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You can ask your own question or look at similar Further Mathematics questions. You have to find a value of k such that AB and BC are perpendicular lines, or their gradients multiply to give -1. The formula for the gradient is

(y2-y1)/(x2-x1)

(2--2)/(7--1) = 4/8 = 1/2

(4-2)/(k-7) = 2/(k-7)

now equate the product to -1 and rearrange

(1/2)*(2/(k-7)) = -1

k-7 = -1

k = 6 Gradient of side AB = (2 - (-2)) / (7 - (-1)) = 1/2

So gradient of (perpendicular) side BC = -1 / (1/2) = -2

Using y = mx + c , with m = -2 , we can sub in the (7,2) x,y coordinates from B to find the constant c. Hence we can use c and the y coordinate of point C to find x coordinate k:

y = -2x + c

2 = -14 + c

c = 16

So, for point C, x = k = (c - y)/2 = (16 -4) / 2 = 6

k = 6  Have a go at drawing it out on graph paper. Plot the two points you've been given for A and B and connect them. Use your protractor to draw a line at a right angle perpendicular to the line AB at point B. Extend that line to where it intersects at y=4 and this point of intersection will give you the value of K.

Hope that makes sense? ABC is a right angle thus AC is the hypotenuse.  Length of AB=sqrt((7+1)^2+(2+2)^2)=sqrt(64+16)=sqrt100=10.

BC=sqrt((k-7)^2+(4-2)^2)=sqrt((k-7)^2+4)

AC=sqrt((k+1)^2+(4+2)^2)=sqrt((k+1)^2+36)

Pythagoras gives: AC^2=AB^2+BC^2 hence

(k+1)^2+36=100+(k-7)^2+4

Expand, simplify etc and you find the value of k. 