Coordinate Geometry Worksheet | PDF | Line (Geometry) | Perpendicular - Free Printable

Here are the step-by-step solutions for the problems in Worksheet D.

Question 1

Part a: Find the equation of line $m$.
1. Find the gradient of line $l$:
The equation of line $l$ is $y = 1 - 2x$, which can be rewritten as $y = -2x + 1$.
The gradient ($m_l$) is $-2$.

2. Find the gradient of line $m$:
Line $m$ is perpendicular to line $l$. Perpendicular lines have gradients that multiply to give $-1$ (negative reciprocals).
Gradient of $m$ ($m_m$) = $\frac{-1}{-2} = \frac{1}{2}$.

3. Find the equation:
Use the point-slope form $y - y_1 = m(x - x_1)$ with point $(6, -1)$ and gradient $\frac{1}{2}$.
$$y - (-1) = \frac{1}{2}(x - 6)$$
$$y + 1 = \frac{1}{2}x - 3$$
Multiply everything by 2 to remove the fraction:
$$2(y + 1) = 1(x - 6)$$
$$2y + 2 = x - 6$$
Rearrange into the form $ax + by + c = 0$:
$$x - 2y - 8 = 0$$

Part b: Find the intersection point.
1. We have two equations:
(1) $y = 1 - 2x$
(2) $x - 2y - 8 = 0 \Rightarrow x = 2y + 8$

2. Substitute equation (1) into equation (2):
$$x = 2(1 - 2x) + 8$$
$$x = 2 - 4x + 8$$
$$5x = 10$$
$$x = 2$$

3. Find $y$ using equation (1):
$$y = 1 - 2(2) = 1 - 4 = -3$$

The coordinates are $(2, -3)$.

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Question 2

Part a: Find an equation of line $l$.
1. Points are $A(1, -3)$ and $B(7, 5)$.
2. Calculate gradient ($m$):
$$m = \frac{5 - (-3)}{7 - 1} = \frac{8}{6} = \frac{4}{3}$$
3. Use point $A(1, -3)$ to find the equation:
$$y - (-3) = \frac{4}{3}(x - 1)$$
$$y + 3 = \frac{4}{3}x - \frac{4}{3}$$
Multiply by 3:
$$3y + 9 = 4x - 4$$
$$4x - 3y - 13 = 0$$

Part b: Show that $C$ is the mid-point of $AB$.
1. Line $m$ is $4x + y - 17 = 0$. Line $l$ is $4x - 3y - 13 = 0$.
2. Subtract equation $l$ from equation $m$ to eliminate $x$:
$$(4x + y - 17) - (4x - 3y - 13) = 0$$
$$4y - 4 = 0 \Rightarrow 4y = 4 \Rightarrow y = 1$$
3. Find $x$ using line $m$:
$$4x + 1 - 17 = 0 \Rightarrow 4x = 16 \Rightarrow x = 4$$
So, $C = (4, 1)$.
4. Calculate the mid-point of $AB$:
Mid-point $x = \frac{1+7}{2} = 4$
Mid-point $y = \frac{-3+5}{2} = 1$
Since the calculated intersection $(4,1)$ matches the mid-point formula result, $C$ is the mid-point.

Part c: Show the perpendicular line passes through the origin.
1. Gradient of line $m$ ($4x + y - 17 = 0 \Rightarrow y = -4x + 17$) is $-4$.
2. Gradient of the perpendicular line is $\frac{-1}{-4} = \frac{1}{4}$.
3. Equation passing through $C(4, 1)$:
$$y - 1 = \frac{1}{4}(x - 4)$$
$$4(y - 1) = x - 4$$
$$4y - 4 = x - 4$$
$$4y = x$$
4. Check if origin $(0,0)$ works:
$$4(0) = 0$$
It works, so the line passes through the origin.

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Question 3

Part a: Find constants $p$ and $q$.
1. $M(q, \frac{9}{2})$ is the mid-point of $A(-2, 7)$ and $B(4, p)$.
2. For x-coordinates:
$$q = \frac{-2 + 4}{2} = \frac{2}{2} = 1$$
3. For y-coordinates:
$$\frac{9}{2} = \frac{7 + p}{2}$$
Multiply by 2:
$$9 = 7 + p \Rightarrow p = 2$$
So, $p = 2$ and $q = 1$.

Part b: Find equation of perpendicular line through A.
1. Coordinates are now $A(-2, 7)$ and $B(4, 2)$.
2. Gradient of $AB$:
$$m_{AB} = \frac{2 - 7}{4 - (-2)} = \frac{-5}{6}$$
3. Perpendicular gradient:
$$m_{\perp} = \frac{-1}{-5/6} = \frac{6}{5}$$
4. Equation through $A(-2, 7)$:
$$y - 7 = \frac{6}{5}(x - (-2))$$
$$y - 7 = \frac{6}{5}(x + 2)$$
Multiply by 5:
$$5(y - 7) = 6(x + 2)$$
$$5y - 35 = 6x + 12$$
$$6x - 5y + 47 = 0$$

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Question 4

Part a: Length of $PQ$.
1. $P(-5, -2)$ and $Q(-1, 6)$.
2. Distance formula:
$$PQ = \sqrt{(-1 - (-5))^2 + (6 - (-2))^2}$$
$$PQ = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80}$$
3. Simplify $\sqrt{80}$:
$$\sqrt{16 \times 5} = 4\sqrt{5}$$
So, $k = 4$.

Part b: Coordinates of mid-point $M$ of $PQ$.
1. $$M_x = \frac{-5 + (-1)}{2} = -3$$
2. $$M_y = \frac{-2 + 6}{2} = 2$$
$M = (-3, 2)$.

Part c: Show $MS$ is perpendicular to $PQ$.
1. Gradient of $PQ$:
$$m_{PQ} = \frac{6 - (-2)}{-1 - (-5)} = \frac{8}{4} = 2$$
2. Gradient of $MS$ (using $M(-3, 2)$ and $S(3, -1)$):
$$m_{MS} = \frac{-1 - 2}{3 - (-3)} = \frac{-3}{6} = -\frac{1}{2}$$
3. Product of gradients:
$$2 \times (-\frac{1}{2}) = -1$$
Therefore, they are perpendicular.

Part d: Area of parallelogram $PQRS$.
1. Since $MS \perp PQ$, $MS$ is the height relative to base $PQ$.
2. Length of base $PQ = \sqrt{80} = 4\sqrt{5}$.
3. Length of height $MS$:
$$MS = \sqrt{(3 - (-3))^2 + (-1 - 2)^2} = \sqrt{6^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$$
4. Area = Base $\times$ Height:
$$Area = (4\sqrt{5}) \times (3\sqrt{5}) = 12 \times 5 = 60$$

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Question 5

Part a: Equation of line $l$.
1. Line $l$ is parallel to $2x - y + 4 = 0$.
Rewrite given line: $y = 2x + 4$. Gradient is $2$.
2. Line $l$ has gradient $2$ and passes through $(-1, -3)$.
$$y - (-3) = 2(x - (-1))$$
$$y + 3 = 2(x + 1)$$
$$y + 3 = 2x + 2$$
$$2x - y - 1 = 0$$

Part b: Equation of line $m$.
1. Line $m$ is perpendicular to $6x + 5y - 2 = 0$.
Rewrite given line: $5y = -6x + 2 \Rightarrow y = -\frac{6}{5}x + \dots$ Gradient is $-\frac{6}{5}$.
2. Perpendicular gradient is $\frac{5}{6}$.
3. Passes through $(4, 4)$.
$$y - 4 = \frac{5}{6}(x - 4)$$
Multiply by 6:
$$6(y - 4) = 5(x - 4)$$
$$6y - 24 = 5x - 20$$
$$5x - 6y + 4 = 0$$

Part c: Intersection of $l$ and $m$.
1. Equations:
(1) $2x - y = 1 \Rightarrow y = 2x - 1$
(2) $5x - 6y = -4$
2. Substitute (1) into (2):
$$5x - 6(2x - 1) = -4$$
$$5x - 12x + 6 = -4$$
$$-7x = -10$$
$$x = \frac{10}{7}$$
3. Find $y$:
$$y = 2(\frac{10}{7}) - 1 = \frac{20}{7} - \frac{7}{7} = \frac{13}{7}$$

Final Answer:
1a) $x - 2y - 8 = 0$
1b) $(2, -3)$
2a) $4x - 3y - 13 = 0$
2b) Intersection is $(4,1)$, Midpoint is $(4,1)$. They match.
2c) Equation is $x - 4y = 0$. Substituting $(0,0)$ gives $0=0$.
3a) $p = 2, q = 1$
3b) $6x - 5y + 47 = 0$
4a) $4\sqrt{5}$
4b) $(-3, 2)$
4c) Gradient $PQ = 2$, Gradient $MS = -0.5$. Product is $-1$.
4d) $60$
5a) $2x - y - 1 = 0$
5b) $5x - 6y + 4 = 0$
5c) $(\frac{10}{7}, \frac{13}{7})$