If the x-intercept of some line L is double t | vedclass

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

Question

(D) The given line is $3x + 4y = 12$. Dividing by $12$,we get $\frac{x}{4} + \frac{y}{3} = 1$. Thus,the $x-$intercept is $4$ and the $y-$intercept is

Vedclass · Accepted Answer

$-3/16$

Vedclass · Answer

(D) The given line is $3x + 4y = 12$. Dividing by $12$,we get $\frac{x}{4} + \frac{y}{3} = 1$. Thus,the $x-$intercept is $4$ and the $y-$intercept is $3$. Let the line $L$ have $x-$intercept $a$ and $y-$intercept $b$. According to the problem,$a = 2 	imes 4 = 8$ and $b = \frac{3}{2}$. The equation of line $L$ in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$,which becomes $\frac{x}{8} + \frac{y}{3/2} = 1$. This simplifies to $\frac{x}{8} + \frac{2y}{3} = 1$. Multiplying by $24$,we get $3x + 16y = 24$. Rewriting in slope-intercept form $y = mx + c$,we have $16y = -3x + 24$,or $y = -\frac{3}{16}x + \frac{24}{16}$. Therefore,the slope of $L$ is $-\frac{3}{16}$.

If the $x-$intercept of some line $L$ is double that of the line $3x + 4y = 12$ and the $y-$intercept of $L$ is half that of the same line,then the slope of $L$ is

Similar Questions

Find the equation of the line passing through the point $(2, -3)$ such that the sum of its intercepts on the axes is $-2$.

Find the equation of the line which is at a perpendicular distance of $5$ units from the origin and the angle made by the perpendicular with the positive $x-$ axis is $30^{\circ}$.

If $k = \frac{a+b}{ab}$ is a non-zero constant,then the point which lies on the straight line $\frac{x}{a} + \frac{y}{b} = 1$ is

If the line $2x + 3y = 5$ and $y = mx + c$ are parallel,then:

If the slope of a line passing through the point $A(3, 2)$ is $3/4$,then the points on the line which are $5$ units away from $A$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module