Distance

Geometry Level 3

Given the line $k: 3x+4y=12$ . The point $P$ has coordinates $(x_p,0)$ and lies at a distance of 3 to the line $k$ .

Calculate the coordinates $x_p = a$ and $x_p = b$ of $P$ .

Give your answer as $a + b$ .

1 solution

Peter van der Linden
Mar 14, 2017

Since $d(P,k) = 3$ we have:

$\frac{|3p + 4\cdot 0 -12|}{\sqrt{3^2 + 4^2}} = 3$ .

$\frac{|3p -12|}{\sqrt{25}} = 3$

$|3p - 12| =15$

$3p - 12 = 15 \vee 3p -12 = -15$

$3p = 27 \vee 3p = -3$

$p = 9 \vee p = -1$

So $a+b=8$

I used a short-cut. The two points will lie an equal distance along the $x$ -axis from $(4,0)$ , so $4$ will be the average of $a$ and $b$ , i.e.,

$\dfrac{a + b}{2} = 4 \Longrightarrow a + b = 8$ .

Brian Charlesworth - 4 years, 2 months ago