Question
Using the distance formula, show taht the given points are collinear:
(1, -1), (5, 2) and (9, 5)
(1, -1), (5, 2) and (9, 5)
✓
Answer
Let A(1, -1), B(5, 2) and C(9, 5) be the given points
Then,
$\text{AB}=\sqrt{(1-5)^2+(-1-2)^2}=\sqrt{(4)^2+(-3)^2}$
$=\sqrt{16+9}=\sqrt{25}=5\text{ units}$
$\text{BC}=\sqrt{(5-9)^2+(2-5)^2}=\sqrt{(-4)^2+(-3)^2}$
$=\sqrt{16+9}=\sqrt{25}=5\text{ units}$
$\text{AC}=\sqrt{(1-9)^2+(-1-5)^2}=\sqrt{(-8)^2+(-6)^2}$
$=\sqrt{64+36}=\sqrt{100}=10\text{ units}$
$\therefore\text{AB}+\text{AC}=5+5=10\text{ units}=\text{BC}$
$\Rightarrow\text{AB}+\text{AC}=\text{BC}$
Hence, the given points are collinear.
Then,
$\text{AB}=\sqrt{(1-5)^2+(-1-2)^2}=\sqrt{(4)^2+(-3)^2}$
$=\sqrt{16+9}=\sqrt{25}=5\text{ units}$
$\text{BC}=\sqrt{(5-9)^2+(2-5)^2}=\sqrt{(-4)^2+(-3)^2}$
$=\sqrt{16+9}=\sqrt{25}=5\text{ units}$
$\text{AC}=\sqrt{(1-9)^2+(-1-5)^2}=\sqrt{(-8)^2+(-6)^2}$
$=\sqrt{64+36}=\sqrt{100}=10\text{ units}$
$\therefore\text{AB}+\text{AC}=5+5=10\text{ units}=\text{BC}$
$\Rightarrow\text{AB}+\text{AC}=\text{BC}$
Hence, the given points are collinear.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
A bucket made up of a metal sheet is in the form of a frustum of a cone of height 16cm and radii of its lower and upper ends are $8\ cm$ and $20\ cm$, respectively. Find the cost of the bucket if the cost of metal sheet used is$ Rs.\ 15\ per\ 100\ cm^2. $
$\big[\text{Use }\pi=3.14\big]$
→→$\big[\text{Use }\pi=3.14\big]$
There are six cards in a box, each bearing a number from 0 to 5. Find the probability of each of the following events, that a card drawn shows,
(1) a natural number.
(2) a number less than 1.
(3) a whole number.
(4) a number is greater than 5.
(1) a natural number.
(2) a number less than 1.
(3) a whole number.
(4) a number is greater than 5.
Two cones with same base radius $8\ cm$ and height $15\ cm$ are joined together along their bases. Find the surface area of the shape formed.
→Find the value of k for which root are real and equal in the following equations:
$(3k + 1)x^2 + 2(k + 1)x + k = 0$
→$(3k + 1)x^2 + 2(k + 1)x + k = 0$
The table below shows the daily expenditure on food of 25 households in a locality:
|
Daily expenditure (in Rs.)
|
100-150
|
150-200
|
200-250
|
250-.300
|
300-350
|
|
Number of households
|
4
|
5
|
12
|
2
|
3
|
Draw a circle with 'O' as centre and radius 4 cm. Take a point P at a distance of 7.5 cm from 'O'. Draw tangents to the circle through the point P.
If $\sin\theta=\frac{\text{a}}{\text{b}},$ show that $(\sec\theta+\tan\theta)=\sqrt{\frac{\text{b}+\text{a}}{\text{b}-\text{a}}}.$
→Solve the following system of equations by the method of cross-multiplication:
→→$\frac{\text{ax}}{\text{b}}-\frac{\text{by}}{\text{a}}=\text{a}+\text{b}$
$\text{ax}-\text{by}=2\text{ab}$