A line meets the x-axis and y-axis at the poi | vedclass

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

Question

(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.Since $(x_1, y_1)$ is the midpoint of $AB$,we have $x_1 = \frac{a+0}{2} = \frac{a}{2}$

Vedclass · Accepted Answer

$y_1x + x_1y = 2x_1y_1$

Vedclass · Answer

(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.Since $(x_1, y_1)$ is the midpoint of $AB$,we have $x_1 = \frac{a+0}{2} = \frac{a}{2}$ and $y_1 = \frac{0+b}{2} = \frac{b}{2}$.This implies $a = 2x_1$ and $b = 2y_1$.The intercept form of the equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.Substituting the values of $a$ and $b$,we get $\frac{x}{2x_1} + \frac{y}{2y_1} = 1$.Multiplying both sides by $2x_1y_1$,we obtain $x y_1 + y x_1 = 2x_1y_1$.

$A$ line meets the $x$-axis and $y$-axis at the points $A$ and $B$ respectively. If the midpoint of $AB$ is $(x_1, y_1)$,then the equation of the line is:

Similar Questions

The equation of the line passing through the point $(2,3)$ such that its $x$-intercept is twice its $y$-intercept is ......... .

The equation of the line,where the length of the perpendicular segment from the origin to the line is $4$ and the inclination of this perpendicular segment with the positive direction of the $X$-axis is $30^{\circ}$,is:

The equation of a straight line passing through the points $(-5, -6)$ and $(3, 10)$ is:

Find the equation of a line which is bisected at the point $(x_1, y_1)$ by the axes.

Find the equation of a line perpendicular to the line $x-2y+3=0$ and passing through the point $(1, -2).$

Vedclass Test Series

Exam Paper Generator

Online Exam Module