1. Find the point-slope form equation of line passing through the point (1, 2) with slope -4.
2. Find the equation of line in slope-intercept form which is passing through the point (-2, 3) with slope 3.
3. Find the equation of line in general form which is passing through the point (-5, -4) with slope 2/3.
4. Find the equation of a line in standard passing through the point (3, -4) and having slope -5.
5. Find the point-slope form equation of the straight line passing through the point (1, 2) and parallel to the line whose equation is x + 2y + 3 = 0.
6. Find the point-slope form equation of the straight line passing through the point (-2, 3) and perpendicular to the line whose equation is x - 2y - 6 = 0.
7. Find the equation of line in point-slope form which passing through the point (1, 4) and perpendicular to the line joining points (2, 5) and (4, 7).
1. Answer :
Given : Point = (1, 2) and slope m = -4.
Equation of line in point-slope form :
y - y_{1} = m(x - x_{1})
Substitute (x_{1} , y_{1}) = (1 , 2) and m = -4.
y - 2 = -4(x - 1)
2. Answer :
Given : Point = (-2, 3) and slope m = 3.
Equation of line in point-slope form :
y - y_{1} = m(x - x_{1})
Substitute (x_{1} , y_{1}) = (-2 , 3) and m = 3.
y - 3 = 3[x - (-2)]
y - 3 = 3(x + 2)
y - 3 = 3x + 6
Hence, the equation of line in slope-intercept form :
y = 3x + 9
3. Answer :
Given : Point = (-5, -4) and slope m = 2/3.
Equation of line in point-slope form :
y - y_{1} = m(x - x_{1})
Substitute (x_{1} , y_{1}) = (-5 , -4) and m = 2/3.
y - (-4) = (-2/3)[x - (-5)]
y + 4 = (-2/3)(x + 5)
3(y + 4) = -2(x + 5)
3y + 12 = -2x - 10
Hence, the equation of line in general form :
2x + 3y + 22 = 0
4. Answer :
Given : Point = (3, -4) and slope m = -5.
Equation of line in point-slope form :
y - y_{1} = m(x - x_{1})
Substitute (x_{1} , y_{1}) = (3 , -4) and m = -5.
y - (-4) = -5(x - 3)
y + 4 = -5x + 15
Hence, the equation of line in standard form :
5x + y = 11
5. Answer :
Write the equation of the line 'x + 2y + 3 = 0' in slope intercept form.
x + 2y + 3 = 0
2y = -x - 3
y = (-1/2)x - 3/2
Slope of the given line is -1/2.
Because the required line is parallel to the given line, the slopes are equal.
Then, slope of the required line is -1/2.
The required line is passing through (1, 2) with slope -1/2.
Equation of line in point-slope form :
y - y_{1} = m(x - x_{1})
Substitute (x_{1} , y_{1}) = (1, 2) and m = -1/2.
y - 2 = (-1/2)(x - 1)
6. Answer :
Write the equation of the line 'x - 2y - 6 = 0' in slope intercept form.
x - 2y - 6 = 0
-2y = -x + 6
2y = x - 6
y = (1/2)x - 3
Slope of the given line is 1/2.
Because the required line is perpendicular to the given line, product of the slopes is equal to -1.
Let 'm' be the slope of the required line.
m x (1/2) = -1
m/2 = -1
m = -2
The required line is passing through (-2, 3) with slope -2.
Equation of line in point-slope form :
y - y_{1} = m(x - x_{1})
Substitute (x_{1} , y_{1}) = (-2, 3) and m = -2.
y - 3 = -2[x - (-2)]
7. Answer :
Let the given points be A(1, 4), B(2, 5) and C(4, 7).
Slope of line BC = (7 - 5)/(4 - 2)
= 2/2
= 1
Let m be the slope of the required line.
Since the required line is perpendicular to BC,
m x 1 = -1
m = -1
The required line is passing through A(1, 4) with slope -1.
Equation of line in point-slope form :
y - y_{1} = m(x - x_{1})
Substitute (x_{1} , y_{1}) = (1, 4) and m = -1.
y - 4 = -1(x - 1)
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Sep 29, 22 04:11 AM
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