 # Find the points on the line x + y = 4 `
Question:

Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10.

Solution:

Let (x1, y1) be any point lying in the equation x+ y = 4

∴ x1 + y1 = 4 …..1

Distance of the point (x1, y1) from the equation 4x + 3y = 10

Let $\left(x_{1}, y_{1}\right)$ be any point lying in the equation $x+y=4$

$\therefore \mathrm{x}_{1}+\mathrm{y}_{1}=4 \ldots .1$

Distance of the point $\left(x_{1}, y_{1}\right)$ from the equation $4 x+3 y=10$

$d=\frac{\left|A x_{0}+B y_{0}+C\right|}{\sqrt{A^{2}+B^{2}}}$

$\Rightarrow 1=\left|\frac{4 x_{1}+3 y_{1}-10}{\sqrt{(4)^{2}+(3)^{2}}}\right|$

On simplification we get

$\Rightarrow 1=\left|\frac{4 \mathrm{x}_{1}+3 \mathrm{y}_{1}-10}{\sqrt{16+9}}\right|$

$\Rightarrow 1=\left|\frac{4 x_{1}+3 y_{1}-10}{5}\right|$

$\Rightarrow 4 x_{1}+3 y_{1}-10=\pm 5$\

$4 x_{1}+3 y_{1}-10=5$ or $4 x_{1}+3 y_{1}-10=-5$

$4 x_{1}+3 y_{1}=5+10$ or $4 x_{1}+3 y_{1}=-5+10$

$4 x_{1}+3 y_{1}=15 \ldots . .2$

$\operatorname{Or} 4 x_{1}+3 y_{1}=5 \ldots . . .3$

From equation 1 , we have $y_{1}=4-x_{1} \ldots \ldots . .4$

Putting the value of $y_{1}$ in equation 2, we get

$4 x_{1}+3\left(4-x_{1}\right)=15$

$\Rightarrow 4 x_{1}+12-3 x_{1}=15$

$\Rightarrow x_{1}=15-12$

$\Rightarrow \mathrm{x}_{1}=3$

Putting the value of $x_{1}$ in equation 4 , we get

$y_{1}=4-3$

$\Rightarrow y_{1}=1$

Putting the value of y1 = 4 – x1 in equation 3, we get

4x1 + 3(4 – x1) = 5

⇒ 4x1 + 12 – 3x1 = 5

⇒ x1 = 5 – 12

⇒ x1 = – 7

Putting the value of x1 in equation 4, we get

y1 = 4 – (-7)

⇒ y1 = 4 + 7

⇒ y1 = 11

Hence, the required points on the given line are (3, 1) and (-7, 11)