The point A has coordinates (1,1) & the point B has coordinates (5,k) ?

Question

The point A has coordinates (1,1) & the point B has coordinates (5,k).
The line AB has equation 3x + 4y = 7
(a) i) Show that k= -2
ii) Hence find coordinates of t he midpoint of AB.

(b) Find the gradient of AB

(c) The line AC is perpendicular to the line AB.
i) Find the gradient of AC.
ii) Hence find an equation of the line AC.
iii) Given that point C lies on the x~axis, find its x~coordinate.

thanks.

Answer 1

(a) i) Given that 3x+4y=7 passes through the point (5, k), you can plug in those values into the equation:



3(5)+4k=7

15+4k=7

4k=-8

k=-2



ii) The coordinates of the midpoint can be found by taking the averages of the coordinates of the endpoints:

x=(5+1)/2=3 y=(1+(-2))/2=-1/2



So the ccoordinates of the midpoint are (3, -1/2)





(b) The gradient, also known as the slope, can be found by taking the difference of the y-values divided by the diffference of the x-values:



(-2-1)/(5-1)=-3/4





(c) i) Being perpendicular, this means that the gradient of AC is equivalent to the negative inverse gradient of AB. So:



-(-3/4)^-1=4/3



ii) As the line goes through the point (1,1) and has the slope 4/3, the equation for the line is:



y-1=4/3(x-1)

y=4/3x-4/3+1

y=4/3x-1/3, or in previous terms, 4x-3y=1



iii) Since point C lies on the x-axis, that means the y-value of that point is 0. Given this point, you can find point C by plugging it into our previous equation:



4x-3(0)=1

4x=1

x=1/4



So the coordinates for point C are (1/4,0)





Hope this helps.

Answer 2

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The point A has coordinates (1,1) & the point B has coordinates (5,k) ?



The point A has coordinates (1,1) & the point B has coordinates (5,k).

The line AB has equation 3x + 4y = 7





(a) i) Show that k= -2



m = -3 / 4



(k-1)/(5-1) = -3/4



(k-1)/4=-3/4......Multiply both sides by 4



k-1=-3.........add 1



k = -2



ii) Hence find coordinates of t he midpoint of AB.



(1,1)..............(5,-2)



(1+5)/2........(1+-2)/2



( 3 , -1/2 )



(b) Find the gradient of AB



m = -3/4



(c) The line AC is perpendicular to the line AB.

i) Find the gradient of AC.



slope is Opposite Reciprocal



m = + 4/3



ii) Hence find an equation of the line AC



4x - 3y = 1.................Email I will show you how to do this last part in ONE step (just write answer)

.

iii) Given that point C lies on the x~axis, find its x~coordinate.



Y must be 0



4x - 3(0) = 1



x = 1/2



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Answer 3

3x + 4y = 7



y = -3/4x + 7/4



So the line has a slope of -3/4



(a)

i)

slope = (y2 - y1)/(x2 - x1)

-3/4 = (k - 1)/(5 - 1)

-3/4 = (k - 1)/4

-12/4 = k - 1

-3 = k - 1

-2 = k



ii)

The coordinates of the minpoint are going to be halfway between 1 and 5 for x and 1 and -2 for y.



(3, -1/2)



(b)



It has a slope of -3/4



(c)

i) It has a slope of 4/3

ii) y = 4/3x + 1

iii) (-3/4, 0)

Answer 4

y-y?= slope/gradient(x-x?) --> y-a million= -5/12(x-2) -->5x+12y-22=0 This equation is likewise desirable for the q(ok,11) as this instantly line is going with the aid of q so, 5k+ 12X11 -22= 0 --->ok = -22 (ans)