Question

Write the equation of a line that passes
through point (-5,3) and is perpendicular to
y=3/5x +2.

Answer 1

Answer:

$y=-\frac{5}{3}x-\frac{16}{3}$

Step-by-step explanation:

Hi there!

What we need to know:

• Linear equations are typically organized in slope-intercept form: $y=mx+b$ where m is the slope and b is the y-intercept (the value of y when the line crosses the y-axis)
• Perpendicular lines always have slopes that are negative reciprocals (ex. 2 and -1/2, 4/3 and -3/4, etc.)

1) Determine the slope (m)

$y=\frac{3}{5} x +2$

From the given equation, we can see that $\frac{3}{5}$ is the slope of the line since it's in the place of m. Because perpendicular lines always have slopes that are negative reciprocals, the line we're currently solving for would have a slope of $-\frac{5}{3}$. Plug this into $y=mx+b$:

$y=-\frac{5}{3}x+b$

2) Determine the y-intercept (b)

$y=-\frac{5}{3}x+b$

Plug in the given point (-5,3)

$3=-\frac{5}{3}(-5)+b\\3=\frac{25}{3}+b$

Subtract $\frac{25}{3}$ from both sides to isolate b

$3-\frac{25}{3}=\frac{25}{3}+b-\frac{25}{3}\\-\frac{16}{3} = b$

Therefore, the y-intercept is $-\frac{16}{3}$. Plug this back into $y=-\frac{5}{3}x+b$

$y=-\frac{5}{3}x+(-\frac{16}{3})\\y=-\frac{5}{3}x-\frac{16}{3}$

I hope this helps!