# 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 $y = {x}^{2} + 1$

##### 1 Answer
Mar 6, 2017

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

#### Explanation:

$\left(a\right)$ To calculate the gradient use the $\textcolor{b l u e}{\text{gradient formula}}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where m represents the gradient and $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \text{ 2 points on the line}$

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

let $\left({x}_{1} , {y}_{1}\right) = \left(- 1 , 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(1 , 6\right)$

$\Rightarrow {m}_{A B} = \frac{6 - 2}{1 - \left(- 1\right)} = \frac{4}{2} = 2$

$\left(b\right)$ the equation of a line in $\textcolor{b l u e}{\text{point-slope form}}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y - {y}_{1} = m \left(x - {x}_{1}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where m represents the gradient and $\left({x}_{1} , {y}_{1}\right) \text{ a point on the line}$

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

$\text{Using "m=2" and } \left({x}_{1} , {y}_{1}\right) = \left(1 , 6\right)$

$\Rightarrow y - 6 = 2 \left(x - 1\right) \leftarrow \textcolor{red}{\text{ in point-slope form}}$

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

$y - 6 = 2 x - 2$

$\Rightarrow y = 2 x + 4 \leftarrow \textcolor{red}{\text{ in slope-intercept form}}$

$\left(c\right)$ The gradient of a line perpendicular to AB is the
$\textcolor{b l u e}{\text{negative inverse of the gradient of AB}}$

$\Rightarrow {m}_{\text{perpendicular}} = - \frac{1}{{m}_{A B}} = - \frac{1}{2}$

$\text{Using "m=-1/2" and } \left({x}_{1} , {y}_{1}\right) = \left(- 2 , 5\right)$

$y - 5 = - \frac{1}{2} \left(x - \left(- 2\right)\right)$

$y - 5 = - \frac{1}{2} \left(x + 2\right)$

$\Rightarrow y - 5 = - \frac{1}{2} x - 1$

$\Rightarrow y = - \frac{1}{2} x + 4 \leftarrow \textcolor{red}{\text{ in slope-intercept form}}$

~~~~~

(d) Point C can be found by taking the two line equations, $y = 2 x + 4 , y = - \frac{1}{2} x + 4$, 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:

$2 x + 4 = - \frac{1}{2} x + 4$

$2 x = - \frac{1}{2} x$

$x = 0$

$y = 2 x + 4 \implies 2 \left(0\right) + 4 = 4$

And so point C is $\left(0 , 4\right)$ 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 ($y = 2 x + 4$) intersect can be found by again setting the two equations equal to each other (they are both solved for $y$ already):

${x}^{2} + 1 = 2 x + 4$

${x}^{2} - 2 x - 3 = 0$

$\left(x - 3\right) \left(x + 1\right) = 0$

$x = 3 , - 1$

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

$y = {x}^{2} + 1 \implies {3}^{2} + 1 = 10 \implies \left(3 , 10\right)$

$y = {x}^{2} + 1 \implies {\left(- 1\right)}^{2} + 1 = 2 \implies \left(- 1 , 2\right)$

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]}