Let two points be A(1, -1) and B(0, 2). If a | vedclass

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

Question

(C) The equation of line $AB$ passing through $(1, -1)$ and $(0, 2)$ is given by $y - 2 = \frac{-1 - 2}{1 - 0}(x - 0)$ $\Rightarrow y - 2 = -3x$ $\Rig

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The equation of line $AB$ passing through $(1, -1)$ and $(0, 2)$ is given by $y - 2 = \frac{-1 - 2}{1 - 0}(x - 0)$ $\Rightarrow y - 2 = -3x$ $\Rightarrow 3x + y - 2 = 0$.The length of the base $AB = \sqrt{(1 - 0)^2 + (-1 - 2)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$.The area of $\Delta PAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = 5$.$\frac{1}{2} 	imes \sqrt{10} 	imes h = 5 \Rightarrow h = \frac{10}{\sqrt{10}} = \sqrt{10}$.The height $h$ is the perpendicular distance from point $P$ to the line $AB$. Since $P$ lies on $3x + y - 4\lambda = 0$,the distance from $P$ to $3x + y - 2 = 0$ is $\frac{|3x' + y' - 2|}{\sqrt{3^2 + 1^2}} = \sqrt{10}$.Since $3x' + y' = 4\lambda$,we have $\frac{|4\lambda - 2|}{\sqrt{10}} = \sqrt{10} \Rightarrow |4\lambda - 2| = 10$.Case $1$: $4\lambda - 2 = 10$ $\Rightarrow 4\lambda = 12$ $\Rightarrow \lambda = 3$.Case $2$: $4\lambda - 2 = -10$ $\Rightarrow 4\lambda = -8$ $\Rightarrow \lambda = -2$.Thus,the possible values for $\lambda$ are $3$ or $-2$. Comparing with the options,the correct value is $3$.

Let two points be $A(1, -1)$ and $B(0, 2)$. If a point $P(x', y')$ is such that the area of $\Delta PAB = 5 \; \text{sq units}$ and it lies on the line $3x + y - 4\lambda = 0$,then a value of $\lambda$ is

Similar Questions

The system of equations $x+2y=3$ and $3x+6y=a-2$ has no solution if:

What is the perpendicular distance of the line $3x + 4y = 10$ from the origin?

The distance of the point $(1, 2)$ from the line $x + y + 5 = 0$ measured along the line parallel to $3x - y = 7$ is equal to

The distance of point $(-2, 3)$ from the line $x - y - 5 = 0$ is

Reduce the equation $x-y=4$ into the normal form $x \cos \omega + y \sin \omega = p$. Find the perpendicular distance from the origin $(p)$ and the angle between the perpendicular and the positive $x$-axis $(\omega)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module