"Write two equations: one parallel, one perpendicular to (5,-2) y= 1/5x-3

Mathematics

2 answers:

guajiro [1.7K]3 years ago

7 0

Answer:

For parallel:

gradient is 1/5

$y = mx + c$

consider (5, -2):

$- 2 = ( \frac{1}{5} \times 5) + c \\ - 2 = 1 + c \\ c = - 3$

${ \boxed{ \bf{equation : y = \frac{1}{5}x - 3 }}}$

For perpendicular:

gradient, m1:

$m _{1} \times m_{2} = - 1 \\ m _{1} \times \frac{1}{5} = - 1 \\ \\ m _{1} = - 5$

gradient = -5

$y = mx + c \\ - 2 = (5 \times - 5) + c \\ - 2 = - 25 + c \\ c = 23$

${ \boxed{ \bf{equation :y = - 5x + 23 }}}$

iren [92.7K]3 years ago

5 0

Answer:

Parallel: $y=\displaystyle \frac{1}{5}x-3$

Perpendicular: $y=-5x+23$

Step-by-step explanation:

Hi there!

What we must know:

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)
Parallel lines always have the same slope
Perpendicular lines always have slopes that are negative reciprocals (ex. 2 and -1/2, -6/7 and 7/6)

Finding the Parallel Line

$y=\displaystyle \frac{1}{5} x-3$

Given this equation, we can identify that its slope (m) is $\displaystyle \frac{1}{5}$ . Because parallel lines always have the same slope, the slope of the line we're currently solving for would be $\displaystyle \frac{1}{5}$ as well. Plug this into $y=mx+b$ :

$y=\displaystyle \frac{1}{5}x+b$

Now, to find the y-intercept, plug in the given point (5,-2) and solve for b:

$-2=\displaystyle \frac{1}{5}(5)+b\\\\-2=1+b\\-2-1=b\\-3=b$

Therefore, the y-intercept is -3. Plug this back into $y=\displaystyle \frac{1}{5}x+b$ :

$y=\displaystyle \frac{1}{5}x+(-3)\\\\y=\displaystyle \frac{1}{5}x-3$

Our final equation is $y=\displaystyle \frac{1}{5}x-3$ .

Finding the Perpendicular Line

$y=\displaystyle \frac{1}{5} x-3$

Again, the slope of this line is $\displaystyle \frac{1}{5}$ . The slopes of perpendicular lines are negative reciprocals, so the slope of the line we're solving for would be -5. Plug this into $y=mx+b$ :

$y=-5x+b$