Triangle ABC has an angle of 90 degrees at B. Point A is on the y axis, AB is part of the line x-2y+8=0 and C is the point (6,2). (i) Sketch the triangle. (ii) Find the equations of AC and BC. I done part (i) and found the equation of line AC but I am not sure how to find the equation of the line BC as the co-ordinates of B are unknown. Kindly help me out, thanks.
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Since point A is on the y-axis (x=0) then plug x=0 into the equation of the line x-2y+8=0 and we get y=4, so A is at (0,4). Since B is a right angle, BC is perpendicular to AB which means that the slope of the line of which BC is a segment has a slope of -2 because the AB line has a slope of ½ (y=½x+4). So the perpendicular has the equation y-2=-2(x-6), slope intercept form, plugging in C(6,2). This gives us y=-2x+14. Point B is where the two lines intersect: ½x+4=-2x+14, equating the y values. So 5x/2=10, x=10×⅖=4, and y=6 (½(4)+4 or -2(4)+14).

(i)

The triangle ABC is in red, and the line containing AB is blue.

(ii) BC is the equation 2x+y-14=0. Slope of AC is (2-4)/(6-0)=-⅓ and the equation is y-4=-⅓x, that is, x+3y-12=0.

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