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### Worksheet on Collinearity of 3 Points

**Problem 1:** Find the value of ‘a’ for which the points A (a, -2), B (3, 1), and C (5, 6) are collinear?

**Solution:**

As given in the question,

The points (x_{1}, y_{1}) is A(a, -2).

The points (x_{2}, y_{2}) are B(3, 1).

The points (x_{3}, y_{3}) are C(5, 6).

The given A, B, and C are collinear.

For this the Slope of AB = Slope of BC.

So, the formula is (y_{2} â€“ y_{1})/(x_{2}â€“ x_{1}) = (y_{3}â€“ y_{2})/(x_{3} â€“ x_{2}).

Now, Substituting the given coordinates points values in a formula. Then it will be,

(1- (-2))/(3 â€“ a) = (6 â€“ 1)/(5 â€“ 3)

(1 + 2)/(3-a) = (5)/(2)

3/(3 â€“ a) = 5/2

3(2) = 5(3-a)

6=15-5a

6-15 =5a

-9 = 5a

a= -9/5

Thus, the value of a is -9/5.

**Problem 2:** Prove that the points P(-2, -4), Q(0, 2), and R(5, 17) are collinear points.

**Solution:**

Given that,

The points (x_{1}, y_{1}) are P(-2, -4).

The points (x_{2}, y_{2}) are Q(0, 2).

The points (x_{3}, y_{3}) are R(5, 17).

The given points P, Q, and R are collinear.

Now, we will prove that the points are collinear are not using the slope method. If the given points are collinear, then the following formula condition is satisfied.

The formula is (y_{2} â€“ y_{1})/(x_{2}â€“ x_{1}) = (y_{3}â€“ y_{2})/(x_{3} â€“ x_{2}).

Now, place the given coordinates values in the above equation. It will be,

(2-(-4))/(0-(-2)) = (17-2)/(5-0)

(2+4) /(2) = (15)/(5)

(6)/2 = 15/5

3=3

L.H.S = R.H.S

Therefore, the given three coordinate points are collinear.

**Problem 3:** Prove that the points (4, -5) and (1, 1) and (-2, 7) are collinear or not.

**Solution:**

As per the given information,

The points on x coordinates and y-coordinates (x_{1}, y_{1}) are (4, -5).

The points onÂ (x_{2}, y_{2}) are (1, 1).

The points on (x_{3}, y_{3}) are (-2, 7).

Now, we will prove whether the given points are collinear or not.

The formula of these is (y_{2} â€“ y_{1})/(x_{2}â€“ x_{1}) = (y_{3}â€“ y_{2})/(x_{3} â€“ x_{2}).

Putting the given points value in a formula. It will be,

(1-(-5))/(1-4) = (7-1)/(-2-1)

(1+5)/(-3) = (6)/(-3)

(6)/(-3) = 6/-3

1=1

Therefore, the L.H.S is equal to R.H.S.

Hence, the given three points are collinear.

**Problem 4**: The distance between two points, shows that points A(5, -2), B(4, -1) and C(1, 2) are collinear.

**Solution:**

In the given question, the points are A(5, -2), B(4,-1), and C(1,2).

So, the points on x coordinates and y-coordinates (x_{1}, y_{1}) are (5, -2).

The points onÂ (x_{2}, y_{2}) are (4, -1).

The points on (x_{3}, y_{3}) are (1, 2).

Now, we will find the distances between the points.

First, we will find the distance between A and B,

The formula is AB = âˆš[(x_{2} – x_{1})Â²Â + (y_{2} – y_{1})Â²]

Substitute the (x1, y1) and (x2, y2) values in the above formula.

The value of AB is âˆš[(4 – 5)Â²+ (-1 + 2)Â²]

= âˆš[(-1)Â² + 1Â²]

= âˆš[1 + 1] = âˆš2

Now, we will find the distance between B and C.

So, the formula for BC is âˆš[(x_{3} – x_{2})Â² + (y_{3} – y_{2})Â²]

=âˆš[(1 – 4)^{2}Â + (2 + 1)^{2}]

= âˆš[(-3)^{2}Â + 3^{2}]

= âˆš[9 + 9]

= âˆš18 = 3âˆš2

Next, find the distance between A and C.

The formula for AC is âˆš[(x_{1} – x_{3})Â² + (y_{1} – y_{3})Â²]

The value of AC = âˆš[(5 – 1)^{2}Â + (-2 – 2)^{2}]

= âˆš[4^{2}Â + (-4)^{2}]

= âˆš[16 + 16]

= âˆš32 = 4âˆš2

So, the value of AB + BC is âˆš2 + 3âˆš2 = 4âˆš2 = AC

Thus, the AB + BC = AC.

So, the three given points A, B, and C are collinear.

Also, Read: Conditions of Collinearity of Three Points

**Problem 5:** Find the value of ‘x’ for which the points A (6, 2), B(3, 4), and C (x, 1) are collinear?

**Solution:**

As given in the question,

The points (x_{1}, y_{1}) are A(6, 2).

The points (x_{2}, y_{2}) are B(6, 4).

The points (x_{3}, y_{3}) are C(x, 1).

The three given points A, B, and C are collinear.

We know that the Slope of line AB = Slope of line BC.

So, the formula is (y_{2} â€“ y_{1})/(x_{2}â€“ x_{1}) = (y_{3}â€“ y_{2})/(x_{3} â€“ x_{2}).

Now, Substituting the given coordinates points values in a formula. Then it will be,

(4- 2)/(6 â€“ 6) = (1â€“ 4)/(x â€“ 4)

(3)/(6-6) = (-3)/(x-4)

3/(0) = -3/(x-4)

3 = -3(x-4)

3= -3x+ 12

3-12 =-3x

-9 = -3x

x= -9/-3

x = 3

Thus, the value of a is 3.

**Problem 6: **Find the equation of a line and how the points (-1,3), (0,2), and (1,1) are collinear?

**Solution:**

In the given question, the points are given.

The points on the x-coordinate and y-coordinate are (-1,3) that is (x,y).

The pointsÂ (x_{1}, y_{1}) are (0, 2).

The points (x_{2}, y_{2}) are (1, 1).

The formula for equation of a line isÂ (y â€“ y_{1}) = (y_{1}â€“ y_{2}) /(x_{1}â€“ x_{2}) * (x â€“ x_{1}).

Now, substitute the value of (x_{1}, y_{1}) and (x_{2}, y_{2}). Then we will get the equation of a line.

(y-2)= (2-1)/(0-1)*(x-0)

y-2 = 1/-1*(x-0)

y-2 = -1(x-0)

y-2 = -x+0

x+y-2=0

So, the equation of a straight line is x+y-2 =0.

Now, substitute the values of x and y. To check Collinearity.

x+y-2 =0

After substituting the value, it will be

-1+3-2=0

3-3=0

0=0

Therefore, the value of LHS is equal to the value of the RHS.

Hence, proved the given three points are collinear.

**Problem 7:** Point A(h, -4) is collinear with points B(3, 3) and C(-3, 1) then find the value of h and also find the value of the slope of a line which are containing the three points.

**Solution:**

As per the given question, the points are (h, -4), (3,3), and (-3,1).

Now, we will find the value of the h and slope of a line.

The points (x_{1}, y_{1}) are A(h, -4).

The points (x_{2}, y_{2}) are B(3, 3).

The points (x_{3}, y_{3}) are C(-3,1).

First, find the value of the h. So, the formula is (y_{2} â€“ y_{1})/(x_{2}â€“ x_{1}) = (y_{3}â€“ y_{2})/(x_{3} â€“ x_{2}).

Now, put the given coordinates points values in a formula. Then it will be,

(3- (-4))/(3 â€“ h) = (1â€“ 3)/(-3 â€“ 3)

(3+4)/(3-h) = (1-3)/(-6)

7/(3-h) = -2/-6

7 = 1/3(3-h)

7×3= 1(3-h)

21 = 3-h

21-3 = -h

18=-h

h=-18

Therefore, the value of the h is -18.

Now, find the slope of a line contacting three given points.

First, find the slope of AB.

So, the Slope of AB Line = (y_{2} â€“ y_{1})/(x_{2}â€“ x_{1})

Replace the (x1, y1) as A(-18, -4) and (x2, y2) as B(3, 3).

The slope AB is = (3 + 4)/(3 +18)

= 7/(21) = 1/3

Now, find the Slope of BC value.

The slope of BC is (y_{3}â€“ y_{2})/(x_{3} â€“ x_{2}).

Place the values in above formula, we get

= (1 -3)/(-3 – 3)

= -2/(-6)

= 1/3

So, the Slope of AB is equal to the Slope of BC. B is the common point.

Thus, the three given points A, B, and C are collinear.

**Problem 8:** If the three points (2, -8) and (0, 0) and (-1, -1). Prove that points are collinear or not.?

**Solution:**

As per the given information,

The points on x coordinates and y-coordinates (x_{1}, y_{1}) are (2, -8).

The points onÂ (x_{2}, y_{2}) are (0, 0).

The points on (x_{3}, y_{3}) are (-1, -1).

Now, we have to prove whether the given points are collinear or not.

The formula of these is (y_{2} â€“ y_{1})/(x_{2}â€“ x_{1}) = (y_{3}â€“ y_{2})/(x_{3} â€“ x_{2}).

Placing the given x-coordinate and y-coordinate points value in a formula. It will be,

(0-(-8))/(0-2) = (-1-0)/(-1-0)

(0+8)/(-2) = (-1)/(-1)(8)/(-2) = 1

-4=1

Thus, the L.H.S is not equal to R.H.S.

Therefore, the three points (2, -8), (0, 0) and (-1, -1) are not collinear.

**Problem 9:** Find the equation of the straight line that passes through the points (1, 2) and (3, -4) and show that the given three points (1,2), (3, -4 ), and (2,- 1) are collinear or not?

**Solution:**

In the given question, the points are given.

First, we will find the value of the equation of a line.

The points on the x-coordinate and y-coordinate are (x_{1}, y_{1}) are (1, 2).

The points (x_{2}, y_{2}) are (3, -4).

The formula for equation of a line isÂ (y â€“ y_{1}) = (y_{1}â€“ y_{2}) /(x_{1}â€“ x_{2}) * (x â€“ x_{1}).

Now, substitute the value of (x_{1}, y_{1}) and (x_{2}, y_{2}). Then we will get the equation of a line.

(y-2)= (2-(-4))/(1-3)*(x-1)

y-2 = 2+4/-2*(x-1)

y-2 = 6/-2*(x-1)

y-2 = -3(x-1)

y-2=-3x+3

3x+y-2-3=0

3x+y-5=0

So, the equation of a straight line is 3x+y-5 =0.

Now, check the given points are collinear are not.

The formula is, (y_{2} â€“ y_{1})/(x_{2}â€“ x_{1}) = (y_{3}â€“ y_{2})/(x_{3} â€“ x_{2}).

Here, the points on (x_{1}, y_{1}) are (1, 2).

The points onÂ (x_{2}, y_{2}) are (3, -4).

The points on (x_{3}, y_{3}) are (2, -1).

Placing the given x-coordinate and y-coordinate points value in a formula. It will be,

(-4-2)/(3-1) = (-1-(-4))/(2-(-3))

(-6)/(2) = (-1+4)/(2+3)

-3 = (3)/(5)

-3=3/5

So, the values of L.H.S and RHS are not equal.

Hence, the given three points are not collinear.

**Problem 10:**Â The point A(4, -4), B(3, 4) and C(z, 1) is collinear then find the value of z and also find the value of the slope of a line which are containing the three points.

**Solution:**

As per the given question, the points are (4, -4), (3,4), and (z,1).

Now, we will find the value of the z and the slope of a line.

The points (x_{1}, y_{1}) are A(4, -4).

The points (x_{2}, y_{2}) are B(3, 4).

The points (x_{3}, y_{3}) are C(z,1).

First, find the value of the h. So, the formula is (y_{2} â€“ y_{1})/(x_{2}â€“ x_{1}) = (y_{3}â€“ y_{2})/(x_{3} â€“ x_{2}).

Now, put the given coordinates points values in a formula. Then it will be,

(4- (-4))/(3 â€“ 4) = (1â€“ 4)/(z â€“ 3)

(4+4)/(3-4) = (1-4)/(z-3)

8/(-1) = -3/(z-3)

8(z-3) = -3*-1

8z-21= 2

8z= 3+21

8z=23

z=24/8

z=3

Therefore, the value of the z is 3.

Now, find the slope of a line contacting three given points.

First, find the slope of AB.

The Slope of AB = (y_{2} â€“ y_{1})/(x_{2}â€“ x_{1})

Replace the (x1, y1) as A(4, -4) and (x2, y2) as B(3, 4).

The slope AB is = (4 +4)/(3 -4)

= 8/(-1) = -8

Now, find the Slope of BC value.

The formula for the slope of BC line is (y_{3}â€“ y_{2})/(x_{3} â€“ x_{2}).

Place the values in above formula, we get

= (1 -4)/(3 – 3)

= -3/(0)

= -3

So, the Slope of AB is not equal to the Slope of BC. B is the common point.

Hence, the given three points A, B, and C are not collinear.