A straight line through the point of intersec | vedclass

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

Question

(B) Given lines are $x+2y=4$ $(i)$ and $2x+y=4$ (ii).Solving equations $(i)$ and (ii),we get the point of intersection as $(\frac{4}{3}, \frac{4}{3})$

Vedclass · Accepted Answer

$2(x+y)=3xy$

Vedclass · Answer

(B) Given lines are $x+2y=4$ $(i)$ and $2x+y=4$ (ii).Solving equations $(i)$ and (ii),we get the point of intersection as $(\frac{4}{3}, \frac{4}{3})$.Let the equation of the line passing through this point be $\frac{x}{a} + \frac{y}{b} = 1$.Since it passes through $(\frac{4}{3}, \frac{4}{3})$,we have $\frac{4}{3a} + \frac{4}{3b} = 1$,which simplifies to $\frac{1}{a} + \frac{1}{b} = \frac{3}{4}$ (iii).Let the mid-point of $AB$ be $(h, k)$. Then $h = \frac{a}{2}$ and $k = \frac{b}{2}$,so $a = 2h$ and $b = 2k$.Substituting these into equation (iii),we get $\frac{1}{2h} + \frac{1}{2k} = \frac{3}{4}$.Multiplying by $2hk$,we get $k + h = \frac{3}{2}hk$,or $2(h+k) = 3hk$.Replacing $(h, k)$ with $(x, y)$,the locus is $2(x+y) = 3xy$.

$A$ straight line through the point of intersection of the lines $x+2y=4$ and $2x+y=4$ meets the coordinate axes at $A$ and $B$. The locus of the mid-point of $AB$ is

Similar Questions

The locus of a point which is at a distance of $2$ units from the line $2x - 3y + 4 = 0$ and at a distance of $\sqrt{13}$ units from the point $(5, 0)$ is:

$A$ line is at a constant distance $c$ from the origin and meets the coordinate axes in $A$ and $B$. The locus of the centre of the circle passing through $O, A, B$ is

If the equation of the locus of a point equidistant from $(a_1, b_1)$ and $(a_2, b_2)$ is $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$,find the value of $c$.

$A$ straight line cuts off intercepts on the coordinate axes,the reciprocal of whose sum is $1/p$. Through which fixed point does this line always pass?

Vedclass Test Series

Exam Paper Generator

Online Exam Module