Question

What is Perpendicular to 5x-3y=30 and passing through (-2,7)

Answer 1

To find a line perpendicular to 5x-3y=30 and passing through (-2,7), we can determine the slope of the given line and then use the negative reciprocal of that slope to write the equation of the perpendicular line.

Step-by-step explanation:

To find a line that is perpendicular to the equation 5x-3y=30 and passes through the point (-2,7), we first need to determine the slope of the given line.

The given equation can be rewritten in slope-intercept form as y = (5/3)x - 10.

The slope of this line is 5/3.

Since the slopes of perpendicular lines are negative reciprocals, the slope of the line perpendicular to 5x-3y=30 is -3/5.

Using the point-slope form of a line, we can write the equation of the line as y - 7 = (-3/5)(x + 2).

Simplifying this equation, we get y = (-3/5)x + 19/5.

Answer 2

`k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\k\ \perp\ l\iff m_1\cdot m_2=-1\\\\k\ ||\ l\iff m_1=m_2`

We have:

`k:5x-3y=30\ \ \ \ |-5x\\\\-3y=-5x+30\ \ \ \ |:(-3)\\\\y=(5)/(3)x-10\to m_1=(5)/(3)\\\\l:y=m_2x+b\\\\k\ \perp l\Rightarrow (5)/(3)m_2=-1\ \ \ \ |\cdot(3)/(5)\\\\m_2=-(3)/(5)`

`l:y=-(3)/(5)x+b`

The line l is passing through (-2, 7). Substitute the coordinates of the point to the equation of a line l:

`7=-(3)/(5)\cdot(-2)+b\\\\(6)/(5)+b=7\ \ \ \ |-(6)/(5)\\\\b=(35)/(5)-(6)/(5)\\\\b=(29)/(5)`

`l:y=-(3)/(5)x+(29)/(5)\ \ \ \ |\cdot5\\\\5y=-3x+29\ \ \ \ |+3x\\\\3x+5y=29`

Answer: 3x + 5y = 29