Let A(a, b),B(3, 4),and C(-6, -8) respectivel | vedclass

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

Question

(C) We know that the centroid $A$ divides the line segment joining the orthocentre $C$ and circumcentre $B$ in the ratio $2:1$.Given $C(-6, -8)$ and $

Vedclass · Accepted Answer

$\frac{17 \sqrt{5}}{7}$

Vedclass · Answer

(C) We know that the centroid $A$ divides the line segment joining the orthocentre $C$ and circumcentre $B$ in the ratio $2:1$.Given $C(-6, -8)$ and $B(3, 4)$,the centroid $A(a, b)$ is given by the section formula:$a = \frac{2(3) + 1(-6)}{2+1} = \frac{6-6}{3} = 0$$b = \frac{2(4) + 1(-8)}{2+1} = \frac{8-8}{3} = 0$So,$A(0, 0)$.Now,the point $P(2a+3, 7b+5)$ becomes $P(2(0)+3, 7(0)+5) = P(3, 5)$.We need to find the distance of $P(3, 5)$ from the line $L: 2x+3y-4=0$ measured parallel to the line $M: x-2y-1=0$.The slope of line $M$ is $m = \frac{1}{2}$. Thus,$	an 	heta = \frac{1}{2}$,which implies $\sin 	heta = \frac{1}{\sqrt{5}}$ and $\cos 	heta = \frac{2}{\sqrt{5}}$.The coordinates of any point on the line passing through $P(3, 5)$ with slope $m = \frac{1}{2}$ are $(3+r \cos 	heta, 5+r \sin 	heta) = (3+\frac{2r}{\sqrt{5}}, 5+\frac{r}{\sqrt{5}})$.Substituting this into the line $L: 2x+3y-4=0$:$2(3+\frac{2r}{\sqrt{5}}) + 3(5+\frac{r}{\sqrt{5}}) - 4 = 0$$6 + \frac{4r}{\sqrt{5}} + 15 + \frac{3r}{\sqrt{5}} - 4 = 0$$17 + \frac{7r}{\sqrt{5}} = 0$$\frac{7r}{\sqrt{5}} = -17$$r = -\frac{17 \sqrt{5}}{7}$Since distance is the magnitude,the required distance is $|r| = \frac{17 \sqrt{5}}{7}$.

Let $A(a, b)$,$B(3, 4)$,and $C(-6, -8)$ respectively denote the centroid,circumcentre,and orthocentre of a triangle. Then,the distance of the point $P(2a+3, 7b+5)$ from the line $2x+3y-4=0$ measured parallel to the line $x-2y-1=0$ is

Similar Questions

The distance between the lines given by $3x + 4y = 9$ and $6x + 8y = 15$ is (in $units$)

$P$ is a point on $x+y+5=0$,whose perpendicular distance from $2x+3y+3=0$ is $\sqrt{13}$. Then the coordinates of $P$ are:

The base of an equilateral triangle is represented by the equation $2x - y - 1 = 0$ and its vertex is $(1, 2)$. Then,the length (in units) of the side of the triangle is:

The distance between the lines $4x + 3y = 11$ and $8x + 6y = 15$ is:

The length of the perpendicular from the point $(a \cos \alpha, a \sin \alpha)$ to the line $y = x \tan \alpha + c, c > 0$ is .....

Vedclass Test Series

Exam Paper Generator

Online Exam Module