# Match the question given under Column $C_1$ with their appropriate answer given under the Column $C_2$ for the following :

 Column $c_1$ Column $c_2$ (a) The coordinates of the points P and Q on the line x+5y=13 which are at a distance of 2 units from the line 12x-5y+26=0 are (i) (3, 1) (-7, 11) (b)  The coordinates of the point on the line x + y = 4, which are at a unit distance from the line 4x + 3y – 10 = 0 are (ii) $\bigg( - \large\frac{1}{3}, \large\frac{11}{3} \bigg), \bigg( \large\frac{4}{3}, \large\frac{7}{3} \bigg)$ (c) The coordinates of the point on the line joining A (–2, 5) and B (3, 1) such that AP = PQ = QB are (iii) $\bigg( 1, \large\frac{12}{5} \bigg) \bigg( -3, \large\frac{16}{5} \bigg)$

Toolbox:
• Section formula : $\large\frac{mx_2+mx_1}{m+n}$$, \large\frac{my_2+ny_1}{m+n} • Mid point formula : \large\frac{x_1+x_2}{2}$$, \large\frac{y_1+y_2}{2}$
Match the following
Column $C_1$ - a matches column $C_2$ - (iii).
Substitute the values for x and y in the equation $x+5y=13$
$1+\bigg( \large\frac{12}{5} \bigg)$$(5) = 13 True -3 + 5 \bigg( \large\frac{16}{5} \bigg)$$=13$ True
Column $C_1$ - b matches $C_2$ - (i)
Substitute the values in $x+y=4$
$3+1=4$ True
also $-7+11=4$ True
Column $C_1$ - c matches Column $C_2$ (ii)
It is given AP = PQ = QB
$\large\frac{AP}{PB}$$= \large\frac{1}{2} By section formula we know, x = \large\frac{mx_2+nx_1}{m+n}$$, \large\frac{my_2+ny_1}{m+n}$
$(x_1, y_1)$ be (-2, 5) and $(x_2, y_2)$ (3, 1)
$\therefore$ coordinates of P are
$\large\frac{1 \times 3 + 2(-2)}{2+1}$ and $\large\frac{1 \times 1 +2 \times 5 }{2+1}$
(i.e) $\bigg( - \large\frac{1}{3}$$, \large\frac{11}{3} \bigg) Since Q is the mid point of P and B the coordinates of Q are \large\frac{x_1+x_2}{2}$$, \large\frac{y_1+y_2}{2}$
(i.e) $\large\frac{-\Large\frac{1}{3}+3}{2}$$, \large\frac{\Large\frac{11}{3}+1}{2} (i.e) \bigg( \large\frac{4}{3}$$, \large\frac{7}{3} \bigg)$
edited Jul 9, 2014