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maxonik [38]
3 years ago
14

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.
Mathematics
2 answers:
Anuta_ua [19.1K]3 years ago
7 0

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 isy = - \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.


Alexeev081 [22]3 years ago
4 0

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)

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