**Question 1 :**

Find the value of ‘a’, if the line through (–2, 3) and (8, 5) is perpendicular to y = ax +2

**Solution :**

Equation of the line passing through the points (-2, 3) and (8, 5).

(y - y_{1})/(y_{2} - y_{1}) = (x - x_{1})/(x_{2} - x_{1})

(y - 3)/(5 - 3) = (x + 2)/(8 + 2)

(y - 3)/2 = (x + 2)/10

10(y - 3) = 2(x + 2)

10y - 30 = 2x + 4

2x + 10y - 30 - 4 = 0

2x + 10y - 34 = 0

Dividing the entire equation by 2, we get

x + 5y - 17 = 0

**Question 2 :**

The hill is in the form of a triangle has its foot at (19, 3) . The inclination of the hill to the ground is 45˚. Find the equation of the hill joining the foot and top.

**Solution :**

Equation of the hill joining the foot and top :

slope (m) = tan 45 = 1

y - y_{1} = m(x - x_{1})

y - 3 = 1(x - 19)

y - 3 = x - 19

x + y - 19 + 3 = 0

x + y - 16 = 0

**Question 3 :**

Find the equation of a line through the given pair of points

(i) (2, 2/3) and (-1/2, -2)

**Solution :**

(y - y_{1})/(y_{2 }- y_{1}) = (x - x_{1})/(x_{2 }- x_{1})

(y - (2/3))/(-2 - (2/3)) = (x - 2)/((-1/2)_{ }- 2)

((3y - 2)/3)/(-8/3) = (x - 2)/(-5/2)

-(3y - 2)/8 = -2(x - 2)/5

5(3y - 2) = 16(x - 2)

15y - 10 = 16x - 32

16x - 15y - 32 + 10 = 0

16x - 15y - 22 = 0

(ii) (2, 3) and (-7,-1)

(y - 3)/(-1 - 3) = (x - 2)/(-7 - 2)

(y - 3)/(-4) = (x - 2)/(-9)

-9(y - 3) = -4(x - 2)

-9y + 27 = -4x + 8

4x - 9y + 27 - 8 = 0

4x - 9y + 19 = 0

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