A line is passing through the point (4,3) and | 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 line passes through $(4,3)$,we have $\frac{4}{a}

Vedclass · Accepted Answer

$3x + 4y = 24$ or $x + y = 7$

Vedclass · Answer

(C) Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$. Given that the line passes through $(4,3)$,we have $\frac{4}{a} + \frac{3}{b} = 1$. Also,the sum of the intercepts is $a + b = 14$,so $b = 14 - a$. Substituting $b$ in the first equation: $\frac{4}{a} + \frac{3}{14 - a} = 1$. $4(14 - a) + 3a = a(14 - a) \implies 56 - 4a + 3a = 14a - a^2$. $a^2 - 15a + 56 = 0$. $(a - 7)(a - 8) = 0$. If $a = 7$,then $b = 14 - 7 = 7$. The equation is $\frac{x}{7} + \frac{y}{7} = 1 \implies x + y = 7$. If $a = 8$,then $b = 14 - 8 = 6$. The equation is $\frac{x}{8} + \frac{y}{6} = 1 \implies 3x + 4y = 24$. Thus,the possible equations are $x + y = 7$ or $3x + 4y = 24$.

$A$ line is passing through the point $(4,3)$ and the sum of its intercepts made on the coordinate axes is $14$. Then an equation of that line 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 $-2$.

The equation of a line is $3x - 4y + 10 = 0$. Find its $x$-intercept and $y$-intercept.

The equation of the line passing through the points $(1, 2)$ and $(2, 5)$ is:

The area of the triangle formed by the line $L$ with the coordinate axes is $12$ sq. units. If $L$ passes through the point $(12, 4)$ and the product $P$ of the $X$-intercept of $L$ and the square of the $Y$-intercept of $L$ is negative,then $P=$

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 $30^{\circ}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module