Respuesta :
make y the subject
y= 5x + 3
perpendicular line needs the reciprocal of the gradient
y = 1/5x + c
substitute in the point
1 = 1/5(5) + c
5/5 = 1 so the answer is y = 1/5x
hope this is helpful!
The required equation of the line is [tex]\bold{y=-\frac{1}{5}x+2}[/tex]
What is the slope-point form of equation of the line?
"The equation of the line having slope m and passing through points (p, q) is (y - q) = m(x - p)"
What is the slope-intercept form of the line?
"y = mx + c, where m is the slope and c is the Y-intercept of the line."
For given question,
We have been given a point (5, 1)
The required line is perpendicular to the line 5x - y = -3
⇒ 5x - y = -3
⇒ - y = -5x - 3
⇒ y = 5x + 3
Comparing above equation with slop-intercept form of the line.
m = 5
Let m1 be the slope of the required line.
We know that the product of the slopes of perpendicular lines is -1.
⇒ m1 × m = -1
⇒ m1 = [tex]\frac{-1}{5}[/tex]
Now we find the equation of the required line by using slope-point form.
Let (p, q) = (5, 1)
By using slope-point form,
⇒ (y - q) = m1(x - p)
⇒ (y - 1) = [tex]-\frac{1}{5}[/tex] (x - 5)
⇒ y - 1 = [tex]-\frac{1}{5}x[/tex] + 1
⇒ y = [tex]-\frac{1}{5}x[/tex] + 2
Therefore, the required equation of the line is [tex]\bold{y=-\frac{1}{5}x+2}[/tex]
Learn more about equation of line here:
brainly.com/question/21511618
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