Find the two points on the line x + y = 4 whi | vedclass

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(C) Let the required point be $P(x_1, y_1)$. Since $P$ lies on $x + y = 4$,we have $y_1 = 4 - x_1$. The perpendicular distance from $P(x_1, y_1)$ to t

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

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(C) Let the required point be $P(x_1, y_1)$. Since $P$ lies on $x + y = 4$,we have $y_1 = 4 - x_1$. The perpendicular distance from $P(x_1, y_1)$ to the line $4x + 3y - 10 = 0$ is given as $1$. Using the formula $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$,we get: $\frac{|4x_1 + 3(4 - x_1) - 10|}{\sqrt{4^2 + 3^2}} = 1$ $\frac{|4x_1 + 12 - 3x_1 - 10|}{5} = 1$ $|x_1 + 2| = 5$ This gives two cases: Case $1$: $x_1 + 2 = 5 \implies x_1 = 3$. Then $y_1 = 4 - 3 = 1$. Point is $(3, 1)$. Case $2$: $x_1 + 2 = -5 \implies x_1 = -7$. Then $y_1 = 4 - (-7) = 11$. Point is $(-7, 11)$. Thus,the required points are $(3, 1)$ and $(-7, 11)$.

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

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