Which of the following equations best expresses the line (drawn to sca

Please note : Figure is drawn to scale:

Certain things should be clear from the fig.

1) The slope is negative

( Lines that move from the third quadrant upwards have positive slope and lines that move from the second quadrant downwards have negative slope, here we have the second case )

2) The \(x\)-intercept is positive
3) The \(y\)- intercept is positive
4) The \(x\)- intercept looks > than the \(y\) intercept

Let's convert the options to general eqn. of a line.

(1) \( y = \frac{x}{3}-1\), \(+ ve\) slope , hence reject

(2) \( y= -3x +3\) , \(- ve \) slope, positive \(x\) intercept , Positive \(y\) intercept -KEEP

(3) \( y = \frac{x}{3}+1\), \(+ ve\) slope , hence reject

(4) \( y = 3x-3 \),\(+ ve\) slope , hence reject

(5) \( y = -\frac{x}{3}+1\) , \(-ve\) slope , positive \(x \) intercept , Positive \(y \) intercept -KEEP

It's between B and E

For B :\(y= -3x +3, \) \(x \) intercept is \(1\) and \(y\) intercept is \( 3 \)

For E :\(y = -\frac{x}{3}+1,\) \( x \) intercept is \(3\) and \(y\) intercept is \(1\)

Hence as seen in the fig, \(x\) intercept should be \(> y\) intercept , E is the eqn.that best describes this line.

Ans E

Hope it's clear.