \dfrac{x}{a}+\dfrac{y}{b}=1\ldots \ldots \left( 3 \right)$
Here, a represents the x intercept and b represents the y intercept. And, the two points of intersection with the x and y axes are given as $\left( a,0 \right)$ and $\left( 0,b \right).$
We convert this by rearranging the terms of the above equation.
$\Rightarrow x=my+c$
Taking the ‘my’ term to the left-hand side,
$\Rightarrow x-my=c$
Now we divide both sides of the equation by c,
$\Rightarrow \dfrac{x}{c}-\dfrac{my}{c}=\dfrac{c}{c}$
Simplifying the right-hand side, we get 1 and taking the $\dfrac{-m}{c}$ term to the denominator,
$\Rightarrow \dfrac{x}{c}+\dfrac{y}{\left( -\dfrac{c}{m} \right)}=1$
This equation is in the intercept form with the x intercept as c and y intercept as $-\dfrac{c}{m}.$
Hence, the two points where the line $x=my+c$ cuts ‘x’ and ‘y’ axes are $\left( c,0 \right),\left( 0,-\dfrac{c}{m} \right).$

So, the correct answer is “Option A”.

Note: Students need to know the general concepts of straight lines and the different ways to represent the equation of a straight line. We know the x intercept for the equation as c from the slope intercept form, hence, we can also solve this question by representing the given equation in the form of equation $\left( 1 \right)$ and finding the y intercept by rearranging the equation to obtain the answer.