91Ó°ÊÓ

We are given an equation of the line $4x+3y=1$and the point on the perpendicular line $(0,0)$.

We will use the slope-point formula to find the equation of the perpendicular line.

• The slopes of the perpendicular lines are negative reciprocals of each other. If $m$is the slope of a given line, then the slope of the line perpendicular to it is$-\frac{1}{m}$.
• The slope-point form of the equation of a line is $y-y_1=m(x-x_1)$where $m$is the slope of the line.

The equation of the given line is $4x+3y=1$

Rewriting this equation into slope-intercept form, we have

$$\begin{gathered} 4x+3y=1 \\ 3y=-4x+1 \\ y=-\frac{4}{3}x+\frac{1}{3} \end{gathered}$$

Comparing this with the slope-intercept form$y=mx+b$, the slope of the given line is$m=-\frac{4}{3}$

The slope of the given line $4x+3y=1$is $m=-\frac{4}{3}$

Since the slope of the perpendicular lines are the negative reciprocals of each other, the slope of the line perpendicular to the given line is$m_{âŠ\yen }=\frac{3}{4}$

The slope of the perpendicular line is $m_{âŠ\yen }=\frac{3}{4}$and the point is$(0,0)$.

Using the slope-point formula, we have

role="math" localid="1644761136824" $$\begin{gathered} y-y_1=m_{âŠ\yen }(x-x_1) \\ y-0=\frac{3}{4}(x-0) \\ y=\frac{3}{4}x \end{gathered}$$

This is the slope-intercept form.

The equation of the line perpendicular to the line $4x+3y=1$, that contains the point $(0,0)$in the slope-intercept form is$y=\frac{3}{4}x$