Write the equation of the line that passes through (1;2) and is perpendicular to y = 6x - 2

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

Or

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