show that the points a(-6,10), b(-4,6) and c(3,-8) are collinear.
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Answers
Step-by-step explanation:
If the points A(-6,10),B(-4,6) and C(3,-8) are collinear, then AB=\frac{2}{9}\times ACAB=
9
2
×AC
Step-by-step explanation:
It is given that points A(-6,10),B(-4,6) and C(3,-8) are collinear.
To prove: AB=2/9AC
Distance formula:
d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}d=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
Using distance formula we get
AB=\sqrt{(-4-(-6))^2+(6-10)^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}AB=
(−4−(−6))
2
+(6−10)
2
=
4+16
=
20
=2
5
AC=\sqrt{(3-(-6))^2+(-8-10)^2}=\sqrt{81+324}=\sqrt{405}=9\sqrt{5}AC=
(3−(−6))
2
+(−8−10)
2
=
81+324
=
405
=9
5
\frac{AB}{AC}=\frac{2\sqrt{5}}{9\sqrt{5}}
AC
AB
=
9
5
2
5
Cancel out common factors.
\frac{AB}{AC}=\frac{2}{9}
AC
AB
=
9
2
Multiply both sides by AC.
AB=\frac{2}{9}\times ACAB=
9
2
×AC
Hence proved.
- If the area of triangle formed by the points (x ,y ), (x , y ) and (x , y ) is zero, then the points are collinear.
