Question

Find the equation of a line that is perpendicular to line g that contains (P, Q).

coordinate plane with line g that passes through the points negative 2 comma 6 and negative 3 comma 2

4x + y = Q + 4P
x − 4y = −4Q + P
−4x + y = Q − 4P
x + 4y = 4Q + P

Answer 1

We want to find the equation of a line that is perpendicular to G and pass through point (P, Q)

The coordinate of G is

(-2,6) and (-3,2)

Point 1 = (x1,y1) = (-2,6)

Point 2 = (x2,y2) = (-3,2)

So, let find the slope of this line G.

Slope can be calculated using

m = ∆y/∆x

m = (y2 - y1) / (x2 - x1)

m = (2-6) / (-3--2), -×- = +

m = -4 / (-3 + 2)

m = -4 / -1

m = 4

So, the gradient or slope of line G is 4

From geometry, since the equation of the line we want to find is perpendicular to the G,

Then, the slope of the line is

m2 = -1 / m1

m2 = - 1 / 4

m2 = -¼.

Then, also we can write equation of a line given the slope and a point using

m = (y - y1) / (x - x1)

Where m = -¼

And the point (x1,y1) = (P, Q)

Then, we have

-¼ = (y - Q) / (x - P)

Cross multiply

-(x - P) = 4(y - Q)

-x + P = 4y - 4Q

Rearrange

-x - 4y = - 4Q - P

Divide through by -1

x + 4y = 4Q + P

Then, the last option is correct

Answer 2

The equation of a line that is perpendicular to line g contains

(P, Q) is $x+4y=4Q+P$

Given:

Coordinate plane with line g that passes through the points negative 2 commas 6 and negative 3 commas 2

The coordinate of G: (-2,6) and (-3,2)

Let:

$(x_{1} ,y_{1} )=(-2,6)\\(x_{2} ,y_{2} )=(-3,2)$

The slope of a line g:

$m=\frac{y_{2} - y_{1} }{x_{2} - x_{1}} \\m=\frac{2-6}{-3-(-2)} \\m=\frac{-4}{-1} \\m=4$

So, the slope of a line g is 4.

To find the slope of a line perpendicular to g,

$m_{1} =-\frac{1}{m} \\m_{1} =-\frac{1}{4}$

The equation of  slope point form of the line is

$(y-y_{1} )=m(x-x_{1} )\\y-Q=-\frac{1}{4}(x-P)\\4y-4Q=-x+P\\x+4y=4Q+P$

Therefore, The equation of a line that is perpendicular to line g contains

(P, Q) is $x+4y=4Q+P$

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