Question

show that the points A - 11 b 5,7 and c( 7,9 )all collinear​

Answer 1

Answer:

To find if three points, A (5,2), B(7,9) and C(9,16) are collinear. We can use two methods.

(A) METHOD USING SLOPES

Slope of straight line AB = y(B) -y(A)/x(B) - x(A) = (9–2)/7–5= 7/2

Slope of straight line BC= y(c) - y(B)/x(B)- x(A) = (16 -9)/(9–7)= 7/2

Slope of straight line AC= y(C)-y(A)/x(C) -x(A)= 16–2/9–5= 14/4= 7/2

Slope of AB = slope of BC= Slope of AC

Since slope of AB = Slope of BC, and point B is common to both lines AB and line BC, therefore the points A,B,and C are collinear. This is further confirmed by the fact that the slope of AC is the same as that of lines AB and BC.

(B) FINDING THE AREA OF THE TRIANGLE FORMED BY THE THREE VERTICES.

Area of a triangle with vertices (x¹, y¹), (x², y²) and (x³,y³) as vertices is given by;

∆ = ½ [ x¹( y² - y³) + x²(y³ -y¹) +x³(y¹ -y²)] = 1[ 5(9–16)+7(16–2)+9(2–9)]=½ [ 5×-7+7×14+9×-7]=½[ -35 +98–63]=½ [ -98+98]= 0

Since the area of the triangle is zero, therefore the points are collinear.

I could not type subscripts, so I used superscripts.