Let L be the line passing through the point P | vedclass

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

Question

(A) Let the intercepts of line $L$ on the $x$-axis and $y$-axis be $a$ and $b$ respectively. The equation of the line is $\frac{x}{a} + \frac{y}{b} =

Vedclass · Accepted Answer

$\left( \frac{4}{5}, \frac{12}{5} \right)$

Vedclass · Answer

(A) Let the intercepts of line $L$ on the $x$-axis and $y$-axis be $a$ and $b$ respectively. The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.Since the segment is bisected at $P(1, 2)$,we have $\frac{a}{2} = 1 \implies a = 2$ and $\frac{b}{2} = 2 \implies b = 4$.Thus,the equation of line $L$ is $\frac{x}{2} + \frac{y}{4} = 1$,which simplifies to $2x + y = 4 \quad (1)$.The slope of line $L$ is $m = -2$. The slope of the line $L_1$ perpendicular to $L$ is $m_1 = -\frac{1}{m} = \frac{1}{2}$.Since $L_1$ passes through $(-2, 1)$,its equation is $y - 1 = \frac{1}{2}(x + 2) \implies 2y - 2 = x + 2 \implies x - 2y = -4 \quad (2)$.Solving equations $(1)$ and $(2)$:From $(1)$,$y = 4 - 2x$. Substituting into $(2)$:$x - 2(4 - 2x) = -4$$x - 8 + 4x = -4$$5x = 4 \implies x = \frac{4}{5}$.Then $y = 4 - 2(\frac{4}{5}) = 4 - \frac{8}{5} = \frac{12}{5}$.The point of intersection is $\left( \frac{4}{5}, \frac{12}{5} \right)$.

Let $L$ be the line passing through the point $P(1, 2)$ such that its intercepted segment between the coordinate axes is bisected at $P$. If $L_1$ is the line perpendicular to $L$ and passing through the point $(-2, 1)$,then the point of intersection of $L$ and $L_1$ is

Similar Questions

If straight lines $ax + by + p = 0$ and $x \cos \alpha + y \sin \alpha - p = 0$ include an angle $\pi / 4$ between them and meet the straight line $x \sin \alpha - y \cos \alpha = 0$ in the same point,then the value of $a^2 + b^2$ is equal to

The origin and the points where the line $L_1$ intersects the $x$-axis and $y$-axis are vertices of a right-angled triangle $T$ whose area is $8$. Also,the line $L_1$ is perpendicular to the line $L_2: 4x - y = 3$. Then,the perimeter of triangle $T$ is:

$A$ is a point on either of two lines $y + \sqrt{3} |x| = 2$ at a distance of $\frac{4}{\sqrt{3}}$ units from their point of intersection. The coordinates of the foot of the perpendicular from $A$ on the bisector of the angle between them are

One vertex of an equilateral triangle is $(2, 3)$ and the line of its opposite side is $x + y = 2$. Find the equations of the other two sides.

The number of integral points (integral point means both the coordinates should be integers) exactly in the interior of the triangle with vertices $(0, 0)$,$(0, 21)$,and $(21, 0)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module