Question

We have points A (-1,2) and B (1,6). Find the following?

a. the gradient,
b. the line equation in point-slope form,
c. the gradient of a line perpendicular to the one in b that runs through point (-2,5),
d. the point where the two lines intersect,
e. the points where the line found in b intersects with the curve `y=x^2+1`

Answer 1

`m=2;y=2x+4;y=-1/2x+4`

Explanation:

`(a)` To calculate the gradient use the `color(blue)"gradient formula"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(m=(y_2-y_1)/(x_2-x_1))color(white)(2/2)|)))`
where m represents the gradient and `(x_1,y_1),(x_2,y_2)" 2 points on the line"`

The 2 points here are A(-1 ,2) and B(1 ,6)

let `(x_1,y_1)=(-1,2)" and " (x_2,y_2)=(1,6)`

`rArrm_(AB)=(6-2)/(1-(-1))=4/2=2`

`(b)` the equation of a line in `color(blue)"point-slope form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))`
where m represents the gradient and `(x_1,y_1)" a point on the line"`

For a point on the line, use either (-1 ,2) or (1 ,6)

`"Using "m=2" and " (x_1,y_1)=(1,6)`

`rArry-6=2(x-1)larrcolor(red)" in point-slope form"`

distributing the bracket and simplifying gives an alternative version of the equation.

`y-6=2x-2`

`rArry=2x+4larrcolor(red)" in slope-intercept form"`

`(c)` The gradient of a line perpendicular to AB is the
`color(blue)"negative inverse of the gradient of AB"`

`rArrm_("perpendicular")=-1/(m_(AB))=-1/2`

`"Using "m=-1/2" and " (x_1,y_1)=(-2,5)`

`y-5=-1/2(x-(-2))`

`y-5=-1/2(x+2)`

`rArry-5=-1/2x-1`

`rArry=-1/2x+4larrcolor(red)" in slope-intercept form"`

~~~~~

(d) Point C can be found by taking the two line equations, `y=2x+4, y=-1/2x+4`, setting them equal to each other (they are both solved for `y` already), finding the common `x` term and then finding the common `y` term:

`2x+4=-1/2x+4`

`2x=-1/2x`

`x=0`

`y=2x+4 => 2(0)+4=4`

And so point C is `(0,4)` and we can verify that by looking at the graph:

graph{(y-(2x+4))(y-(-1/2x+4))=0 [-9.96, 10.04, -2, 9.4]}

(e) The point where the curve `y=x^2+1` and L1 (`y=2x+4`) intersect can be found by again setting the two equations equal to each other (they are both solved for `y` already):

`x^2+1=2x+4`

`x^2-2x-3=0`

`(x-3)(x+1)=0`

`x=3, -1`

We can plug these points back into either the curve or the line to find the associated `y` values:

`y=x^2+1=>3^2+1=10=>(3,10)`

`y=x^2+1=>(-1)^2+1=2=>(-1,2)`

And we can again see that in the graph:

graph{(y-(x^2+1))(y-(2x+4))=0 [-11.82, 16.66, -0.76, 13.48]}