If a straight line passing through the point | vedclass

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

Question

(B) Let the line be $\frac{x}{a} + \frac{y}{b} = 1.$ Since the line passes through $P(-3, 4)$ and this point bisects the intercepted portion between t

Vedclass · Accepted Answer

$4x-3y+24=0$

Vedclass · Answer

(B) Let the line be $\frac{x}{a} + \frac{y}{b} = 1.$ Since the line passes through $P(-3, 4)$ and this point bisects the intercepted portion between the axes,the coordinates of the intercepts are $A(a, 0)$ and $B(0, b).$ By the midpoint formula,$P = \left( \frac{a+0}{2}, \frac{0+b}{2} \right) = \left( \frac{a}{2}, \frac{b}{2} \right).$ Given $P(-3, 4),$ we have $\frac{a}{2} = -3 \implies a = -6$ and $\frac{b}{2} = 4 \implies b = 8.$ Substituting these values into the intercept form equation: $\frac{x}{-6} + \frac{y}{8} = 1.$ Multiplying by $24,$ we get $-4x + 3y = 24,$ or $4x - 3y + 24 = 0.$

If a straight line passing through the point $P(-3, 4)$ is such that its intercepted portion between the coordinate axes is bisected at $P,$ then its equation is

Similar Questions

The equation of the line perpendicular to $x = c$ is

$A$ straight line makes an angle of $135^\circ$ with the $x$-axis and cuts the $y$-axis at a distance of $-5$ from the origin. The equation of the line is

In what ratio does the line $y - x + 2 = 0$ divide the line segment joining the points $(3, -1)$ and $(8, 9)$?

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 $\alpha$ is the angle made by the perpendicular drawn from the origin to the line $12x - 5y + 13 = 0$ with the positive $X$-axis in the anti-clockwise direction,then $\alpha =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module