Coordinate Geometry Worksheet D from Solomon Press, presenting five mathematical problems related to coordinate geometry, including line equations, perpendicularity, midpoints, and area calculations, with a diagram of a parallelogram.

Let’s solve each question step by step.

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Question 1

Given: Line $ l: y = 1 - 2x $

This is in slope-intercept form $ y = mx + c $, so slope of line $ l $ is $ m_l = -2 $.

Line $ m $ is perpendicular to $ l $, so its slope $ m_m = \frac{1}{2} $ (negative reciprocal).

Line $ m $ passes through point $ (6, -1) $.

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a) Find equation of $ m $ in form $ ax + by + c = 0 $, with integers $ a, b, c $.

Use point-slope form:

$ y - (-1) = \frac{1}{2}(x - 6) $

→ $ y + 1 = \frac{1}{2}x - 3 $

Multiply both sides by 2 to eliminate fraction:

→ $ 2y + 2 = x - 6 $

Bring all terms to one side:

→ $ -x + 2y + 8 = 0 $

Multiply by -1 to make leading coefficient positive (optional, but standard):

→ $ x - 2y - 8 = 0 $

✔ So, equation of $ m $: $ x - 2y - 8 = 0 $

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b) Find coordinates where $ l $ and $ m $ intersect.

Solve the system:

- $ l: y = 1 - 2x $
- $ m: x - 2y - 8 = 0 $

Substitute $ y = 1 - 2x $ into $ m $:

→ $ x - 2(1 - 2x) - 8 = 0 $

→ $ x - 2 + 4x - 8 = 0 $

→ $ 5x - 10 = 0 $

→ $ x = 2 $

Then $ y = 1 - 2(2) = 1 - 4 = -3 $

✔ So, intersection point is $ \boxed{(2, -3)} $

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Question 2

Line $ l $ passes through $ A(1, -3) $ and $ B(7, 5) $

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a) Find equation of line $ l $.

Slope $ m = \frac{5 - (-3)}{7 - 1} = \frac{8}{6} = \frac{4}{3} $

Use point-slope form with point $ A(1, -3) $:

→ $ y + 3 = \frac{4}{3}(x - 1) $

Multiply by 3:

→ $ 3y + 9 = 4x - 4 $

→ $ -4x + 3y + 13 = 0 $

Or rearranged: $ 4x - 3y - 13 = 0 $

✔ Equation of $ l $: $ \boxed{4x - 3y - 13 = 0} $

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b) Show that $ C $ (intersection of $ l $ and $ m $) is midpoint of $ AB $.

Line $ m: 4x + y - 17 = 0 $

We already have $ l: 4x - 3y - 13 = 0 $

Solve the system:

Equation 1: $ 4x + y = 17 $

Equation 2: $ 4x - 3y = 13 $

Subtract Eq2 from Eq1:

→ $ (4x + y) - (4x - 3y) = 17 - 13 $

→ $ 4y = 4 $ → $ y = 1 $

Plug into Eq1: $ 4x + 1 = 17 $ → $ 4x = 16 $ → $ x = 4 $

So, $ C = (4, 1) $

Now find midpoint of $ AB $:

$ A(1, -3), B(7, 5) $

Midpoint: $ \left( \frac{1+7}{2}, \frac{-3+5}{2} \right) = (4, 1) $

✔ So, $ C $ is indeed the midpoint of $ AB $.

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c) Show that the line perpendicular to $ m $ passing through $ C $ also passes through origin.

First, slope of $ m: 4x + y - 17 = 0 $ → $ y = -4x + 17 $, so slope = -4

Perpendicular slope = $ \frac{1}{4} $

Line through $ C(4, 1) $ with slope $ \frac{1}{4} $:

→ $ y - 1 = \frac{1}{4}(x - 4) $

→ $ y = \frac{1}{4}x - 1 + 1 = \frac{1}{4}x $

So, equation: $ y = \frac{1}{4}x $

Does this pass through origin? Plug in $ x=0 $, $ y=0 $: yes!

✔ So, it passes through origin.

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Question 3

Point $ A(-2, 7) $, point $ B(4, p) $, midpoint $ M(q, \frac{9}{2}) $

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a) Find values of constants $ p $ and $ q $.

Midpoint formula:

$ M_x = \frac{-2 + 4}{2} = \frac{2}{2} = 1 $ → so $ q = 1 $

$ M_y = \frac{7 + p}{2} = \frac{9}{2} $

→ $ 7 + p = 9 $ → $ p = 2 $

✔ So, $ \boxed{p = 2}, \boxed{q = 1} $

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b) Find equation of straight line perpendicular to $ AB $ which passes through $ A $. Give in form $ ax + by + c = 0 $.

Points $ A(-2, 7) $, $ B(4, 2) $

Slope of $ AB = \frac{2 - 7}{4 - (-2)} = \frac{-5}{6} $

So, perpendicular slope = $ \frac{6}{5} $

Equation through $ A(-2, 7) $:

→ $ y - 7 = \frac{6}{5}(x + 2) $

Multiply by 5:

→ $ 5y - 35 = 6x + 12 $

→ $ -6x + 5y - 47 = 0 $

Multiply by -1: $ 6x - 5y + 47 = 0 $

✔ Final answer: $ \boxed{6x - 5y + 47 = 0} $

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Question 4

Vertices: $ P(-5, -2), Q(-1, 6), R(7, 7), S(3, -1) $

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a) Find length of $ PQ $ in form $ k\sqrt{5} $.

Distance formula:

$ PQ = \sqrt{(-1 - (-5))^2 + (6 - (-2))^2} = \sqrt{(4)^2 + (8)^2} = \sqrt{16 + 64} = \sqrt{80} $

$ \sqrt{80} = \sqrt{16 \cdot 5} = 4\sqrt{5} $

✔ So, $ k = 4 $, answer: $ \boxed{4\sqrt{5}} $

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b) Coordinates of midpoint $ M $ of $ PQ $.

$ M = \left( \frac{-5 + (-1)}{2}, \frac{-2 + 6}{2} \right) = \left( \frac{-6}{2}, \frac{4}{2} \right) = (-3, 2) $

✔ Answer: $ \boxed{(-3, 2)} $

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c) Show that $ MS $ is perpendicular to $ PQ $.

We have $ M(-3, 2) $, $ S(3, -1) $

Slope of $ MS = \frac{-1 - 2}{3 - (-3)} = \frac{-3}{6} = -\frac{1}{2} $

Slope of $ PQ = \frac{6 - (-2)}{-1 - (-5)} = \frac{8}{4} = 2 $

Product of slopes: $ 2 \times (-\frac{1}{2}) = -1 $ → perpendicular.

✔ So, $ MS \perp PQ $

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d) Find area of parallelogram $ PQRS $.

We can use vector cross product or base × height.

Since we know vectors:

Vector $ \vec{PQ} = Q - P = (-1 - (-5), 6 - (-2)) = (4, 8) $

Vector $ \vec{PS} = S - P = (3 - (-5), -1 - (-2)) = (8, 1) $

Area = magnitude of cross product = $ |4 \cdot 1 - 8 \cdot 8| = |4 - 64| = |-60| = 60 $

✔ Area = $ \boxed{60} $

*(Note: In 2D, cross product magnitude = |x1y2 - x2y1|)*

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Question 5

Line $ l $ is parallel to $ 2x - y + 4 = 0 $, passes through $ (-1, -3) $

Line $ m $ is perpendicular to $ 6x + 5y - 2 = 0 $, passes through $ (4, 4) $

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a) Find equation of line $ l $.

Line $ 2x - y + 4 = 0 $ → $ y = 2x + 4 $, slope = 2

So, line $ l $ has slope 2, through $ (-1, -3) $

→ $ y + 3 = 2(x + 1) $

→ $ y + 3 = 2x + 2 $

→ $ -2x + y + 1 = 0 $

Or: $ 2x - y - 1 = 0 $

✔ Equation of $ l $: $ \boxed{2x - y - 1 = 0} $

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b) Find equation of line $ m $ in form $ ax + by + c = 0 $.

Line $ 6x + 5y - 2 = 0 $ → slope = $ -\frac{6}{5} $

Perpendicular slope = $ \frac{5}{6} $

Through point $ (4, 4) $:

→ $ y - 4 = \frac{5}{6}(x - 4) $

Multiply by 6:

→ $ 6y - 24 = 5x - 20 $

→ $ -5x + 6y - 4 = 0 $

Multiply by -1: $ 5x - 6y + 4 = 0 $

✔ Equation of $ m $: $ \boxed{5x - 6y + 4 = 0} $

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c) Find coordinates of intersection of $ l $ and $ m $ as exact fractions.

Solve:

- $ l: 2x - y = 1 $ → $ y = 2x - 1 $
- $ m: 5x - 6y = -4 $

Substitute $ y = 2x - 1 $ into second:

→ $ 5x - 6(2x - 1) = -4 $

→ $ 5x - 12x + 6 = -4 $

→ $ -7x = -10 $

→ $ x = \frac{10}{7} $

Then $ y = 2(\frac{10}{7}) - 1 = \frac{20}{7} - \frac{7}{7} = \frac{13}{7} $

✔ Intersection point: $ \boxed{\left( \frac{10}{7}, \frac{13}{7} \right)} $

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## ✔ FINAL ANSWERS SUMMARY:

1.

a) $ x - 2y - 8 = 0 $
b) $ (2, -3) $

2.

a) $ 4x - 3y - 13 = 0 $
b) Midpoint of AB is (4,1), same as C — shown
c) Perpendicular line through C is $ y = \frac{1}{4}x $, passes through origin — shown

3.

a) $ p = 2, q = 1 $
b) $ 6x - 5y + 47 = 0 $

4.

a) $ 4\sqrt{5} $
b) $ (-3, 2) $
c) Slopes multiply to -1 — shown perpendicular
d) Area = 60

5.

a) $ 2x - y - 1 = 0 $
b) $ 5x - 6y + 4 = 0 $
c) $ \left( \frac{10}{7}, \frac{13}{7} \right) $

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All problems solved with full working!