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 intercept form of the line be $\frac{x}{a} + \frac{y}{b} = 1$. The intercepts are $(a, 0)$ and $(0, b)$.The point $(-5, 3)$ divides the li

Vedclass · Accepted Answer

$9x - 20y + 105 = 0$

Vedclass · Answer

(C) Let the intercept form of the line be $\frac{x}{a} + \frac{y}{b} = 1$. The intercepts are $(a, 0)$ and $(0, b)$.The point $(-5, 3)$ divides the line segment joining $(a, 0)$ and $(0, b)$ in the ratio $4:3$.Using the section formula, the coordinates of the point are $\left( \frac{4(0) + 3(a)}{4+3}, \frac{4(b) + 3(0)}{4+3} \right) = \left( \frac{3a}{7}, \frac{4b}{7} \right)$.Equating these to $(-5, 3)$:$\frac{3a}{7} = -5 \implies a = -\frac{35}{3}$$\frac{4b}{7} = 3 \implies b = \frac{21}{4}$Substituting $a$ and $b$ into the intercept form: $\frac{x}{-35/3} + \frac{y}{21/4} = 1$.$-\frac{3x}{35} + \frac{4y}{21} = 1$.Multiplying by $105$ (the $LCM$ of $35$ and $21$): $-9x + 20y = 105$.$9x - 20y + 105 = 0$.

The equation of the straight line passing through the point $(-5, 3)$ such that the portion of the line intercepted between the axes is divided by the point in the ratio $4:3$ (from the $X$-axis to the $Y$-axis) is...

Similar Questions

Find the equation of the line passing through the point $(-2, 3)$ with a slope of $-4$.

$A$ line passes through $(2,2)$ and is perpendicular to the line $3x+y=3$. Its $y$-intercept is

The line $(3x - y + 5) + \lambda (2x - 3y - 4) = 0$ will be parallel to the $y$-axis,if $\lambda$ =

Find the equation of the line,which cuts off intercepts on the axes whose sum and product are $1$ and $-6,$ respectively.

If a straight line passes through the point $(-5, 4)$ and makes an intercept of length $\frac{2}{\sqrt{5}}$ between the lines $x+2y+1=0$ and $x+2y-1=0$,then the equation of that line is

Vedclass Test Series

Exam Paper Generator

Online Exam Module