Among the statements:(S1) : If A(5, -1) and B | vedclass

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

Question

(D) Solution of statement - $1$:Let the third vertex be $C(h, k)$.The orthocentre $O$ is $(0, 0)$.Since $AO \perp BC$, the slope of $AO$ is $m_{AO} =

Vedclass · Accepted Answer

Both are correct

Vedclass · Answer

(D) Solution of statement - $1$:Let the third vertex be $C(h, k)$.The orthocentre $O$ is $(0, 0)$.Since $AO \perp BC$, the slope of $AO$ is $m_{AO} = \frac{0 - (-1)}{0 - 5} = -\frac{1}{5}$.The slope of $BC$ is $m_{BC} = \frac{k - 3}{h - (-2)} = \frac{k - 3}{h + 2}$.Since $m_{AO} \cdot m_{BC} = -1$, we have $(-\frac{1}{5}) \cdot (\frac{k - 3}{h + 2}) = -1$ $\Rightarrow k - 3 = 5(h + 2)$ $\Rightarrow 5h - k + 13 = 0$ ....$(1)$Since $BO \perp AC$, the slope of $BO$ is $m_{BO} = \frac{0 - 3}{0 - (-2)} = -\frac{3}{2}$.The slope of $AC$ is $m_{AC} = \frac{k - (-1)}{h - 5} = \frac{k + 1}{h - 5}$.Since $m_{BO} \cdot m_{AC} = -1$, we have $(-\frac{3}{2}) \cdot (\frac{k + 1}{h - 5}) = -1$ $\Rightarrow 3(k + 1) = 2(h - 5)$ $\Rightarrow 3k + 3 = 2h - 10$ $\Rightarrow 2h - 3k = 13$ ....$(2)$Solving equations $(1)$ and $(2)$, we get $h = -4$ and $k = -7$.So, the third vertex is $(-4, -7)$. Thus, statement $1$ is correct.Solution of statement - $2$:Given $2a, b, c$ are in $A.P.$, so $2b = 2a + c \Rightarrow 2a - 2b + c = 0$.The line equation is $ax + by + c = 0$.Comparing this with $2a - 2b + c = 0$, we can see that for $x = 2$ and $y = -2$, the equation $a(2) + b(-2) + c = 0$ holds true.Thus, the lines $ax + by + c = 0$ are concurrent at $(2, -2)$. Thus, statement $2$ is correct.

Similar Questions

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 through the origin $O(0, 0)$ meets the lines $4x + 3y - 10 = 0$ and $8x + 6y + 5 = 0$ at points $A$ and $B$ respectively. Then $O$ divides the segment $AB$ in the ratio:

If $L_1$ is a line passing through the point $P(4, -3)$ and perpendicular to the line $3x - 4y + k = 0$,then the distance of $P$ from the line $5x - 3y - 2 = 0$ measured along the line $L_1$ is

$A$ straight line $L$ passes through the point of intersection of the lines $x - y + 1 = 0$ and $2x + y - 7 = 0$. If $L$ intersects the positive $x$-axis at $A(a, 0)$ and the positive $y$-axis at $B(0, b)$, then the minimum area of the triangle $OAB$ (where $O$ is the origin) 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