The equation of the straight line passing thr | vedclass

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

Question

(C) Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$. Given that the sum of intercepts is $a + b = -1$,so $b = -1 - a

Vedclass · Accepted Answer

$\frac{x}{-2} + \frac{y}{1} = 1$ and $\frac{x}{2} + \frac{y}{-3} = 1$

Vedclass · Answer

(C) Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$. Given that the sum of intercepts is $a + b = -1$,so $b = -1 - a$. The line passes through $(4, 3)$,so $\frac{4}{a} + \frac{3}{b} = 1$. Substituting $b = -(1 + a)$,we get $\frac{4}{a} - \frac{3}{1 + a} = 1$. $4(1 + a) - 3a = a(1 + a) \Rightarrow 4 + 4a - 3a = a + a^2$. $a^2 = 4 \Rightarrow a = 2$ or $a = -2$. If $a = 2$,then $b = -1 - 2 = -3$. The equation is $\frac{x}{2} + \frac{y}{-3} = 1$. If $a = -2$,then $b = -1 - (-2) = 1$. The equation is $\frac{x}{-2} + \frac{y}{1} = 1$.

The equation of the straight line passing through the point $(4, 3)$ and making intercepts on the coordinate axes whose sum is $-1$ is

Similar Questions

If the straight line $2x - 3y + 17 = 0$ is perpendicular to the line passing through the points $(7, 17)$ and $(15, \beta)$,then $\beta$ equals

The equation of the line passing through the point $(-3, 2)$ and parallel to the $x$-axis is:

$A$ line $L$ passes through the point $P(1, 2)$ and makes an angle of $60^{\circ}$ with the positive $X$-axis. $A$ and $B$ are two points lying on $L$ at a distance of $4$ units from $P$. If $O$ is the origin,then the area of $\triangle OAB$ is

The equation of a line passing through $(3, -4)$ and perpendicular to the line $3x + 4y = 5$ is

The equation of the line passing through the point $(-3, 1)$ and bisecting the angle between the coordinate axes in the second quadrant is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module