Question

What is the equation of a line that passes through the point (8, 1) and is perpendicular to the line whose equation is y=−2/3x+5 ? Enter your answer in the box.

Answer 1

Answer:

$y = \frac{3}{2} x-11$

Step-by-step explanation:

We are to find the equation a line that passes through the point (8, 1) and which is perpendicular to a line whose equation is$y = - \frac{2}{3} x+5$.

We know that the slope of line which is perpendicular to another line is the negative reciprocal of the slope of the other line so it will be $\frac{3}{2}$.

Then, we will find the y-intercept of the line using the standard equation of a line:

$y=mx+c$

$1=\frac{3}{2} (8)+c$

$c=-11$

Therefore, the equation of the line will be $y = \frac{3}{2} x-11$.

Answer 2

Answer:

Answer: the equation is y'=$\frac{3}{2}x'-11$

Step-by-step explanation:

A line is given whose equation is given as $y=-\frac{2x}{3}+5$------(1)

and we have to find the equation of another line which is perpendicular to this.

Let the equation of the line be y'=mx'+c------(2)

This line passes through a point (8,1) so we put the values of x & y in the equation.

⇒ 1 = m×8+c

⇒ 8m+c = 1------(3)

We know that if two lines are perpendicular to each other then multiplication of their slopes is equal to (-1).

Therefore m×$(-\frac{2}{3})$=(-1)

⇒$\frac{2}{3} m=1$

⇒m=$\frac{3}{2}$

Now we put the value of m in equation (3) to get the value of c

8×$(\frac{3}{2})$+c = 1

12+c=1

c = (-11)

Therefore the equation will be y'=$\frac{3}{2}x'+(-11)$