Question

Given the line $3x + 5y = 15$ and a point on this line equidistant from the coordinate axes. Such a point exists in which of the following quadrants?
A. None of the quadrants.
B. Quadrant I only
C. Quadrant I, II only
D. Quadrant I, II, III only
E. Each of the quadrants

Answer 1

In order to solve this question first we will assume the point and its coordinates further we will use the the property as distance of point from the $x$ axis is given by $\left| y \right|$ and from y axis is given by $\left| x \right|$by using it we will get the relation between $x$ and $y$ and we will again make the system equation by using the concept as the point lies of the equation satisfy its corresponding equation.

Complete step-by-step answer:
Let the point on the given line $3x + 5y = 15$ , which is equidistant from the coordinate axis be$A\left( {x,y} \right).$
So we get
$\left| x \right| = \left| y \right| \to (1)$
Also since $A$ lies on $3x + 5y = 15$, $A$ must satisfy the equation of line.
So we get
$3x + 5y = 15 \to (2)$
From equation $1$ we already have $x = y$ or $x = - y$
If $x = {\text{ }}y$, substitute value of $x$ in equation $2$

$$3x + 5x = 15 \\\ \Rightarrow 8x = 15 \\\ \Rightarrow x = 15/8 \\\$$

Therefore $x{\text{ }} = {\text{ }}y{\text{ }} = {\text{ }}15/8$
Hence $A = (15/8,15/8)$
Both $x$and $y$are positive that means it should be present in first quadrant
If $x{\text{ }} = {\text{ }} - y$ substitute value of $x$ in equation 2

$$3x - 5x = 15 \\\ \Rightarrow - 2x = 15 \\\ \Rightarrow x = - 15/2 \\\$$

$\therefore x = - y$
So $y = 15/2$
Therefore $x = - 15/2$ and $y = 15/2$
Hence $A = ( - 15/2,15/2)$
Here $x$is negative whereas $y$is positive that reflects it should be present on the second quadrant.
Therefore $A$ lies either in the first quadrant or in the second quadrant

So, the correct answer is “Option C”.

Note: The axes of a two-dimensional Cartesian system divide the plane into four infinite regions, each bounded by two half-axes, called quadrants. In first quadrant both $x$and $y$would be positive, for Second quadrant $x$ would be negative and$y$ would be positive for third quadrant both $x$and $y$ would be negative for fourth quadrant $x$ would be positive and $y$ would be negative.