A line cuts the X-axis at A(5,0) and the Y-ax | vedclass

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

Question

(A) The equation of line $AB$ is $\frac{x}{5} + \frac{y}{-3} = 1$, which simplifies to $3x - 5y = 15$.Since line $PQ$ is perpendicular to $AB$, its eq

Vedclass · Accepted Answer

$x^{2}+y^{2}-5x+3y=0$

Vedclass · Answer

(A) The equation of line $AB$ is $\frac{x}{5} + \frac{y}{-3} = 1$, which simplifies to $3x - 5y = 15$.Since line $PQ$ is perpendicular to $AB$, its equation is of the form $5x + 3y = \lambda$.The coordinates of $P$ (where $y=0$) are $(\frac{\lambda}{5}, 0)$ and the coordinates of $Q$ (where $x=0$) are $(0, \frac{\lambda}{3})$.The equation of line $AQ$ passing through $A(5,0)$ and $Q(0, \frac{\lambda}{3})$ is $\frac{x}{5} + \frac{y}{\lambda/3} = 1$, which gives $\frac{x}{5} + \frac{3y}{\lambda} = 1$. Thus, $\frac{1}{\lambda} = \frac{1}{3y}(1 - \frac{x}{5})$.The equation of line $BP$ passing through $B(0,-3)$ and $P(\frac{\lambda}{5}, 0)$ is $\frac{x}{\lambda/5} + \frac{y}{-3} = 1$, which gives $\frac{5x}{\lambda} - \frac{y}{3} = 1$. Thus, $\frac{1}{\lambda} = \frac{1}{5x}(\frac{y}{3} + 1)$.Equating the two expressions for $\frac{1}{\lambda}$:$\frac{1}{3y}(1 - \frac{x}{5}) = \frac{1}{5x}(\frac{y}{3} + 1)$$5x(1 - \frac{x}{5}) = 3y(\frac{y}{3} + 1)$$5x - x^{2} = y^{2} + 3y$$x^{2} + y^{2} - 5x + 3y = 0$.

$A$ line cuts the $X$-axis at $A(5,0)$ and the $Y$-axis at $B(0,-3)$. $A$ variable line $PQ$ is drawn perpendicular to $AB$ cutting the $X$-axis at $P$ and the $Y$-axis at $Q$. If $AQ$ and $BP$ meet at $R$, then the locus of $R$ is

Similar Questions

The locus of the midpoint of the portion of the line $x \cos \alpha + y \sin \alpha = p$ intercepted by the coordinate axes,where $p$ is a constant,is

Let $A$ be the point $(0,4)$ and $B$ be a moving point on the $x$-axis. Let $M$ be the midpoint of $AB$ and let the perpendicular bisector of $AB$ meet the $y$-axis at $R$. The locus of the midpoint $P$ of $MR$ is

$A$ straight line moves such that the sum of the reciprocals of its intercepts on two perpendicular lines is constant. Then the line always passes through:

Let $A(2,3), B(3,-6), C(5,-7)$ be three points. If $P$ is a point satisfying the condition $PA^2+PB^2=2PC^2$,then a point that lies on the locus of $P$ is

$A$ variable straight-line $L$ with negative slope passes through the point $(4,9)$ and cuts the positive coordinate axes in $A$ and $B$. If $O$ is the origin, then the minimum value of $OA+OB$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module