Find the points on the x-axis,whose distances | vedclass

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Question

(A) The given equation of the line is $\frac{x}{3}+\frac{y}{4}=1$. Multiplying by $12$,we get $4x + 3y - 12 = 0$. Let the point on the $x$-axis be $(a

Vedclass · Accepted Answer

$(-2, 0)$ and $(8, 0)$

Vedclass · Answer

(A) The given equation of the line is $\frac{x}{3}+\frac{y}{4}=1$. Multiplying by $12$,we get $4x + 3y - 12 = 0$. Let the point on the $x$-axis be $(a, 0)$. The perpendicular distance $d$ from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. Here,$A = 4, B = 3, C = -12, x_1 = a, y_1 = 0$,and $d = 4$. Substituting these values,we get $4 = \frac{|4a + 3(0) - 12|}{\sqrt{4^2 + 3^2}}$. $4 = \frac{|4a - 12|}{5} \implies |4a - 12| = 20$. This gives two cases: $4a - 12 = 20 \implies 4a = 32 \implies a = 8$. $4a - 12 = -20 \implies 4a = -8 \implies a = -2$. Thus,the required points are $(-2, 0)$ and $(8, 0)$.

Find the points on the $x$-axis,whose distances from the line $\frac{x}{3}+\frac{y}{4}=1$ are $4$ units.

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