The points on the line x + y = 4 which lie at | vedclass

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(A) Let the point be $(h, k)$. Since it lies on the line $x + y = 4$,we have $h + k = 4$ or $k = 4 - h$ $(i)$.The perpendicular distance from $(h, k)$

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$(3, 1), (-7, 11)$

Vedclass · Answer

(A) Let the point be $(h, k)$. Since it lies on the line $x + y = 4$,we have $h + k = 4$ or $k = 4 - h$ $(i)$.The perpendicular distance from $(h, k)$ to the line $4x + 3y - 10 = 0$ is given by $d = \frac{|4h + 3k - 10|}{\sqrt{4^2 + 3^2}} = 1$.Thus,$|4h + 3k - 10| = 5$,which implies $4h + 3k - 10 = 5$ or $4h + 3k - 10 = -5$.Case $1$: $4h + 3k = 15$. Substituting $k = 4 - h$,we get $4h + 3(4 - h) = 15$ $\Rightarrow 4h + 12 - 3h = 15$ $\Rightarrow h = 3$. Then $k = 4 - 3 = 1$. So,the point is $(3, 1)$.Case $2$: $4h + 3k = 5$. Substituting $k = 4 - h$,we get $4h + 3(4 - h) = 5$ $\Rightarrow 4h + 12 - 3h = 5$ $\Rightarrow h = -7$. Then $k = 4 - (-7) = 11$. So,the point is $(-7, 11)$.Therefore,the required points are $(3, 1)$ and $(-7, 11)$.

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

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