The slope of a line which cuts off intercepts | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the intercepts cut by the line on the axes be $a$ and $a$ (since they are of equal length). Using the intercept form of the equation of a line

Vedclass · Accepted Answer

-$1$

Vedclass · Answer

(A) Let the intercepts cut by the line on the axes be $a$ and $a$ (since they are of equal length). Using the intercept form of the equation of a line: $\frac{x}{a} + \frac{y}{a} = 1$. Multiplying by $a$,we get $x + y = a$,or $x + y - a = 0$. The equation is in the form $Ax + By + C = 0$,where $A = 1$ and $B = 1$. The slope $m$ of the line is given by $m = -\frac{A}{B} = -\frac{1}{1} = -1$. Alternatively,if the intercepts are equal in magnitude but opposite in sign,the slope could be $1$. However,given the options,the standard interpretation is equal intercepts $a$,leading to a slope of $-1$.

The slope of a line which cuts off intercepts of equal lengths on the axes is:

Similar Questions

$A$ pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve $x^2+y^2=4$ with $x+y=a$. The set containing the value of $a$ is

The area (in sq. units) of the triangle formed by the lines $x^2-3xy+y^2=0$ and $x+y+1=0$ is

The line $3x + 4y - 5 = 0$ cuts the curve $2x^2 + 3y^2 = 5$ at $A$ and $B$. If $O$ is the origin,then $\angle AOB =$

The line $x+2y=k$ meets the curve $2x^2-2xy+3y^2+2x-y-1=0$ at two points $A$ and $B$. Let $O$ be the origin. If the line segments $OA$ and $OB$ are perpendicular to each other,then $k=$

The condition that the lines joining the origin to the points of intersection of the two curves $x^2+y^2+gx+c=0$ and $x^2+y^2+2fy-c=0$ are at right angles is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module