Answer:
[tex]y = \frac{3}{2} x-11[/tex]
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[tex]y = - \frac{2}{3} x+5[/tex].
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 [tex]\frac{3}{2}[/tex].
Then, we will find the y-intercept of the line using the standard equation of a line:
[tex]y=mx+c[/tex]
[tex]1=\frac{3}{2} (8)+c[/tex]
[tex]c=-11[/tex]
Therefore, the equation of the line will be [tex]y = \frac{3}{2} x-11[/tex].
Answer:
Answer: the equation is y'=[tex]\frac{3}{2}x'-11[/tex]
Step-by-step explanation:
A line is given whose equation is given as [tex]y=-\frac{2x}{3}+5[/tex]------(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×[tex](-\frac{2}{3})[/tex]=(-1)
⇒[tex]\frac{2}{3} m=1[/tex]
⇒m=[tex]\frac{3}{2}[/tex]
Now we put the value of m in equation (3) to get the value of c
8×[tex](\frac{3}{2})[/tex]+c = 1
12+c=1
c = (-11)
Therefore the equation will be y'=[tex]\frac{3}{2}x'+(-11)[/tex]
