**10th grade math solution exercise 5.4 part 4:**

Here we are going to see solutions of some problems that we find in the exercise number 5.4

**Question 9 :**

Find the equation of the median from the vertex R in a Δ PQR with vertices at P(1, -3), Q(-2, 5) and R(-3, 4).

**Solution : **

Let us draw a rough diagram based on the given information.

By using the concept of the equation of the straight line, prove that the given three points are collinear.

(i) (4, 2), (7, 5) and (9, 7) (ii) (1, 4), (3, -2) and (-3, 16)

Midpoint of AB = (x₁ + x₂)/2, (y₁ + y₂)/2

A(4, 1) B(3, 2)

= (4 + 3)/2 , (1 + 2)/2

= C (7/2, 3/2)

From the angle of inclination of the required line, we have to find the slope.

m = tan θ ==> tan 30° ==> 1/√3

Equation of the required line :

(y - y₁) = m (x - x₁)

(y - (3/2)) = (1/√3) (x - (7/2))

√3 (y - (3/2)) = 1 (x - (7/2))

√3 (2y - 3) = 1 (2x - 7)

2√3 y - 3√3 = 2x - 7

2x - 2√3y - 7 + 3√3 = 0

Hence equation of the required line is 2x - 2√3y + (3√3 - 7) = 0

**Question 10 :**

By using the concept of the equation of the straight line, prove that the given three points are collinear.

(i) (4, 2), (7, 5) and (9, 7) (ii) (1, 4), (3, -2) and (-3, 16)

**Solution : **

(i) (4, 2), (7, 5) and (9, 7)

To prove the given points are collinear, first we have to find the equation of the line passing through the first two points.

Now we have to apply the remaining point in the equation that we have found.

If the remaining point satisfies the equation of the line, then we can decide that the three points are collinear. Otherwise we can say the three points are not collinear.

Equation of the line passing through two given points.

(y - y₁) /(y₂-y₁) = (x - x₁) /(x₂-x₁)

(x₁, y₁) ==> (4, 2) and (x₂, y₂) ==> (7, 5)

(y-2)/(5 - 2) = (x - 4)/(7 - 4)

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

(y-2) = (x - 4)

x - 4 - y + 2 = 0

x - y - 2 = 0

Apply x = 9 and y = 7

9 - 7 - 2 = 0

2 - 2 = 0

0 = 0

Hence the given points are collinear.

(ii) (1, 4), (3, -2) and (-3, 16)

Equation of the line passing through two given points.

(y - y₁) /(y₂-y₁) = (x - x₁) /(x₂-x₁)

(x₁, y₁) ==> (1, 4) and (x₂, y₂) ==> (3, -2)

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

(y - 4)/(-6) = (x - 1)/(2)

2 (y - 4) = -6 (x - 1)

2y - 8 = -6x + 6

6x + 2y - 8 - 6 = 0

6x + 2y - 14 = 0

Apply x = -3 and y = 16

6(-3) + 2(16) - 14 = 0

-18 + 32 - 14 = 0

0 = 0

Hence the given points are collinear.

After having gone through the stuff given above, we hope that the students would have understood "10th grade math solution exercise 5.4 part4".

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