If a variable line drawn through the point of | vedclass

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

Question

(b) The equation of a line passing through the intersection of straight lines $\frac{x}{\alpha } + \frac{y}{\beta } = 1$ and $\frac{x}{\beta } + \frac

Vedclass · Accepted Answer

$\alpha \beta (x + y) = 2xy(\alpha + \beta )$

Vedclass · Answer

(b) The equation of a line passing through the intersection of straight lines $\frac{x}{\alpha } + \frac{y}{\beta } = 1$ and $\frac{x}{\beta } + \frac{y}{\alpha } = 1$ is
$\left( {\frac{x}{\alpha } + \frac{y}{\beta } - 1} \right) + \lambda \left( {\frac{x}{\beta } + \frac{y}{\alpha } - 1} \right) = 0$

or $x\,\left( {\frac{1}{\alpha } + \frac{\lambda }{\beta }} \right) + y\left( {\frac{1}{\beta } + \frac{\lambda }{\alpha }} \right) - \lambda - 1 = 0$

This meets the axes at

$A{\rm{ }}\left( {\frac{{\lambda + 1}}{{\frac{1}{\alpha } + \frac{\lambda }{\beta }}},0} \right)$ and $B{\rm{ }}\left( {0,\frac{{\lambda + 1}}{{\frac{1}{\beta } + \frac{\lambda }{\alpha }}}} \right)$.

Let $(h, k)$ be the mid point of $AB$,

then $h = \frac{1}{2}.\frac{{\lambda + 1}}{{\frac{1}{\alpha } + \frac{\lambda }{\beta }}},k = \frac{1}{2}.\frac{{\lambda + 1}}{{\frac{1}{\beta } + \frac{\lambda }{\alpha }}}$

Eliminating $\lambda $from these two, we get

$2hk(\alpha + \beta ) = \alpha \beta (h + k)$.

 The locus of $(h,k)$is $2xy(\alpha + \beta ) = \alpha \beta (x + y)$.

If a variable line drawn through the point of intersection of straight lines $\frac{x}{\alpha } + \frac{y}{\beta } = 1$and $\frac{x}{\beta } + \frac{y}{\alpha } = 1$ meets the coordinate axes in $A$ and $B$, then the locus of the mid point of $AB$ is

Similar Questions

The co-ordinates of three points $A(-4, 0) ; B(2, 1)$ and $C(3, 1)$ determine the vertices of an equilateral trapezium $ABCD$ . The co-ordinates of the vertex $D$ are :

The equations of two equal sides of an isosceles triangle are $7x - y + 3 = 0$ and $x + y - 3 = 0$ and the third side passes through the point $(1, -10)$. The equation of the third side is

In a triangle $ABC,$ side $AB$ has the equation $2 x + 3 y = 29$ and the side $AC$ has the equation , $x + 2 y = 16$ . If the mid - point of $BC$ is $(5, 6)$ then the equation of $BC$ is :

The equation of straight line passing through $( - a,\;0)$ and making the triangle with axes of area ‘$T$’ is

Two vertices of a triangle are $(5, - 1)$ and $( - 2,3)$. If orthocentre is the origin then coordinates of the third vertex are