Question

On the plane xy, there are 4 points:
A(a, b), B(-1,0), C(2,1), & D(0,2).

If ∠ABC = 90' & point D is on the side AC, then a & b = ?

Answer 1

1)|slope(AC)|=|slope(DC)|
2)AB^2+BC^2=AC^2

Answer 2

alternatively for (2), to check that the angle=90degrees, you can use the dot product
(A-B).(B-C)=0

Answer 3

solve equations 1,2 for a,b

Answer 4

Why is |slope (AC)| = |slope(DC)|?
Also, is No. 1 like this?
\[|\sqrt{a^2 - 4a + b^2 - 2b + 5}| = |\sqrt{5}|\]
I don't know how to do no. 1 after that.

Answer 5

Since angle ABC = 90 degrees
Hence AB must be Perpendicular to BC
For perpendicular lines we know that:
Slope of AB * Slope of BC = -1
i.e. (b-0)/[a-(-1)] *(0-1)/(-1-2) =-1
i.e. b/(a+1) *(-1)/(-3) =-1
i.e. b/(a+1) *(1)/(3) =-1
i.e. b/(a+1) =-3
i.e. b = -3(a+1)------(1)

Again, since given triangle is right angled, hence by Pythagoras Theorem and distance formula we have;
\[AC^2= AB^2+BC^2\]
A(a, b), B (-1,0), C (2, 1)
\[(a-2)^2+(b-1)^2= (a+1)^2+(b-0)^2+(-1-2)^2+(0-1)^2\]
\[a^2-4a+4+b^2-2b+1= a^2+2a+1+b^2+9+1\]
\[-4a+4-2b+1= 2a+1+9+1\]
\[-4a-2b+5= 2a+11\]
\[-2b= 4a+2a+11-5\]
\[-2b= 6a+6\]
\[b= -3a-3\]
\[b= -3(a+1)\]-----(2

From equations (1 and (2 we have