Question

Find the coordinates of a point on the line
$x+y=5$, whose distance from the line $6x+8y+1=0$ is $3.5$ units

Answer 1

Let the coordinates of a point $P$ on the line $x+y=5$ be $P(h,k)$.

Given, that $P$ is at a distance of $3.5$ units from $6x+8y+1=0$ distance of a point $\left(x_1{y}_1\right)$ from a line $ax+by+c=0$ is given by

$d=\left|\frac{{ax}_1+{by}_1+c}{\sqrt{a^2+b^2}}\right|⟶\left(1\right)$

Using equation $\left(1\right)$ we can write

$3.5=\frac{6h+8k+1}{\sqrt{6^2+8^2}}=\frac{6h+8k+1}{\sqrt{100}}=\frac{6h+8k+1}{10}$

$\therefore$ $6h+8k+1=35$

$6h+8k=34⟶\left(2\right)$

Now $P(h,k)$ lies on $x+y=5$ line, hence

$h+k=5⟶\left(3\right)$

Solving $\left(2\right)$ & $\left(3\right)$ simultaneously

Eqn $\left(2\right)\times 1\Rightarrow 6h+8k=34$

$\underset{–}{Eqn\left(3\right)\times 8\Rightarrow 8h+8k=40}$

(Subtracting) $-2h=-6$

$h=3$

and $k=2$

$\because$ the corodinates of $P$ are $(3,2)$