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

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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#

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

##### 1 Answer

#### Answer:

#### 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"#

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(d) Point C can be found by taking the two line equations,

And so point C is

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

(e) The point where the curve

We can plug these points back into either the curve or the line to find the associated

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]}