O(0,0), A(-3,-1) and B(-1,-3) are the vertice | vedclass

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

Question

(B) The vertices are $O(0,0), A(-3,-1)$,and $B(-1,-3)$.Slope of $OB = \frac{-3-0}{-1-0} = 3$.Equation of line $OB$ is $y = 3x$,or $3x - y = 0$.Since $

Vedclass · Accepted Answer

$21x-7y+32=0$

Vedclass · Answer

(B) The vertices are $O(0,0), A(-3,-1)$,and $B(-1,-3)$.Slope of $OB = \frac{-3-0}{-1-0} = 3$.Equation of line $OB$ is $y = 3x$,or $3x - y = 0$.Since $AD \perp OB$,the slope of $AD$ is $-\frac{1}{3}$.Equation of line $AD$ passing through $A(-3,-1)$ is $y - (-1) = -\frac{1}{3}(x - (-3))$,which simplifies to $y + 1 = -\frac{1}{3}(x + 3)$,or $x + 3y + 6 = 0$.To find $D$,the intersection of $OB$ $(y=3x)$ and $AD$ $(x+3y+6=0)$:$x + 3(3x) + 6 = 0$ $\Rightarrow 10x = -6$ $\Rightarrow x = -\frac{3}{5}$.$y = 3(-\frac{3}{5}) = -\frac{9}{5}$. So,$D = (-\frac{3}{5}, -\frac{9}{5})$.Point $P$ divides $AD$ in the ratio $3:4$. Using the section formula:$P = \left( \frac{3(-\frac{3}{5}) + 4(-3)}{3+4}, \frac{3(-\frac{9}{5}) + 4(-1)}{3+4} \right) = \left( \frac{-\frac{9}{5} - 12}{7}, \frac{-\frac{27}{5} - 4}{7} \right) = \left( \frac{-69}{35}, \frac{-47}{35} \right)$.Line $L$ is parallel to $OB$ $(y=3x)$,so its equation is $y = 3x + k$.Since $L$ passes through $P(-\frac{69}{35}, -\frac{47}{35})$:$-\frac{47}{35} = 3(-\frac{69}{35}) + k \Rightarrow k = \frac{-47 + 207}{35} = \frac{160}{35} = \frac{32}{7}$.Equation $L: y = 3x + \frac{32}{7}$ $\Rightarrow 7y = 21x + 32$ $\Rightarrow 21x - 7y + 32 = 0$.

$O(0,0), A(-3,-1)$ and $B(-1,-3)$ are the vertices of a $\triangle OAB$. $P$ is a point on the perpendicular $AD$ drawn from $A$ on $OB$ such that $\frac{AP}{PD}=\frac{3}{4}$. Then the equation of the line $L$ parallel to $OB$ and passing through $P$ is:

Similar Questions

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

$A$ line,with a slope greater than $1$,passes through the point $A(4, 3)$ and intersects the line $x - y - 2 = 0$ at the point $B$. If the length of the line segment $AB$ is $\frac{\sqrt{29}}{3}$,then $B$ also lies on the line:

The line $2x + 3y = 12$ meets the $x$-axis at $A$ and the $y$-axis at $B$. The line through $(5, 5)$ perpendicular to $AB$ meets the $x$-axis,$y$-axis,and the line $AB$ at $C, D, E$ respectively. If $O$ is the origin,then the area of the quadrilateral $OCEB$ is:

$A$ straight line passing through the origin $O$ intersects the lines $10x - 8y - 10 = 0$ and $\frac{x}{4} - \frac{y}{5} + 1 = 0$ at right angles at points $P$ and $Q$ respectively. Then the ratio in which $O$ divides the line segment $PQ$ is

Consider the fourteen lines in the plane given by $y=x+r$ and $y=-x+r$,where $r \in \{0, 1, 2, 3, 4, 5, 6\}$. The number of squares formed by these lines,whose sides are of length $\sqrt{2}$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module