The line passing through the points (a, b) an | vedclass

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

Question

(C) Let the given points be $A(a, b)$ and $B(-a, -b)$. The slope $m$ of the line passing through $A$ and $B$ is given by: $m = \frac{y_2 - y_1}{x_2 -

Vedclass · Accepted Answer

$(a^2, ab)$

Vedclass · Answer

(C) Let the given points be $A(a, b)$ and $B(-a, -b)$. The slope $m$ of the line passing through $A$ and $B$ is given by: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-b - b}{-a - a} = \frac{-2b}{-2a} = \frac{b}{a}$. The equation of the line passing through $(a, b)$ with slope $\frac{b}{a}$ is: $y - b = \frac{b}{a}(x - a)$ $ay - ab = bx - ab$ $bx = ay$ Now,we check the given options to see which point satisfies the equation $bx = ay$: For option $(C)$,substituting $x = a^2$ and $y = ab$: $b(a^2) = a(ab)$ $a^2b = a^2b$ Since the equation is satisfied,the line passes through $(a^2, ab)$.

The line passing through the points $(a, b)$ and $(-a, -b)$ passes through which of the following points?

Similar Questions

$PS$ is the median of the triangle with vertices at $P(2, 2)$,$Q(6, -1)$,and $R(7, 3)$. The intercepts on the coordinate axes of the line passing through point $(1, -1)$ and parallel to $PS$ are respectively:

$A$ straight line $L$ at a distance of $4$ units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of $60^o$ with the line $x + y = 0$. Then an equation of the line $L$ is

Find the equation of the line passing through the point $(2, 3)$ and perpendicular to the line joining the points $(-5, 6)$ and $(-6, 5)$.

The equation of the line which cuts off the intercepts $2a \sec \theta$ and $2a \csc \theta$ on the axes is

What is the equation of the line passing through $(-1, 1)$ and perpendicular to the line $2x + 3y + 4 = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module