A line cuts off equal intercepts on the coord | vedclass

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(B) Let the intercepts on the $X$ and $Y$ axes be $a$ and $a$ respectively. The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{a} =

Vedclass · Accepted Answer

$135$

Vedclass · Answer

(B) Let the intercepts on the $X$ and $Y$ axes be $a$ and $a$ respectively. The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{a} = 1$,which simplifies to $x + y = a$ or $y = -x + a$.Comparing this with the slope-intercept form $y = mx + c$,the slope $m = -1$.Since the slope $m = 	an 	heta$,we have $	an 	heta = -1$.Since $	an 	heta$ is negative,the angle $	heta$ lies in the second quadrant.$	heta = 180^{\circ} - 45^{\circ} = 135^{\circ}$.Alternatively,if the intercepts are equal in magnitude but opposite in sign,the line could be $x - y = a$,giving a slope of $1$,which corresponds to an angle of $45^{\circ}$. However,standard interpretation of 'equal intercepts' usually implies the same sign,leading to $135^{\circ}$.

$A$ line cuts off equal intercepts on the coordinate axes. The angle made by this line with the positive direction of the $X$-axis is: (in $^{\circ}$)

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