List-I		List-II
P	Two vertices of a triangle are (5, –1) and (–2, 3). If orthocentre is the origin, then coordinates of the third vertex are	1	(–4, –7)
Q	A point onthe line x + y = 4 which lies at a unit distance from the line 4x+ 3y = 10, is	2	(–7, 11)
R	Orthocentre of the triangle made by the lines x + y – 1 = 0, x – y + 3 = 0, 2x + y = 7 is	3	(1, –2)
S	If a, b, c are in A.P., then lines ax + by = c are concurrent at	4	(–1, 2)
		5	(4, –7)

$AH\perp BC\Rightarrow \left(\frac{k}{h}\right)\left(\frac{3+1}{-2-5}\right)=-1$

$4k=7h.......\left(\mathrm{i}\right)$

$BH\perp AC\Rightarrow \left(\frac{0+1}{0-5}\right)\left(\frac{k-3}{h+2}\right)=-1$

$\mathrm{K}-3=5(\mathrm{h}+2).............\left(\mathrm{ii}\right)$

$\Rightarrow 7h-12=20h+40$

$13h=-52$

$h=-4$

$\mathrm{k}=-7$

$a(-4,-7)$

$x+y-4=0......\left(\mathrm{i}\right)$

$3=4x+3y-10=0.........\left(\mathrm{ii}\right)$

Let (h, 4 – h) be the point on (i),

$\text{ then }\left|\frac{4h+3(4-h)-10}{5}\right|=1$

i.e., $h+2=\pm 5$

required point is either $(3,1)$ or $(–7, 11)$

Orthocentre of the triangle is the point of intersection

of the lines $x + y – 1 = 0$

and $x – y + 3 = 0$

i.e., $(–1, 2)$

Since a , b, c are in A.P.

$b=\frac{a+c}{2}$

the family of lines is $ax+\frac{a+c}{2}y=c$

$\text{ i.e., }a\left(x+\frac{y}{2}\right)+c\left(\frac{y}{2}-1\right)=0$

point of concurrency is $(–1, 2)$