Question

$\begin{array}{cc}Column1& Column2\\ \text{a. Two vertices of a triangle are}(5,-1)\text{and}(-2,3).& \text{p.}(-4,-7)\\ \text{If orthocenter}\text{is}\text{the origin then coordinates of}& \\ \text{the third vertex is},& \\ \text{b. A point on the line}x+y=4\text{which lies at a unit}& \text{q.}(-7,-11)\\ \text{distance from the line}4x+3y=10\text{is}& \\ \text{c. Orthocentre of the triangle formed by the lines}& \text{r.}(2,-2)\\ x+y-1=0,x-y+3=0,2x+y=7\text{is}& \\ \text{d. If}2a,b,c\text{are in A.P.,then lines}ax+by+c=0\text{are}& \text{s.}(-1,2)\\ \text{concurrent at}& \end{array}$
Then which of the following is correct ?

• A. $a\to s,b\to r,c\to p,d\to q$
• B. $a\to p,b\to q,c\to s,d\to r$
• C. $a\to q,b\to p,c\to r,d\to s$
• D. $a\to r,b\to s,c\to p,d\to q$

Answer 1

The correct option is B $a\to p,b\to q,c\to s,d\to r$

$a)$

$AH\perp BC$
or $\left(\frac{k}{h}\right)\left(\frac{3+1}{-2-5}\right)=-1$
$\Rightarrow 4k=7h...\left(1\right)$
$BH\perp AC$
or $\left(\frac{0+1}{0-5}\right)\left(\frac{k-3}{h+2}\right)=-1$
$\Rightarrow k-3=5(h+2)...\left(2\right)$
$\Rightarrow 7h-12=20h+40$
$\Rightarrow 13h=-52$
$\Rightarrow h=-4,k=-7$
Hence, point $A$ is $(-4,-7)$

$b)$
$x+y-4=0...\left(1\right)$
​​​​​​​$4x+3y-10=0...\left(2\right)$
Let $(h,4-h)$ be the point on $\left(1\right)$. Then,
$\left|\frac{4h+3(4-h)-10}{5}\right|=1$
$\Rightarrow h+2=\pm 5$
$\Rightarrow h=3,h=-7$
Hence the required point is either $(3,1)$ or $(-7,11)$

$\left(c\right)$
Since lines $x+y-1=0$ and $x-y+3=0$ are perpendicular, the orthocentre of the triangle is the point of intersection of these lines, i.e., $(-1,2)$

$\left(d\right)$
Since $2a,b,c$ are in A.P., we have
$b=\frac{2a+c}{2}$
$\Rightarrow 2a-2b+c=0$
Comparing with the line $ax+by+c=0$, we have $x=2$ and $y=-2$. Hence, the lines are concurrent at $(2,-2)$