**10th grade math solution exercise 5.4 part 3:**

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

**Question 7 :**

Find the equation of the straight line which passes through the midpoint of the line segment joining (4, 2) and (3, 1) whose angle of inclination is 30° .

**Solution : **

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

From the above diagram, we come to know that we have to find the line passing through the point C.

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 8 :**

Find the equation of the straight line passing through the points

(i) (-2, 5) and (3, 6)

**Solution : **

Equation of the line passing through two given points.

(y - y₁) /(y₂-y₁) = (x - x₁) /(x₂-x₁)

(x₁, y₁) ==> (-2, 5) and (x₂, y₂) ==> (3, 6)

(y-5)/(6 - 5) = (x - (-2))/(3 - (-2))

(y-5)/1 = (x+2)/(3 +2)

(y-5)/1 = (x +2)/5

5(y - 5) = 1(x + 2)

5y - 25 = x + 2

x - 5y + 2 + 25 = 0

x - 5y + 27 = 0

(ii) (0, -6) and (-8, 2)

Equation of the line passing through two given points.

(y - y₁) /(y₂-y₁) = (x - x₁) /(x₂-x₁)

(x₁, y₁) ==> (0, -6) and (x₂, y₂) ==> (-8, 2)

(y-(-6))/(2 - (-6)) = (x - 0)/(-8 - 0)

(y+6)/(2 + 6) = x/(-8)

(y+6)/8 = x/(-8)

(y + 6) = -x

x + y + 6 = 0

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