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Wyatt's Notes — DSE

Geometries

Coordinate geometry provides a bridge between algebraic equations and geometric figures, enabling The use of algebraic methods to solve geometric problems and vice versa.

2D Coordinate Geometry

The Rectangular Coordinate System

In the Cartesian plane, every point PP is uniquely identified by an ordered pair (x,y)(x, y)Where xx is the horizontal coordinate (abscissa) and yy is the vertical coordinate (ordinate). The axes Divide the plane into four quadrants, numbered counterclockwise from the positive xx-axis.

Distance Formula

The distance between two points P1(x1,y1)P_1(x_1, y_1) and P2(x2,y2)P_2(x_2, y_2) is derived from the Pythagorean identity applied to the right triangle formed By the horizontal and vertical differences:

d=(x2−x1)2+(y2−y1)2\begin{aligned} d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \end{aligned}
Examples- Distance between $A(3, 4)$ and $B(7, 1)$: $d = \sqrt{(7-3)^2 + (1-4)^2} = \sqrt{16 + 9} = 5$ - Distance between $O(0, 0)$ and $P(5, 12)$: $d = \sqrt{5^2 + 12^2} = 13$ - Verify that the points $A(0, 0)$$B(3, 4)$$C(6, 8)$ are collinear: $AB = 5$$BC = 5$$AC = 10$. Since $AB + BC = AC$The points are collinear.

Midpoint Formula

The midpoint MM of the line segment joining P1(x1,y1)P_1(x_1, y_1) and P2(x2,y2)P_2(x_2, y_2) is:

M=(x1+x22,  y1+y22)\begin{aligned} M = \left( \frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2} \right) \end{aligned}

The midpoint is the average of the respective coordinates of the two endpoints.

Examples- Midpoint of $A(2, 6)$ and $B(8, -4)$: $M = \left( \frac{2+8}{2}, \frac{6+(-4)}{2} \right) = (5, 1)$ - If $M(4, -1)$ is the midpoint of $A(1, 3)$ and $B$Then $B = (2 \times 4 - 1, 2 \times (-1) - 3) = (7, -5)$

Gradient (Slope)

The gradient mm of the line passing through P1(x1,y1)P_1(x_1, y_1) and P2(x2,y2)P_2(x_2, y_2)Where x1≠x2x_1 \neq x_2Measures the rate of change of yy with respect to xx:

m=y2−y1x2−x1=tan⁡θ\begin{aligned} m = \frac{y_2 - y_1}{x_2 - x_1} = \tan\theta \end{aligned}

Where θ\theta is the angle the line makes with the positive xx-axis. A vertical line (where x1=x2x_1 = x_2) has an undefined gradient. Key gradient values include:

GradientInterpretation
m>0m > 0Line slopes upward from left to right
m<0m < 0Line slopes downward from left to right
m=0m = 0Horizontal line (y=cy = c)
UndefinedVertical line (x=cx = c)
Examples- Gradient of the line through $A(1, 2)$ and $B(4, 8)$: $m = \frac{8-2}{4-1} = 2$ - A line with gradient $\frac{3}{4}$ passing through $(1, 5)$: using $y - y_1 = m(x - x_1)$The equation is $y - 5 = \frac{3}{4}(x - 1)$I.e., $3x - 4y + 17 = 0$

Division of a Line Segment

If point P(x,y)P(x, y) divides the line segment joining A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) in the ratio m:nm : n (internally), then:

x=mx2+nx1m+n,y=my2+ny1m+n\begin{aligned} x = \frac{mx_2 + nx_1}{m + n}, \qquad y = \frac{my_2 + ny_1}{m + n} \end{aligned}

The midpoint formula is the special case where m=n=1m = n = 1.

Equations of Straight Lines

A straight line in the Cartesian plane can be expressed in several equivalent forms.

Slope-Intercept Form

y=mx+c\begin{aligned} y = mx + c \end{aligned}

Where mm is the gradient and cc is the yy-intercept. This form is most useful when the gradient And yy-intercept are known or can be determined.

Point-Slope Form

Given a point (x1,y1)(x_1, y_1) on the line and the gradient mm:

y−y1=m(x−x1)\begin{aligned} y - y_1 = m(x - x_1) \end{aligned}

This is the most versatile form for deriving the equation of a line from partial information.

Two-Point Form

Given two distinct points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) on the line:

y−y1x−x1=y2−y1x2−x1\begin{aligned} \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1} \end{aligned}

Intercept Form

If the line has xx-intercept aa and yy-intercept bb (both non-zero):

xa+yb=1\begin{aligned} \frac{x}{a} + \frac{y}{b} = 1 \end{aligned}

General Form

Every straight line can be written in the general form:

Ax+By+C=0\begin{aligned} Ax + By + C = 0 \end{aligned}

Where A$$B$$C are constants and AA and BB are not both zero. The gradient is m=−ABm = -\frac{A}{B} and the yy-intercept is c=−CBc = -\frac{C}{B} (when B≠0B \neq 0).

Examples- Find the equation of the line through $(2, 3)$ and $(6, -1)$: $m = \frac{-1 - 3}{6 - 2} = -1$. Equation: $y - 3 = -1(x - 2)$I.e., $x + y - 5 = 0$ - Find the equation of the line with gradient $\frac{1}{2}$ and $y$-intercept $-3$: $y = \frac{1}{2}x - 3$I.e., $x - 2y - 6 = 0$ - Convert $3x + 4y - 12 = 0$ to intercept form: $\frac{x}{4} + \frac{y}{3} = 1$. The $x$-intercept is $4$ and the $y$-intercept is $3$.

Parallel and Perpendicular Lines

Parallel Lines

Two non-vertical lines with gradients m1m_1 and m2m_2 are parallel if and only if their gradients Are equal:

m1=m2\begin{aligned} m_1 = m_2 \end{aligned}

In the general form Ax+By+C=0Ax + By + C = 0Two lines A1x+B1y+C1=0A_1x + B_1y + C_1 = 0 and A2x+B2y+C2=0A_2x + B_2y + C_2 = 0 Are parallel if and only if:

A1A2=B1B2≠C1C2\begin{aligned} \frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2} \end{aligned}

If all three ratios are equal, the lines are coincident (the same line).

Perpendicular Lines

Two non-vertical lines with gradients m1m_1 and m2m_2 are perpendicular if and only if the product Of their gradients equals −1-1:

m1⋅m2=−1\begin{aligned} m_1 \cdot m_2 = -1 \end{aligned}

In the general form, two lines are perpendicular if and only if:

A1A2+B1B2=0\begin{aligned} A_1 A_2 + B_1 B_2 = 0 \end{aligned}
Examples- Find the equation of the line through $(1, -2)$ parallel to $2x - 3y + 5 = 0$: The gradient of the given line is $m = \frac{2}{3}$. The parallel line is $y + 2 = \frac{2}{3}(x - 1)$I.e., $2x - 3y - 8 = 0$. - Find the equation of the line through $(3, 1)$ perpendicular to $4x + y - 7 = 0$: The gradient of the given line is $m_1 = -4$So $m_2 = \frac{1}{4}$. The perpendicular line is $y - 1 = \frac{1}{4}(x - 3)$I.e., $x - 4y + 1 = 0$. - Determine whether $3x + 2y - 1 = 0$ and $6x + 4y + 5 = 0$ are parallel: Since $\frac{3}{6} = \frac{2}{4} = \frac{1}{2} \neq \frac{-1}{5}$The lines are parallel but not coincident.

Intersection of Two Lines

The point of intersection of two non-parallel lines A1x+B1y+C1=0A_1x + B_1y + C_1 = 0 and A2x+B2y+C2=0A_2x + B_2y + C_2 = 0 is found by solving the system simultaneously. The coordinates (x,y)(x, y) of The intersection satisfy both equations.

Examples- Find the intersection of $x + y = 5$ and $2x - y = 1$: Adding gives $3x = 6$So $x = 2$$y = 3$. The intersection is $(2, 3)$.

Circles

Standard Form of the Equation of a Circle

A circle with centre C(a,b)C(a, b) and radius rr has equation:

(x−a)2+(y−b)2=r2\begin{aligned} (x - a)^2 + (y - b)^2 = r^2 \end{aligned}

Expanding this yields the general form:

x2+y2−2ax−2by+(a2+b2−r2)=0\begin{aligned} x^2 + y^2 - 2ax - 2by + (a^2 + b^2 - r^2) = 0 \end{aligned}

General Form of the Equation of a Circle

x2+y2+Dx+Ey+F=0\begin{aligned} x^2 + y^2 + Dx + Ey + F = 0 \end{aligned}

Where the centre is (−D2,−E2)\left(-\frac{D}{2}, -\frac{E}{2}\right) and the radius is:

r=(D2)2+(E2)2−F\begin{aligned} r = \sqrt{\left(\frac{D}{2}\right)^2 + \left(\frac{E}{2}\right)^2 - F} \end{aligned}

For a real circle to exist, we require (D2)2+(E2)2−F>0\left(\frac{D}{2}\right)^2 + \left(\frac{E}{2}\right)^2 - F > 0.

Examples- Find the centre and radius of $x^2 + y^2 - 6x + 4y - 12 = 0$: Completing squares, $(x-3)^2 + (y+2)^2 = 25$. Centre $(3, -2)$Radius $5$. - Find the equation of the circle with centre $(-1, 4)$ and radius $3$: $(x+1)^2 + (y-4)^2 = 9$. - Find the equation of the circle with diameter endpoints $A(2, 3)$ and $B(8, 7)$: Centre is $M(5, 5)$Radius is $\frac{1}{2}\sqrt{(8-2)^2 + (7-3)^2} = \frac{1}{2}\sqrt{52} = \sqrt{13}$. Equation: $(x-5)^2 + (y-5)^2 = 13$.

Finding the Equation of a Circle from Conditions

The equation of a circle can be determined when given:

  • The centre and the radius (direct substitution)
  • The centre and a point on the circle (use the distance formula to find rr)
  • Three points on the circle (solve the general form system)
  • The endpoints of a diameter (centre is the midpoint, radius is half the distance)

Intersection of a Line and a Circle

To find the points of intersection of the line y=mx+cy = mx + c and the circle (x−a)2+(y−b)2=r2(x-a)^2 + (y-b)^2 = r^2Substitute the line equation into the circle equation to obtain a Quadratic in xx (or yy). The discriminant Δ=b2−4ac\Delta = b^2 - 4ac determines the nature of Intersection:

DiscriminantIntersection
Δ>0\Delta > 0Two distinct points (secant)
Δ=0\Delta = 0One point (tangent)
Δ<0\Delta < 0No real intersection

Tangent to a Circle

A tangent to a circle at a point P(x1,y1)P(x_1, y_1) on the circle (x−a)2+(y−b)2=r2(x-a)^2 + (y-b)^2 = r^2 is Perpendicular to the radius at PP. The gradient of the radius CPCP (where CC is the centre) is:

mradius=y1−bx1−a\begin{aligned} m_{\mathrm{radius}} = \frac{y_1 - b}{x_1 - a} \end{aligned}

Therefore the gradient of the tangent is:

mtangent=−x1−ay1−b(y1≠b)\begin{aligned} m_{\mathrm{tangent}} = -\frac{x_1 - a}{y_1 - b} \quad (y_1 \neq b) \end{aligned}

The equation of the tangent at P(x1,y1)P(x_1, y_1) on the circle x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0 can be Obtained by replacing:

x2→xx1,y2→yy1,x→x+x12,y→y+y12\begin{aligned} x^2 \to xx_1, \quad y^2 \to yy_1, \quad x \to \frac{x + x_1}{2}, \quad y \to \frac{y + y_1}{2} \end{aligned}

This gives:

xx1+yy1+D(x+x1)2+E(y+y1)2+F=0\begin{aligned} xx_1 + yy_1 + \frac{D(x + x_1)}{2} + \frac{E(y + y_1)}{2} + F = 0 \end{aligned}
Examples- Find the equation of the tangent to $x^2 + y^2 = 25$ at $P(3, 4)$: The radius gradient is $\frac{4}{3}$So the tangent gradient is $-\frac{3}{4}$. Equation: $y - 4 = -\frac{3}{4}(x - 3)$I.e., $3x + 4y - 25 = 0$. Alternatively, using the formula: $3x + 4y = 25$. - Determine whether the line $3x - 4y + 10 = 0$ is tangent to $(x-1)^2 + (y+2)^2 = 9$: Substitute $y = \frac{3x+10}{4}$ into the circle. The resulting quadratic has discriminant $\Delta = 0$Confirming tangency.

2D Measurements

Triangles

For a triangle with base bb and height hh:

Area=12bh\begin{aligned} \mathrm{Area} &= \frac{1}{2}bh \end{aligned}

Using the Heron”s formula) for a triangle with side lengths a$$b$$c and Semi-perimeter s=a+b+c2s = \frac{a+b+c}{2}:

Area=s(s−a)(s−b)(s−c)\begin{aligned} \mathrm{Area} = \sqrt{s(s-a)(s-b)(s-c)} \end{aligned}

The area of a triangle with vertices (x_1, y_1)$$(x_2, y_2)$$(x_3, y_3) is given by:

Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\begin{aligned} \mathrm{Area} = \frac{1}{2}\left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \end{aligned}

Quadrilaterals

FigureAreaPerimeter
Rectangle (l×wl \times w)lwlw2(l+w)2(l+w)
Parallelogram (b$$h)bhbh2(a+b)2(a+b)
Rhombus (diagonals d_1$$d_2)12d1d2\frac{1}{2}d_1 d_24s4s
Trapezium (parallel sides a$$b; height hh)12(a+b)h\frac{1}{2}(a+b)hSum of all sides
Square (side ss)s2s^24s4s

Circles

Circumference=2πr=πdArea=πr2\begin{aligned} \mathrm{Circumference} &= 2\pi r = \pi d \\ \mathrm{Area} &= \pi r^2 \end{aligned}

Where rr is the radius and d=2rd = 2r is the diameter.

Arcs and Sectors

For an arc subtending an angle θ\theta (in radians) at the centre of a circle of radius rr:

Arclength=l=rθAreaofsector=A=12r2θ\begin{aligned} \mathrm{Arc length} &= l = r\theta \\ \mathrm{Area of sector} &= A = \frac{1}{2}r^2\theta \end{aligned}

When the angle is given in degrees (θ∘\theta^\circ), first convert to radians: θ=θ∘×π180\theta = \theta^\circ \times \frac{\pi}{180}.

The area of a segment (the region between a chord and the corresponding arc) is:

Areaofsegment=12r2(θ−sin⁡θ)\begin{aligned} \mathrm{Area of segment} = \frac{1}{2}r^2(\theta - \sin\theta) \end{aligned}
Examples- Find the area and perimeter of a sector of radius $7\mathrm{ cm}$ with angle $120^\circ$: $\theta = \frac{2\pi}{3}$ rad. Area $= \frac{1}{2}(49)\left(\frac{2\pi}{3}\right) = \frac{49\pi}{3} \approx 51.3\mathrm{ cm}^2$. Arc length $= 7 \times \frac{2\pi}{3} = \frac{14\pi}{3} \approx 14.7\mathrm{ cm}$. Perimeter $= 14 + \frac{14\pi}{3} \approx 28.7\mathrm{ cm}$. - A chord of length $8\mathrm{ cm}$ subtends an angle of $90^\circ$ at the centre. Find the area of the minor segment: $r = \frac{8}{\sqrt{2}} = 4\sqrt{2}\mathrm{ cm}$$\theta = \frac{\pi}{2}$. Area $= \frac{1}{2}(32)\left(\frac{\pi}{2} - 1\right) = 16\left(\frac{\pi}{2} - 1\right) = 8\pi - 16 \approx 9.13\mathrm{ cm}^2$.

3D Geometry

The 3D Coordinate System

In three-dimensional space, each point is identified by an ordered triple (x,y,z)(x, y, z). The three Coordinate axes --- xx-axis, yy-axis, and zz-axis --- are mutually perpendicular and intersect at The origin O(0,0,0)O(0, 0, 0). The three planes xyxy-plane, yzyz-plane, and zxzx-plane divide the space Into eight octants.

Distance Formula in 3D

The distance between two points P1(x1,y1,z1)P_1(x_1, y_1, z_1) and P2(x2,y2,z2)P_2(x_2, y_2, z_2) in 3D space is:

d=(x2−x1)2+(y2−y1)2+(z2−z1)2\begin{aligned} d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \end{aligned}

The midpoint of the segment P1P2P_1P_2 is:

M=(x1+x22,  y1+y22,  z1+z22)\begin{aligned} M = \left( \frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2},\; \frac{z_1 + z_2}{2} \right) \end{aligned}
Examples- Distance between $A(1, 2, 3)$ and $B(4, 6, -3)$: $d = \sqrt{3^2 + 4^2 + (-6)^2} = \sqrt{9 + 16 + 36} = \sqrt{61}$ - Show that $A(0, 0, 0)$$B(1, 0, 0)$$C(0, 1, 0)$$D(0, 0, 1)$ form a **trirectangular tetrahedron**: $AB = AC = AD = 1$ (edges from the origin) and $BC = BD = CD = \sqrt{2}$ (edges of the base triangle $BCD$). It is not a regular tetrahedron — a regular tetrahedron would require all six edges to be equal. For a regular tetrahedron, use vertices such as $(1,1,1)$$(1,-1,-1)$$(-1,1,-1)$$(-1,-1,1)$.

Equation of a Plane

The general equation of a plane in 3D is:

Ax+By+Cz+D=0\begin{aligned} Ax + By + Cz + D = 0 \end{aligned}

Where the vector n=(A,B,C)\mathbf{n} = (A, B, C) is a normal vector to the plane. The plane can also be Described in several forms:

  • Point-normal form: Given a point P0(x0,y0,z0)P_0(x_0, y_0, z_0) on the plane and normal n=(A,B,C)\mathbf{n} = (A, B, C):
A(x−x0)+B(y−y0)+C(z−z0)=0\begin{aligned} A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \end{aligned}
  • Intercept form: If the plane has xx-intercept a$$y-intercept bbAnd zz-intercept cc (all non-zero):
xa+yb+zc=1\begin{aligned} \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \end{aligned}

Angles Between Lines and Planes

Angle Between a Line and a Plane

The angle ϕ\phi between a line with direction vector d=(l,m,n)\mathbf{d} = (l, m, n) and a plane with Normal n=(A,B,C)\mathbf{n} = (A, B, C) is defined as the complement of the angle between d\mathbf{d} and n\mathbf{n}:

sin⁡ϕ=∣Al+Bm+Cn∣A2+B2+C2l2+m2+n2\begin{aligned} \sin\phi = \frac{|Al + Bm + Cn|}{\sqrt{A^2 + B^2 + C^2}\sqrt{l^2 + m^2 + n^2}} \end{aligned}

Angle Between Two Planes

The angle θ\theta between two planes with normals n1=(A1,B1,C1)\mathbf{n_1} = (A_1, B_1, C_1) and n2=(A2,B2,C2)\mathbf{n_2} = (A_2, B_2, C_2) is:

cos⁡θ=∣A1A2+B1B2+C1C2∣A12+B12+C12A22+B22+C22\begin{aligned} \cos\theta = \frac{|A_1 A_2 + B_1 B_2 + C_1 C_2|}{\sqrt{A_1^2 + B_1^2 + C_1^2}\sqrt{A_2^2 + B_2^2 + C_2^2}} \end{aligned}

Angle Between Two Lines in 3D

For two lines with direction vectors d1=(l1,m1,n1)\mathbf{d_1} = (l_1, m_1, n_1) and d2=(l2,m2,n2)\mathbf{d_2} = (l_2, m_2, n_2):

cos⁡θ=∣l1l2+m1m2+n1n2∣l12+m12+n12l22+m22+n22\begin{aligned} \cos\theta = \frac{|l_1 l_2 + m_1 m_2 + n_1 n_2|}{\sqrt{l_1^2 + m_1^2 + n_1^2}\sqrt{l_2^2 + m_2^2 + n_2^2}} \end{aligned}

Distance from a Point to a Plane

The perpendicular distance from point P(x0,y0,z0)P(x_0, y_0, z_0) to the plane Ax+By+Cz+D=0Ax + By + Cz + D = 0 is:

d=∣Ax0+By0+Cz0+D∣A2+B2+C2\begin{aligned} d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \end{aligned}
Examples- Find the distance from $P(1, 2, -1)$ to the plane $2x - y + 2z + 3 = 0$: $d = \frac{|2(1) - 1(2) + 2(-1) + 3|}{\sqrt{4 + 1 + 4}} = \frac{|2 - 2 - 2 + 3|}{3} = \frac{1}{3}$. - Find the angle between the planes $x + y + z = 1$ and $2x - y + z = 0$: $\cos\theta = \frac{|1\cdot 2 + 1\cdot(-1) + 1\cdot 1|}{\sqrt{3}\sqrt{6}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3}$. So $\theta = \arccos\left(\frac{\sqrt{2}}{3}\right) \approx 61.9^\circ$. - Find the equation of the plane through $(1, 0, 2)$$(0, 1, -1)$And $(2, 1, 1)$: direction vectors $\overrightarrow{AB} = (-1, 1, -3)$ and $\overrightarrow{AC} = (1, 1, -1)$. Cross product: $\overrightarrow{AB} \times \overrightarrow{AC} = (2, -4, -2)$So normal $\mathbf{n} = (1, -2, -1)$. The plane is $x - 2y - z = -1$.
Wrap-up Questions1. **Question:** Find the equation of the perpendicular bisector of the line segment joining $A(3, -1)$ and $B(7, 5)$. ### DetailsAnswer- Midpoint: $M = \left(\frac{3+7}{2}, \frac{-1+5}{2}\right) = (5, 2)$. - Gradient of $AB$: $m_{AB} = \frac{5 - (-1)}{7 - 3} = \frac{3}{2}$. - Gradient of perpendicular bisector: $m = -\frac{2}{3}$. - Equation: $y - 2 = -\frac{2}{3}(x - 5)$I.e., $2x + 3y - 16 = 0$.
  1. Question: Find the equation of the circle passing through the three points A(0,0)A(0, 0) B(4,0)B(4, 0)And C(0,3)C(0, 3).
Answer
  • Let the circle be x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0.
  • Substituting A(0,0)A(0,0): F=0F = 0.
  • Substituting B(4,0)B(4,0): 16+4D=0  ⟹  D=−416 + 4D = 0 \implies D = -4.
  • Substituting C(0,3)C(0,3): 9+3E=0  ⟹  E=−39 + 3E = 0 \implies E = -3.
  • Equation: x2+y2−4x−3y=0x^2 + y^2 - 4x - 3y = 0I.e., (x−2)2+(y−32)2=254(x-2)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{25}{4}.
  • Centre (2,32)\left(2, \frac{3}{2}\right)Radius 52\frac{5}{2}.
  1. Question: The line y=2x+ky = 2x + k is tangent to the circle x2+y2−4x−2y+1=0x^2 + y^2 - 4x - 2y + 1 = 0. Find The value(s) of kk.
Answer
  • Substitute y=2x+ky = 2x + k into (x−2)2+(y−1)2=4(x-2)^2 + (y-1)^2 = 4:
  • (x−2)2+(2x+k−1)2=4(x-2)^2 + (2x + k - 1)^2 = 4.
  • Expanding: x2−4x+4+4x2+4(k−1)x+(k−1)2=4x^2 - 4x + 4 + 4x^2 + 4(k-1)x + (k-1)^2 = 4.
  • 5x2+(4k−8)x+(k2−2k+1)=05x^2 + (4k - 8)x + (k^2 - 2k + 1) = 0.
  • For tangency, Δ=0\Delta = 0: (4k−8)2−4(5)(k2−2k+1)=0(4k-8)^2 - 4(5)(k^2 - 2k + 1) = 0.
  • 16k2−64k+64−20k2+40k−20=016k^2 - 64k + 64 - 20k^2 + 40k - 20 = 0.
  • −4k2−24k+44=0  ⟹  k2+6k−11=0-4k^2 - 24k + 44 = 0 \implies k^2 + 6k - 11 = 0.
  • k=−6±36+442=−3±14k = \frac{-6 \pm \sqrt{36 + 44}}{2} = -3 \pm \sqrt{14}.
  1. Question: A sector of a circle of radius 12cm12\mathrm{ cm} has an arc length of 16πcm16\pi\mathrm{ cm}. Find the area of the sector and the area of the corresponding segment if the Chord length is 20cm20\mathrm{ cm}.
Answer
  • Arc length l=rθl = r\theta: 16π=12θ  ⟹  θ=4π316\pi = 12\theta \implies \theta = \frac{4\pi}{3} rad.
  • Area of sector: A=12r2θ=12(144)(4π3)=96πcm2A = \frac{1}{2}r^2\theta = \frac{1}{2}(144)\left(\frac{4\pi}{3}\right) = 96\pi \mathrm{ cm}^2.
  • Area of triangle formed by the radii and chord: Using the chord length c=20c = 20The triangle has sides 12$$12$$20. Semi-perimeter s=22s = 22. By Heron’s formula: Area =22×10×10×2=4400=2011cm2= \sqrt{22 \times 10 \times 10 \times 2} = \sqrt{4400} = 20\sqrt{11} \mathrm{ cm}^2.
  • Alternatively, the perpendicular from centre to chord: h=122−102=211h = \sqrt{12^2 - 10^2} = 2\sqrt{11}. Triangle area =12×20×211=2011= \frac{1}{2} \times 20 \times 2\sqrt{11} = 20\sqrt{11}.
  • Area of segment =96π−2011≈247.3cm2= 96\pi - 20\sqrt{11} \approx 247.3 \mathrm{ cm}^2.
  1. Question: Find the angle between the line joining A(1,2,3)A(1, 2, 3) and B(4,5,6)B(4, 5, 6) and the plane x+y+z=1x + y + z = 1.
Answer
  • Direction vector of the line: d=(4−1,5−2,6−3)=(3,3,3)\mathbf{d} = (4-1, 5-2, 6-3) = (3, 3, 3)Simplified to (1,1,1)(1, 1, 1).
  • Normal of the plane: n=(1,1,1)\mathbf{n} = (1, 1, 1).
  • The angle α\alpha between d\mathbf{d} and n\mathbf{n}: cos⁡α=∣1+1+1∣33=33=1\cos\alpha = \frac{|1+1+1|}{\sqrt{3}\sqrt{3}} = \frac{3}{3} = 1So α=0∘\alpha = 0^\circ.
  • The angle ϕ\phi between the line and the plane is the complement: ϕ=90∘−0∘=90∘\phi = 90^\circ - 0^\circ = 90^\circ.
  • The line is parallel to the plane (perpendicular to the normal).
  1. Question: Points A(1, 2)$$B(4, 6)And C(7,4)C(7, 4) are vertices of a triangle. Find the Equation of the altitude from AA to BCBCThe area of triangle ABCABCAnd the length of the Altitude from AA.
Answer
  • Gradient of BCBC: mBC=4−67−4=−23m_{BC} = \frac{4-6}{7-4} = -\frac{2}{3}.
  • Gradient of altitude from AA: m=32m = \frac{3}{2} (negative reciprocal).
  • Equation of altitude: y−2=32(x−1)y - 2 = \frac{3}{2}(x - 1)I.e., 3x−2y+1=03x - 2y + 1 = 0.
  • Area of △ABC\triangle ABC: 12∣1(6−4)+4(4−2)+7(2−6)∣=12∣2+8−28∣=12(18)=9\frac{1}{2}|1(6-4) + 4(4-2) + 7(2-6)| = \frac{1}{2}|2 + 8 - 28| = \frac{1}{2}(18) = 9 square units.
  • Length of BCBC: (7−4)2+(4−6)2=9+4=13\sqrt{(7-4)^2 + (4-6)^2} = \sqrt{9 + 4} = \sqrt{13}.
  • Altitude from AA: 2×AreaBC=1813=181313\frac{2 \times \mathrm{Area}}{BC} = \frac{18}{\sqrt{13}} = \frac{18\sqrt{13}}{13}.
  1. Question: Find the equation of the circle which touches the xx-axis at (3,0)(3, 0) and passes Through (1,4)(1, 4).
Answer
  • Since the circle touches the xx-axis at (3,0)(3, 0)The centre lies on the vertical line x=3x = 3 so the centre is C(3,r)C(3, r) where rr is the radius.
  • The circle equation is (x−3)2+(y−r)2=r2(x-3)^2 + (y-r)^2 = r^2.
  • Substituting (1,4)(1, 4): (1−3)2+(4−r)2=r2(1-3)^2 + (4-r)^2 = r^2.
  • 4+16−8r+r2=r2  ⟹  20−8r=0  ⟹  r=524 + 16 - 8r + r^2 = r^2 \implies 20 - 8r = 0 \implies r = \frac{5}{2}.
  • Centre: (3,52)(3, \frac{5}{2})Radius: 52\frac{5}{2}.
  • Equation: (x−3)2+(y−52)2=254(x-3)^2 + \left(y - \frac{5}{2}\right)^2 = \frac{25}{4}I.e., x2+y2−6x−5y+9=0x^2 + y^2 - 6x - 5y + 9 = 0.
  1. Question: The vertices of a triangle are A(2, 1, -1)$$B(0, 3, -4)And C(5,0,2)C(5, 0, 2). Find (a) the length of each side, (b) the area of the triangle, and (c) the equation of the plane Containing the triangle.
Answer
  • (a) Side lengths:
  • AB=(0−2)2+(3−1)2+(−4+1)2=4+4+9=17AB = \sqrt{(0-2)^2 + (3-1)^2 + (-4+1)^2} = \sqrt{4 + 4 + 9} = \sqrt{17}
  • BC=(5−0)2+(0−3)2+(2+4)2=25+9+36=70BC = \sqrt{(5-0)^2 + (0-3)^2 + (2+4)^2} = \sqrt{25 + 9 + 36} = \sqrt{70}
  • CA=(2−5)2+(1−0)2+(−1−2)2=9+1+9=19CA = \sqrt{(2-5)^2 + (1-0)^2 + (-1-2)^2} = \sqrt{9 + 1 + 9} = \sqrt{19}
  • (b) Using the cross product method: AB→=(−2,2,−3)\overrightarrow{AB} = (-2, 2, -3) AC→=(3,−1,3)\overrightarrow{AC} = (3, -1, 3).
  • AB→×AC→=(2⋅3−(−3)(−1),  (−3)⋅3−(−2)⋅3,  (−2)(−1)−2⋅3)=(3,−3,−4)\overrightarrow{AB} \times \overrightarrow{AC} = (2 \cdot 3 - (-3)(-1),\; (-3) \cdot 3 - (-2) \cdot 3,\; (-2)(-1) - 2 \cdot 3) = (3, -3, -4).
  • Area =12∣AB→×AC→∣=129+9+16=342= \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2}\sqrt{9 + 9 + 16} = \frac{\sqrt{34}}{2}.
  • (c) The normal vector is (3,−3,−4)(3, -3, -4). Using point A(2,1,−1)A(2, 1, -1): 3(x−2)−3(y−1)−4(z+1)=03(x-2) - 3(y-1) - 4(z+1) = 0 I.e., 3x−3y−4z−8=03x - 3y - 4z - 8 = 0.
  1. Question: Two circles C1:x2+y2−4x+2y−4=0C_1: x^2 + y^2 - 4x + 2y - 4 = 0 and C2:x2+y2+2x−6y+6=0C_2: x^2 + y^2 + 2x - 6y + 6 = 0 intersect at points AA and BB. Find the equation of the common Chord ABAB and the coordinates of AA and BB.
Answer
  • Common chord: Subtract the two equations: (−4x+2y−4)−(2x−6y+6)=0(-4x + 2y - 4) - (2x - 6y + 6) = 0.
  • −6x+8y−10=0-6x + 8y - 10 = 0I.e., 3x−4y+5=03x - 4y + 5 = 0.
  • From C1C_1: (x−2)2+(y+1)2=9(x-2)^2 + (y+1)^2 = 9. Substitute y=3x+54y = \frac{3x+5}{4}:
  • (x−2)2+(3x+54+1)2=9(x-2)^2 + \left(\frac{3x+5}{4} + 1\right)^2 = 9.
  • (x−2)2+(3x+94)2=9(x-2)^2 + \left(\frac{3x+9}{4}\right)^2 = 9.
  • Multiply by 1616: 16(x−2)2+(3x+9)2=14416(x-2)^2 + (3x+9)^2 = 144.
  • 16(x2−4x+4)+9x2+54x+81=14416(x^2 - 4x + 4) + 9x^2 + 54x + 81 = 144.
  • 25x2−10x+145=14425x^2 - 10x + 145 = 144.
  • 25x2−10x+1=025x^2 - 10x + 1 = 0.
  • Discriminant =100−100=0= 100 - 100 = 0So the circles are tangent (touch at exactly one point).
  • x = \frac{10}{50} = \frac{1}{5}$$y = \frac{3(1/5) + 5}{4} = \frac{28}{20} = \frac{7}{5}.
  • The circles touch at (15,75)\left(\frac{1}{5}, \frac{7}{5}\right).
  1. Question: A rectangle ABCDABCD has vertices A(1, 1)$$B(5, 1)And C(5,4)C(5, 4). Find the Coordinates of DD and the area of the rectangle.
Answer
  • ABAB is horizontal: AB=5−1=4AB = 5 - 1 = 4. BCBC is vertical: BC=4−1=3BC = 4 - 1 = 3.
  • Verify: AB2+BC2=16+9=25=AC2AB^2 + BC^2 = 16 + 9 = 25 = AC^2So ∠ABC=90∘\angle ABC = 90^\circ. ✓
  • The diagonal is ACAC with midpoint M=(1+52,1+42)=(3,2.5)M = \left(\frac{1+5}{2}, \frac{1+4}{2}\right) = (3, 2.5).
  • DD is such that MM is also the midpoint of CDCD: D=(2Mx−Cx,2My−Cy)=(6−5,5−4)=(1,4)D = (2M_x - C_x, 2M_y - C_y) = (6 - 5, 5 - 4) = (1, 4).
  • Area =AB×BC=4×3=12= AB \times BC = 4 \times 3 = 12 square units.
  1. Question: A solid metal cone has base radius 6cm6\mathrm{ cm} and slant height 10cm10\mathrm{ cm}. A sector is cut from a circular sheet of metal to form the curved surface. Find (a) the angle of the sector, (b) the radius of the circular sheet, and (c) the area of the sector.
Answer
  • (a) The arc length of the sector equals the circumference of the base: l=2π(6)=12πl = 2\pi(6) = 12\pi.
  • The sector radius equals the slant height: r=10cmr = 10\mathrm{ cm}.
  • l=rθ  ⟹  12π=10θ  ⟹  θ=6π5l = r\theta \implies 12\pi = 10\theta \implies \theta = \frac{6\pi}{5} rad =216∘= 216^\circ.
  • (b) The radius of the circular sheet is the slant height: 10cm10\mathrm{ cm}.
  • (c) Area of sector: A=12r2θ=12(100)(6π5)=60πcm2A = \frac{1}{2}r^2\theta = \frac{1}{2}(100)\left(\frac{6\pi}{5}\right) = 60\pi \mathrm{ cm}^2.
  1. Question: Prove that the points A(-2, 0)$$B(0, 4)$$C(6, 2)And D(4,−2)D(4, -2) are the Vertices of a rectangle. Find the equation of the circle passing through all four vertices.
Answer
  • Midpoint of AC=(−2+62,0+22)=(2,1)AC = \left(\frac{-2+6}{2}, \frac{0+2}{2}\right) = (2, 1).
  • Midpoint of BD=(0+42,4+(−2)2)=(2,1)BD = \left(\frac{0+4}{2}, \frac{4+(-2)}{2}\right) = (2, 1).
  • Since the diagonals bisect each other, ABCDABCD is a parallelogram.
  • AB = \sqrt{4 + 16} = 2\sqrt{5}$$BC = \sqrt{36 + 4} = 2\sqrt{10} AC=64+4=217AC = \sqrt{64 + 4} = 2\sqrt{17}.
  • AB2+BC2=20+40=60=AC2AB^2 + BC^2 = 20 + 40 = 60 = AC^2. The parallelogram has a right angle, so ABCDABCD is a rectangle.
  • The circle through all four vertices (circumcircle) has centre at the intersection of the diagonals (2,1)(2, 1) and radius =12AC=17= \frac{1}{2}AC = \sqrt{17}.
  • Equation: (x−2)2+(y−1)2=17(x-2)^2 + (y-1)^2 = 17I.e., x2+y2−4x−2y−12=0x^2 + y^2 - 4x - 2y - 12 = 0.

:::tip Diagnostic Test Ready to test your understanding of Geometries? The contains the hardest questions within the DSE specification for this topic, each with a full worked solution.

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DSE Exam Technique

Showing Working

For geometry problems in DSE Paper 1:

  1. When using the distance or midpoint formula, write the formula before substituting.
  2. When finding the equation of a line, state the gradient and a point, then apply the point-slope form.
  3. For circle problems, show the completing-the-square steps to find centre and radius.
  4. For 3D problems, identify the right-angled triangles used and label the diagram.
  5. For area problems, show which formula is being used and justify the values substituted.

Significant Figures

Length answers to 3 significant figures unless exact. Angle answers to nearest degree or 3 s.f. As appropriate.

Common DSE Question Types

  1. Circle equations (finding centre, radius, tangent equations).
  2. 2D area problems (shoelace formula, Heron’s formula).
  3. 3D distance and angle problems.
  4. Arc length and sector area calculations.
  5. Parallel/perpendicular line problems.

Additional Worked Examples

Worked Example 13: Two circles touching externally

Circle C1C_1 has centre (1,2)(1, 2) and radius 33. Circle C2C_2 has centre (7,2)(7, 2) and radius rr. If C1C_1 and C2C_2 touch externally, find rr.

Solution

For external tangency, the distance between centres equals the sum of the radii:

d=(7−1)2+(2−2)2=6d = \sqrt{(7 - 1)^2 + (2 - 2)^2} = 6

3+r=6  ⟹  r=33 + r = 6 \implies r = 3

Worked Example 14: Area of a regular hexagon

A regular hexagon has side length aa. Find its area in terms of aa.

Solution

A regular hexagon can be divided into 6 equilateral triangles, each with side aa.

Area of one equilateral triangle: 34a2\dfrac{\sqrt{3}}{4}a^2.

Total area: 6×34a2=332a26 \times \dfrac{\sqrt{3}}{4}a^2 = \dfrac{3\sqrt{3}}{2}a^2.

Worked Example 15: Shortest distance between skew lines (3D)

In 3D, find the distance between the point P(1,2,3)P(1, 2, 3) and the line through A(4,5,6)A(4, 5, 6) and B(7,8,9)B(7, 8, 9).

Solution

Direction vector of the line: d=(3,3,3)\mathbf{d} = (3, 3, 3)Simplified to (1,1,1)(1, 1, 1).

Vector AP→=(1−4,2−5,3−6)=(−3,−3,−3)\overrightarrow{AP} = (1 - 4, 2 - 5, 3 - 6) = (-3, -3, -3).

The cross product AP→×d=(−3,−3,−3)×(1,1,1)\overrightarrow{AP} \times \mathbf{d} = (-3, -3, -3) \times (1, 1, 1).

Since AP→\overrightarrow{AP}is parallel to d\mathbf{d} (one is −3-3 times the other), the point PP lies on the line.

The distance is 00.

Verification: PP lies on the line through AA in direction (1,1,1)(1,1,1) since (1,2,3)=(4,5,6)+(−1)(1,1,1)(1, 2, 3) = (4, 5, 6) + (-1)(1, 1, 1). So PP is on the line and the distance is 00.

Worked Example 16: 3D tetrahedron volume

Find the volume of the tetrahedron with vertices A(0, 0, 0)$$B(2, 0, 0)$$C(0, 3, 0)$$D(0, 0, 4).

Solution

The volume of a tetrahedron with vertices at the origin and on the coordinate axes is:

V=16∣xB⋅yC⋅zD∣=16(2)(3)(4)=4V = \frac{1}{6}|x_B \cdot y_C \cdot z_D| = \frac{1}{6}(2)(3)(4) = 4

Alternatively, using the scalar triple product:

V=16∣AB→⋅(AC→×AD→)∣=16∣(2,0,0)⋅((0,3,0)×(0,0,4))∣V = \frac{1}{6}|\overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD})| = \frac{1}{6}|(2, 0, 0) \cdot ((0, 3, 0) \times (0, 0, 4))|

(0,3,0)×(0,0,4)=(12,0,0)(0, 3, 0) \times (0, 0, 4) = (12, 0, 0)

V=16∣(2)(12)∣=16(24)=4V = \frac{1}{6}|(2)(12)| = \frac{1}{6}(24) = 4

Worked Example 17: Angle between two lines in 3D

Find the angle between the lines joining A(1,0,0)A(1, 0, 0) to B(0,1,0)B(0, 1, 0) and AA to C(0,0,1)C(0, 0, 1).

Solution

\overrightarrow{AB} = (-1, 1, 0)$$\overrightarrow{AC} = (-1, 0, 1).

cos⁡θ=∣(−1)(−1)+(1)(0)+(0)(1)∣1+1+0⋅1+0+1=122=12\cos\theta = \frac{|(-1)(-1) + (1)(0) + (0)(1)|}{\sqrt{1 + 1 + 0} \cdot \sqrt{1 + 0 + 1}} = \frac{1}{\sqrt{2}\sqrt{2}} = \frac{1}{2}

θ=60∘\theta = 60^\circ

Worked Example 18: Equation of tangent to a circle with general form

Find the equation of the tangent to x2+y2−4x+6y−12=0x^2 + y^2 - 4x + 6y - 12 = 0 at the point (5,1)(5, 1).

Solution

First verify (5,1)(5, 1) lies on the circle: 25+1−20+6−12=025 + 1 - 20 + 6 - 12 = 0. Confirmed.

Using the tangent formula for the general circle equation x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0:

Replace x^2 \to x \cdot 5$$y^2 \to y \cdot 1$$x \to \dfrac{x + 5}{2}$$y \to \dfrac{y + 1}{2}:

5x+y−4⋅x+52+6⋅y+12−12=05x + y - 4\cdot\frac{x + 5}{2} + 6\cdot\frac{y + 1}{2} - 12 = 0

5x+y−2x−10+3y+3−12=05x + y - 2x - 10 + 3y + 3 - 12 = 0

3x+4y−19=03x + 4y - 19 = 0

DSE Exam-Style Questions

DSE Practice 1. Find the equation of the circle which passes through (0,0)(0, 0) and (0,4)(0, 4) and has its centre on the line x−y+2=0x - y + 2 = 0.

Solution

Let the centre be C(h,k)C(h, k). Since CC lies on x−y+2=0x - y + 2 = 0: h−k+2=0  ⟹  k=h+2h - k + 2 = 0 \implies k = h + 2.

CA2=CB2CA^2 = CB^2 where A=(0,0)A = (0, 0) and B=(0,4)B = (0, 4):

h2+k2=h2+(k−4)2  ⟹  k2=k2−8k+16  ⟹  8k=16  ⟹  k=2h^2 + k^2 = h^2 + (k - 4)^2 \implies k^2 = k^2 - 8k + 16 \implies 8k = 16 \implies k = 2

h=k−2=0h = k - 2 = 0.

Centre: (0,2)(0, 2). r2=02+22=4r^2 = 0^2 + 2^2 = 4.

Equation: x2+(y−2)2=4x^2 + (y - 2)^2 = 4Or x2+y2−4y=0x^2 + y^2 - 4y = 0.

DSE Practice 2. In \triangle ABC$$A = (1, -2)$$B = (5, 4)And C=(8,1)C = (8, 1). Find the equation of the median from AA and the equation of the altitude from BB.

Solution

Median from AA: Midpoint of BCBC: M=(5+82,4+12)=(6.5,2.5)M = \left(\dfrac{5+8}{2}, \dfrac{4+1}{2}\right) = (6.5, 2.5).

Gradient of AMAM: m=2.5−(−2)6.5−1=4.55.5=911m = \dfrac{2.5 - (-2)}{6.5 - 1} = \dfrac{4.5}{5.5} = \dfrac{9}{11}.

Equation: y+2=911(x−1)  ⟹  11y+22=9x−9  ⟹  9x−11y−31=0y + 2 = \dfrac{9}{11}(x - 1) \implies 11y + 22 = 9x - 9 \implies 9x - 11y - 31 = 0.

Altitude from BB: Gradient of ACAC: mAC=1−(−2)8−1=37m_{AC} = \dfrac{1 - (-2)}{8 - 1} = \dfrac{3}{7}.

Gradient of altitude: m⊥=−73m_\perp = -\dfrac{7}{3}.

Equation through B(5,4)B(5, 4): y−4=−73(x−5)  ⟹  3y−12=−7x+35  ⟹  7x+3y−47=0y - 4 = -\dfrac{7}{3}(x - 5) \implies 3y - 12 = -7x + 35 \implies 7x + 3y - 47 = 0.

DSE Practice 3. A sector of a circle of radius 10cm10\mathrm{ cm} has perimeter 30cm30\mathrm{ cm}. Find the area of the sector.

Solution

Perimeter of sector =2r+rθ=20+10θ=30= 2r + r\theta = 20 + 10\theta = 30.

10θ=10  ⟹  θ=110\theta = 10 \implies \theta = 1 rad.

Area=12r2θ=12(100)(1)=50cm2\mathrm{Area} = \frac{1}{2}r^2\theta = \frac{1}{2}(100)(1) = 50\mathrm{ cm}^2

DSE Practice 4. Two circles C1:(x−1)2+y2=9C_1: (x - 1)^2 + y^2 = 9 and C2:(x−5)2+y2=25C_2: (x - 5)^2 + y^2 = 25 intersect at AA and BB. Find the length of the common chord ABAB.

Solution

Centres: O_1 = (1, 0)$$O_2 = (5, 0). Radii: r_1 = 3$$r_2 = 5.

Distance between centres: d=4d = 4.

The common chord is perpendicular to O1O2O_1O_2 (the xx-axis) and passes through the midpoint.

By symmetry, the common chord is the line x=cx = c for some cc.

Substituting x=cx = c into C1C_1: (c−1)2+y2=9  ⟹  y2=9−(c−1)2(c - 1)^2 + y^2 = 9 \implies y^2 = 9 - (c - 1)^2.

Substituting x=cx = c into C2C_2: (c−5)2+y2=25  ⟹  y2=25−(c−5)2(c - 5)^2 + y^2 = 25 \implies y^2 = 25 - (c - 5)^2.

Setting equal: 9−(c−1)2=25−(c−5)29 - (c - 1)^2 = 25 - (c - 5)^2.

9−c2+2c−1=25−c2+10c−259 - c^2 + 2c - 1 = 25 - c^2 + 10c - 25

8+2c=10c  ⟹  8c=8  ⟹  c=18 + 2c = 10c \implies 8c = 8 \implies c = 1.

y2=9−0=9  ⟹  y=±3y^2 = 9 - 0 = 9 \implies y = \pm 3. Points: (1,3)(1, 3) and (1,−3)(1, -3).

AB=6AB = 6.

DSE Practice 5. Find the angle between the planes 2x−y+2z=32x - y + 2z = 3 and x+2y−2z=1x + 2y - 2z = 1.

Solution

Normal to first plane: n1=(2,−1,2)\mathbf{n_1} = (2, -1, 2). Normal to second plane: n2=(1,2,−2)\mathbf{n_2} = (1, 2, -2).

cos⁡θ=∣2(1)+(−1)(2)+2(−2)∣4+1+4⋅1+4+4=∣2−2−4∣3⋅3=49\cos\theta = \frac{|2(1) + (-1)(2) + 2(-2)|}{\sqrt{4 + 1 + 4} \cdot \sqrt{1 + 4 + 4}} = \frac{|2 - 2 - 4|}{3 \cdot 3} = \frac{4}{9}

θ=arccos⁡ ⁣(49)≈63.6∘\theta = \arccos\!\left(\frac{4}{9}\right) \approx 63.6^\circ

DSE Practice 6. A pyramid has a rectangular base ABCDABCD with AB = 8$$BC = 6. The vertex VV is directly above the centre of the base, and VA=10VA = 10. Find the angle between the face VABVAB and the base ABCDABCD.

Solution

The centre of the rectangle is O=(4,3,0)O = (4, 3, 0) (taking AA at the origin).

VA = 10$$OA = \sqrt{4^2 + 3^2} = 5.

VO=VA2−OA2=100−25=75=53VO = \sqrt{VA^2 - OA^2} = \sqrt{100 - 25} = \sqrt{75} = 5\sqrt{3}.

Height of the pyramid: h=53h = 5\sqrt{3}.

The midpoint of ABAB is M=(4,0,0)M = (4, 0, 0).

VM=VO2+OM2=75+9=84=221VM = \sqrt{VO^2 + OM^2} = \sqrt{75 + 9} = \sqrt{84} = 2\sqrt{21}.

The angle between face VABVAB and the base is the angle between VMVM and OMOM:

cos⁡ϕ=OMVM=3221=2114\cos\phi = \frac{OM}{VM} = \frac{3}{2\sqrt{21}} = \frac{\sqrt{21}}{14}

ϕ=arccos⁡ ⁣(2114)≈69.2∘\phi = \arccos\!\left(\frac{\sqrt{21}}{14}\right) \approx 69.2^\circ

DSE Practice 7. The line 3x+4y−25=03x + 4y - 25 = 0 intersects the circle x2+y2=25x^2 + y^2 = 25 at AA and BB. Find the area of the minor segment cut off by the chord ABAB.

Solution

Distance from the centre (0,0)(0, 0) to the line:

d=∣25∣5=5d = \frac{|25|}{5} = 5

Since d=r=5d = r = 5The line is tangent to the circle (not a chord). There is no minor segment — the line touches the circle at exactly one point.

The point of contact: the foot of the perpendicular from (0,0)(0,0) to 3x+4y=253x + 4y = 25.

3x+4y=253x + 4y = 25 and y=34xy = \dfrac{3}{4}x (the perpendicular from the origin).

3x + 4 \cdot \dfrac{3}{4}x = 25 \implies 6x = 25 \implies x = \dfrac{25}{6}$$y = \dfrac{25}{8}.

Point of tangency: (256,258)\left(\dfrac{25}{6}, \dfrac{25}{8}\right).

Since the line is tangent (not a secant), the area of the minor segment is 00.

Common Pitfalls

  • Confusing the Pythagorean theorem converse with the theorem itself. The converse states that if a2+b2=c2a^2 + b^2 = c^2 then the triangle is right-angled at CC.

  • Using the wrong formula for 3D distance. The distance between (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2) is (x2−x1)2+(y2−y1)2+(z2−z1)2\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}.

  • Midpoint formula errors. The midpoint of (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is (x1+x22,y1+y22)\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right), not the average of the squares.

  • 2D distance: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}; extend to 3D by adding (z2−z1)2(z_2-z_1)^2.

  • Midpoint: M=(x1+x22,y1+y22)M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right).

  • A triangle is right-angled iff the sum of squares of two sides equals the square of the third (Pythagorean converse).

  • The shortest distance from a point to a line is the perpendicular distance.

Worked Examples

Example 1: 3D distance

Problem. Find the distance between A(1,2,3)A(1, 2, 3) and B(4,−1,5)B(4, -1, 5).

Solution. AB=(4−1)2+(−1−2)2+(5−3)2=9+9+4=22AB = \sqrt{(4-1)^2 + (-1-2)^2 + (5-3)^2} = \sqrt{9 + 9 + 4} = \sqrt{22}

■\blacksquare

Example 2: Proving a triangle is right-angled

Problem. The vertices of a triangle are A(0,0)A(0, 0), B(4,0)B(4, 0), C(0,3)C(0, 3). Show that △ABC\triangle ABC is right-angled.

Solution. AB2=16,AC2=9,BC2=16+9=25AB^2 = 16, \quad AC^2 = 9, \quad BC^2 = 16 + 9 = 25

AB2+AC2=16+9=25=BC2AB^2 + AC^2 = 16 + 9 = 25 = BC^2. By the converse of Pythagoras’ theorem, △ABC\triangle ABC is right-angled at AA.

■\blacksquare

  1. Forgetting to check that solutions satisfy the original equation (especially with squaring both sides or dividing by variables).

  2. Misreading the question, particularly with ‘hence’ vs ‘hence or otherwise’ — the former requires using previous work.

    Stashed changes:docs/docs_dse/Maths/compulsory/geometries.md

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

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