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The line passes through the point (1,2)
Then the equation for the line is:
d1:y-y1 = m1 (x-x1)
d1:y- 2 = m1 (x-1)
But we know that the line d1 is perpendicular to the line:
d2: y= 6x - 2
Sinde d1 and d2 are perpendiculr, then the product of the slopes equals -1
m1*m2 = -1
m1 * 6 = -1
==> m1 = -1/6
Then the equation for the line d1 is:
d1: y-2 = (-1/6)(x-1)
d1: y= (-1/6)x + 1/6 + 2
d1: y + (1/6)x - 7/6 = 0
Multiply by 6:
d1: 6y + x - 7 = 0
If the 2 lines are perpendicular, then the product of the values of their slopes is -1.
Since the equation is put in the standard form, y = mx + n, we'll find it's slope:
y = 6x - 2, so the slope is m1 = 6
That means that the second slope is:
m1*m2 = -1
m2 = -1/m1
m2 = -1/6
Now, we'll write the equation of a line that passe through a given point and it has a slope, whose value is known.
y - y1 = m2*(x - x1)
The coordinates x1 and y1 are:
y - 2 = m2*(x - 1) and m2 = -1/6
y - 2 = -(x-1)/6
We'll add 2 both sides:
y = -x/6 + 1/6 + 2
The equation of the line that passe through the point (1,2) and it's perpendicular to the line y = 6x-2 is:
y = -x/6 + 13/2
A line perpendicular to ax+by+c = 0 is of the form bx-ay+k = 0
The constant k is determined by the condition that the line passes through a point .
Here y = 6x-2 Or
6x-y+2 = 0 is the line.
A perpendicular line to 6x-y +2 is of the form x - (-6)y +k = 0. Or
x+6y+k = 0. Since this passes through (1 , 2), it satisfies x+6y +k = 0.
1+6*2+k = 0
13+k = 0
So k = -13.
So the required line is x+6y -13 = 0.
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