Question

A point on the straight line 3x+5y = 15 which is equidistant from the coordinates axes lies in
[a] 1st and 2nd quadrants
[b] 4th quadrant
[c] 1st , 2nd and 4th quadrants
[d] 1st quadrant

Answer 1

Hint: Distance of point P(x,y) from the x-axis is given by $\left| y \right|$ and from y-axis is given by $\left| x \right|$. Assume the point on the line (3x+5y) = 15, which is equidistant from the coordinate axis be P(x,y).
Hence find the relation between x and y using the above property. Also, since the point lies on 3x+5y = 15, it must satisfy its equation. This will give a system of equations. Solve the system for x and y. Hence find the coordinates of P. Determine in which quadrant P lies.

Complete step-by-step answer:
Let the point on the line (3x+5y) = 15, which is equidistant from the coordinate axis be P(x,y).
Hence we have
$\left| x \right|=\left| y \right|\text{ (i)}$
Also since P lies on 3x+5y = 15, P must satisfy the equation of the line.
Hence we have $3x+5y=15\text{ (ii)}$
From equation (i) we have x = y or x = -y.
If x = y, we have from equation (ii)
$\begin{aligned} & 3x+5x=15 \\\ & \Rightarrow 8x=15 \\\ & \Rightarrow x=\dfrac{15}{8} \\\ \end{aligned}$
Hence $P\equiv \left( \dfrac{15}{8},\dfrac{15}{8} \right)$
If x = -y, we have from equation (ii)
$\begin{aligned} & 3x-5x=15 \\\ & \Rightarrow -2x=15 \\\ & \Rightarrow x=\dfrac{-15}{2} \\\ \end{aligned}$
Hence $P\equiv \left( \dfrac{-15}{2},\dfrac{15}{2} \right)$
Hence P lies either in the first quadrant or in the second quadrant.

Note: Alternative solution.
We know that points equidistant from the axis either lie on the line y = x or on the line y = -x.

Point of intersection of y = x and 3x+5y = 15 is $\left( \dfrac{15}{8},\dfrac{15}{8} \right)$
Point of intersection y = - x and 3x+5y = 15 is $\left( \dfrac{-15}{2},\dfrac{15}{2} \right)$
Hence P lies in the first or the third quadrant.
This can be viewed graphically as follows