Question

Two straight lines passing through the point A(3, 2) cut the line $2y=x+3$ and x-axis perpendicularly at P and Q respectively. The equation of the line PQ is

• A. $7x+y-21=0$
• B. $x+7y+21=0$
• C. $2x+y-8=0$
• D. $x+2y+8=0$

Answer 1

Answer: (A)

$7x+y-21=0$

Given, $2y=x+\mathrm{3...........}\left(i\right)$

A line passes through $A(3,2)$ and slope $-2.$ [Since, line is perpendicular to the line having slope $\frac{1}{2}$ ]

$\therefore y-2=-2(x-3)$

or, $2x+y-8=\mathrm{0............}\left(ii\right)$

On solving equations (i) and(ii), we get

$P=(\frac{13}{5},\frac{14}{5})$

Now, a line passes through $A(3,2)$ and having slope $\frac{1}{0}$ [Since, line is perpendicular to $X-$axis]

$\therefore y-2=\frac{1}{0}(x-3)$

or, $x=3$

Coordinate of $Q=(3,0)$

Now, equation of line $PQ=y-0=\frac{0-\frac{14}{5}}{3-\frac{13}{5}}(x-3)$

or, $7x+y-21=0$

Hence, A is the correct option.