Question

A point on the straight line, 3x + 5t = 15 which is equidistant from the coordinate axes lies only on:
A. ${{1}^{st}}$ and ${{2}^{nd}}$ quadrants
B. ${{4}^{th}}$ quadrant
C. ${{1}^{st}}$, ${{2}^{nd}}$ and ${{4}^{th}}$ quadrant
D. ${{1}^{st}}$ quadrant

Answer 1

To solve this problem, it is given that the straight line, 3x + 5t = 15 is equidistant from the coordinate axes lies. So, we can assume two values such that x = t and x = -t. So, to find the intersecting points first we can substitute the value of x = t in the equation and then substitute x = -t in the equation. So we will get two intersecting points. By finding the points we will get the quadrants.

Complete step by step answer:
The given straight line is 3x + 5t = 15. It is also given that the straight line is equidistant from the coordinate axes. So, we can assume two values such that x = t and x = -t. By this we can find the intersecting points.
Case 1: We can take x = t. So, we get,
3x + 5x = 15
$\Rightarrow$8x = 15
$\Rightarrow$x = $\dfrac{15}{8}$
So, we get the coordinate points as $\left( \dfrac{15}{8},\dfrac{15}{8} \right)$.
So, this point lies in the ${{1}^{st}}$ quadrant.
Case 2: We can take x = -t. So, we get,
-3t + 5t = 15
$\Rightarrow$2t = 15
$\Rightarrow$t =$\dfrac{15}{2}$
So, we get the coordinate point as $\left( \dfrac{-15}{2},\dfrac{15}{2} \right)$.
So, this point lies in the ${{2}^{nd}}$ quadrant.

So, the correct answer is “Option A”.

Note: We have taken x = t and x = -t because these values will always be equidistant. Alternative method to find the solution is by drawing the graph. Then, we will be able to find the quadrant in which the straight line lies.