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1Exercise 10.1
Find the slope of the line joining the points (3, -2) and (-1, 4).
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Slope m = (y₂ - y₁)/(x₂ - x₁) = (4 - (-2))/(-1 - 3)
= 6/(-4) = -3/2
2Exercise 10.1
Write the equation of the line passing through the point (2, 3) with slope 4, in point-slope form and slope-intercept form.
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Point-slope: y - 3 = 4(x - 2)
Simplify: y - 3 = 4x - 8 ⇒ y = 4x - 5
Slope-intercept: y = 4x - 5
3Exercise 10.1
Find the equation of the line passing through (-1, 1) and (2, -4) using the two-point form.
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m = (-4 - 1)/(2 - (-1)) = -5/3
Using (x₁,y₁) = (-1,1):
y - 1 = (-5/3)(x + 1)
3(y - 1) = -5(x + 1)
3y - 3 = -5x - 5
5x + 3y + 2 = 0
4Exercise 10.1
Find the distance of the point (3, 4) from the line 3x + 4y = 5.
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Distance d = |ax₁ + by₁ + c|/√(a² + b²)
Line: 3x + 4y - 5 = 0
= |3(3) + 4(4) - 5|/√(9 + 16)
= |9 + 16 - 5|/5
= 20/5 = 4 units
5Exercise 10.1
Find the angle between the lines y = 2x + 1 and y = (1/2)x - 3.
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m₁ = 2, m₂ = 1/2
tan θ = |(m₁ - m₂)/(1 + m₁m₂)|
= |(2 - 1/2)/(1 + 1)|
= |(3/2)/2| = 3/4
θ = tan⁻¹(3/4)
6Exercise 10.1
Find the foot of the perpendicular from the point (1, 2) to the line 2x - y + 1 = 0.
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For line ax + by + c = 0, foot of perpendicular from (x₁,y₁):
x' = x₁ - a(ax₁ + by₁ + c)/(a² + b²)
y' = y₁ - b(ax₁ + by₁ + c)/(a² + b²)
Here a=2, b=-1, c=1; (x₁,y₁)=(1,2).
ax₁+by₁+c = 2(1) - 2 + 1 = 1
a²+b² = 5
x' = 1 - 2(1)/5 = 3/5
y' = 2 - (-1)(1)/5 = 2 + 1/5 = 11/5
Foot = (3/5, 11/5)