If 2x + 3y = 5 is the perpendicular bisector | vedclass

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(A) Let $l_1 \equiv 2x + 3y = 5$.Since the line $AB$ is perpendicular to $l_1$,the slope of $l_1$ is $m_1 = -\frac{2}{3}$.Therefore,the slope of $AB$

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$\left(\frac{21}{13}, \frac{49}{39}\right)$

Vedclass · Answer

(A) Let $l_1 \equiv 2x + 3y = 5$.Since the line $AB$ is perpendicular to $l_1$,the slope of $l_1$ is $m_1 = -\frac{2}{3}$.Therefore,the slope of $AB$ is $m_{AB} = -\frac{1}{m_1} = \frac{3}{2}$.The equation of line $AB$ passing through $A\left(1, \frac{1}{3}\right)$ with slope $\frac{3}{2}$ is:$\left(y - \frac{1}{3}\right) = \frac{3}{2}(x - 1)$$\Rightarrow 2y - \frac{2}{3} = 3x - 3$$\Rightarrow 3x - 2y = 3 - \frac{2}{3} = \frac{7}{3}$$\Rightarrow 9x - 6y = 7$ $(i)$The equation of line $l_1$ is $2x + 3y = 5$. Multiplying by $2$,we get $4x + 6y = 10$ (ii).Adding $(i)$ and (ii): $13x = 17 \Rightarrow x = \frac{17}{13}$.Substituting $x$ in $2x + 3y = 5$: $2\left(\frac{17}{13}\right) + 3y = 5$ $\Rightarrow 3y = 5 - \frac{34}{13} = \frac{31}{13}$ $\Rightarrow y = \frac{31}{39}$.The intersection point $P$ (mid-point of $AB$) is $\left(\frac{17}{13}, \frac{31}{39}\right)$.Let $B = (x_2, y_2)$. Since $P$ is the mid-point of $AB$:$\frac{1 + x_2}{2} = \frac{17}{13}$ $\Rightarrow 1 + x_2 = \frac{34}{13}$ $\Rightarrow x_2 = \frac{21}{13}$$\frac{1/3 + y_2}{2} = \frac{31}{39}$ $\Rightarrow \frac{1}{3} + y_2 = \frac{62}{39}$ $\Rightarrow y_2 = \frac{62}{39} - \frac{13}{39} = \frac{49}{39}$Thus,$B = \left(\frac{21}{13}, \frac{49}{39}\right)$.

If $2x + 3y = 5$ is the perpendicular bisector of the line segment joining the points $A\left(1, \frac{1}{3}\right)$ and $B$,then $B$ is equal to

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