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

Answer:

#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]}