Question

The distance of a point of $x-$axis from straight lie $3x+4y+16=0$ is $5$ units. Find its co-ordinates.

• A. $(3,0)$
• B. $(-41/3,0)$
• C. $A$ and $B$ both
• D. None of these

Answer 1

The correct option is C $A$ and $B$ both

Given straight line $3x+4y+16=0$

Let the point on x-axis be $(k,o)$

Given distance from $(k,0)$ to $3x+4y+16=0$ is 5 units

$\frac{|3x+4\left(0\right)+16|}{\sqrt{3^2+4^2}}=5$

$|3x+16|=25$

$3k+16+\pm 25$

$3k+16=25$ $3k+16=-25$

$3k=9$ $k=-\frac{41}{3}$

$k=3$

$k=3,-\frac{41}{3}$

$(3,0)(-\frac{41}{3},0)$