Question

Find the equation of the line that contains the given point and is parallel to the given line? (4,7);y=-3x-1

parallel/perpendicular lines

Answer 1

(1) : The eqn. of parallel line : `y=-3x+19.`

(2) : The eqn of `bot` line :`3y=x+17.`

Explanation:

Recall that the Parallel Lines have the same slope.

The slope of the given line : `y=-3x-1,` is `-3.`

Therefore, the line parallel to this must also have the slope `-3.`

The reqd. parallel line passes through the point `(4,7).`

Using the Slope-Point Form, the reqd. eqn. must be,

`y-7=-3(x-4)=-3x+12, or, y=-3x+19.`

Next, the slope of the reqd. `bot` line, must be, `-1/-3=1/3.`

Hence, its eqn. is, `y-7=1/3(x-4), or, 3y-21=x-4, i.e., 3y=x+17.`

Answer 2

`y = -3x -5`

Explanation:

The general equation for a line is `y = mx +b`.

Since the line is parallel to the line `y = -3x-1` it will have the the same gradient `m`.

We have the point `( 4 , 7 )` and the gradient `-3` so we have altogether:

`7=-3(4) + b => 7=-12+b`

We now solve for `b`:

Adding `12` to both sides:

`19 = b`

So our equation is: `y = -3x +19`

Graph of `y = -3x +19` and `y=-3x-1`