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

$\begin {array} {1 1} (A)\;(-7, 11) or (-3, 1) & \quad (B)\; (7,11) or (3, 1) \\ (C)\;(-7, 11) or (3, 1) & \quad (D)\;(-7, 11) or (3, 1) \end {array}$

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  • The distance from a point $(x_1, y_1)$ to a line $ax+by+c$ is $ \bigg| \large\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \bigg|$
Step 1
The coordinates of the arbitrary point on $x+y=4$ can be obtained by putting $x=t$ (or $y=t)$ and then obtaining $y$(or $ x)$ from the equation of the line, where $t$ is a parameter.
Substituting $x=t$ in $x+y=4$ we get
$ \therefore $ The coordinates of an arbitrary point on the given line are $P(t, 4-t).$
Let $P(t, 4-t)$ be the required point.
Step 2
It is given that the distance of $P$ from the line $ 4x+3y-10=0$ is unity.
$ \therefore \bigg| \large\frac{4t+3(4-t)-10}{\sqrt{4^2+3^2}} \bigg|$$=1$
(i.e) $|t+2|=5$
(i.e) $t+2= \pm 5$
$ \therefore t = -7\: or \: t = 3$
Hence the required points are (-7,11) or ( 3,1)
answered Jun 30, 2014 by thanvigandhi_1

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