Equation of the line using slope point form :
(y - y_{1}) = m(x - x_{1})
Here we have to consider the given point as (x_{1}, y_{1}) and slope as 'm'.
Example 1 :
Find the equation of the straight line whose.
(i) Slope is -4 and passing through (1, 2).
(ii) Slope is 2/3 and passing through (5, -4).
Solution :
Part (i) :
To find the equation of the straight line, we have to use the formula of slope point form.
Slope (m) = -4, x_{1} = 1 and y_{1} = 2.
Equation of the line :
(y - y_{1}) = m(x - x_{1})
(y – 2) = -4(x – 1)
y – 2 = -4x + 4
4x + y – 2 - 4 = 0
4x + y - 6 = 0
Part (ii) :
To find the equation of the straight line we have to use the formula of slope point form.
Slope (m) = 2/3, x_{1} = 5 and y_{1} = -4.
Equation of the line :
(y - y_{1}) = m(x - x_{1})
y – (-4) = (2/3)(x – 5)
(y + 4) = (2/3)(x – 5)
3(y + 4) = 2(x – 5)
3y + 12 = 2x – 10
2x – 3y - 10 - 12 = 0
2x – 3y – 22 = 0
Example 2 :
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 degree.
Solution :
First we have to find midpoint of the line segment joining the points (4, 2) and (3, 1).
Midpoint = (x_{1} + x_{2})/2, (y_{1} + y_{2})/2
= (4 + 3)/2, (2 + 1)/2
= (7/2, 3/2)
angle of inclination = 30°.
θ = 30°
Slope (m) = tan θ
m = tan 30°
m = 1/√3
Equation of the line :
y - y_{1} = m(x - x_{1})
y - 3/2 = (1/√3)(x - 7/2)
2y - 3 = (1/√3)(2x - 7)
2√3y - 3√3 = 2x - 7
2x - 2√3y - 7 + 3√3 = 0
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