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... B (b,0) and C (3,-4) are collinear.
3)Find the equation of the:
a) tangent to the circle with center (-1,2) at the point (3,1)
b) perpendicular bisector bisector of (AB) for A (2,6) and B(5,-2).
4) a) For P( -1,2,3) and Q (1,-2,-3), find:
i) the distance of PQ ii) the midpoint of (PQ)
b) Find K if P is ( 1,3,-1), Q is (2,1,K), and PQ=√30
Added 3+ months ago:
1)) show that A (1,-2), B(4,4) and C (5,6) are collinear.
2) Find b given that A (-6,2) B (b,0) and C (3,-4) are collinear.
3)Find the equation of the:
a) tangent to the circle with center (-1,2) at the point (3,1)
b) perpendicular bisector bisector of (AB) for A (2,6) and B(5,-2).
4) a) For P( -1,2,3) and Q (1,-2,-3), find:
i) the distance of PQ ii) the midpoint of (PQ)
b) Find K if P is ( 1,3,-1), Q is (2,1,K), and PQ=√30
Please post questions separately. I will do one: show that A (1,-2), B(4,4) and C (5,6) are collinear. Collinear means on the same line. So I will show that sgments AB and BC have the same slope.
Slope is rise over run. Run is horizontal distance, left to right. Run is always positive because we always go left to right. Rise is the vertical change in that same distance. A negative rise means it drops. Segment AB runs 3 units from 1 to 4 and rises 6 units from -2 to 4 so the slope is 6/3=2. Segment BC runs 1 unit from 4 to 5 and rises 2 units from 4 to 6 so the slope is 2/1=2. Therefore the segments are the same line.
The equation of a line is y = mx where m is the slope. That line passes through the origin. If you want it to pass through some point (a, b) you subtract the coordinates like this: y - b = m(x - a). That is the point-slope form, and you can rewrite it in other forms if it is convenient. You may do this with any point on the line. They all reduce to the same equation.
AB: y + 2 = 2(x - 1)
y = 2x - 4
BC: y - 6 = 2(x - 5)
y = 2x - 4 <-- SAME ANSWER
