# Emergency!!!

Question: points A (-1,-2),B(7,2) and C (k, 4), where K is a constant, are the vertices of triangle ABC. Angle ABC is a right angle. Calculate the value of K?

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+4Expand, simplify etc and you find the value of k.
marias
17 June 2018
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tausif045
17 June 2018
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?
Sammi M.
20 June 2018
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Miriam F.
30 June 2018
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03 July 2018
Gradient of side AB = (2 - (-2)) / (7 - (-1)) = 1/2So gradient of (perpendicular) side BC = -1 / (1/2) = -2Using 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 + c2 = -14 + cc = 16So, for point C, x = k = (c - y)/2 = (16 -4) / 2 = 6k = 6
Calem C.
12 July 2018
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coolsubh13
25 July 2018
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)gradient for AB(2--2)/(7--1) = 4/8 = 1/2gradient for BC (4-2)/(k-7) = 2/(k-7)now equate the product to -1 and rearrange(1/2)*(2/(k-7)) = -1k-7 = -1k = 6
Aleksandra J.
27 July 2018
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sanharshal
29 July 2018