Find the gradient of the line joining \( A(1,2) \) and \( B(5,10) \).
\( m = \dfrac{10-2}{5-1} = \dfrac{8}{4} = 2 \).
Coordinate Geometry
12 exam-style questions on straight lines, circles and midpoints. Tap Show answer for full working.
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Find the equation of the line with gradient 3 that passes through \( (2,-1) \).
\( y - (-1) = 3(x-2) \) ⇒ \( y + 1 = 3x - 6 \) ⇒ \( y = 3x - 7 \).
Find the midpoint of the line segment joining \( (3,-4) \) and \( (-1,6) \).
\( \left( \dfrac{3+(-1)}{2}, \dfrac{-4+6}{2} \right) = (1, 1) \).
Find the distance between the points \( (0,0) \) and \( (3,4) \).
\( \sqrt{3^2 + 4^2} = 5 \).
Find the equation of the circle with centre \( (2,-1) \) and radius 5.
\( (x-2)^2 + (y+1)^2 = 25 \).
The line \( l \) has equation \( 2x + 3y - 6 = 0 \). Find the gradient of \( l \) and the coordinates of the points where \( l \) meets the axes.
\( 3y = -2x + 6 \) ⇒ \( y = -\dfrac{2}{3}x + 2 \). Gradient \( -\dfrac{2}{3} \). \( x \)-intercept: \( (3,0) \); \( y \)-intercept: \( (0,2) \).
Find the equation of the perpendicular bisector of the line joining \( A(1,3) \) and \( B(5,7) \).
Midpoint \( (3,5) \). Gradient of AB is \( 1 \), so perpendicular gradient is \( -1 \). Equation: \( y-5 = -1(x-3) \) ⇒ \( y = -x + 8 \).
Find the centre and radius of the circle \( x^2 + y^2 - 2x - 4y - 11 = 0 \). Show that the point \( (1,6) \) lies on the circle.
Complete the square: \( (x-1)^2 + (y-2)^2 = 16 \). Centre \( (1,2) \), radius 4. Sub \( (1,6) \): \( 0 + 16 = 16 \), so on the circle.
Find the points of intersection of the line \( y = x + 1 \) and the circle \( x^2 + y^2 = 25 \).
Sub \( y=x+1 \): \( x^2+(x+1)^2=25 \) ⇒ \( 2x^2+2x+1-25=0 \) ⇒ \( 2x^2+2x-24=0 \) ⇒ \( x^2+x-12=0 \) ⇒ \( (x+4)(x-3)=0 \). \( x=-4 \) or \( 3 \). Points \( (-4,-3) \) and \( (3,4) \).
Find the equation of the tangent to the circle \( (x-1)^2 + (y+2)^2 = 10 \) at the point \( (2,1) \).
Centre \( C(1,-2) \). Radius vector from C to point has gradient \( \dfrac{1-(-2)}{2-1}=3 \). Tangent perpendicular, gradient \( -1/3 \). Equation: \( y-1 = -\dfrac{1}{3}(x-2) \) ⇒ \( 3y-3 = -x+2 \) ⇒ \( x+3y-5=0 \).
The circle \( x^2 + y^2 + Dx + Ey + F = 0 \) passes through \( (0,0) \), \( (4,0) \) and \( (0,2) \). Find \( D \), \( E \) and \( F \).
\( (0,0) \): \( F=0 \). \( (4,0) \): \( 16 + 4D = 0 \) ⇒ \( D=-4 \). \( (0,2) \): \( 4 + 2E = 0 \) ⇒ \( E=-2 \). Equation: \( x^2 + y^2 - 4x - 2y = 0 \).
Find the shortest distance from the point \( (5,1) \) to the line \( 3x - 4y + 2 = 0 \).
Distance \( = \dfrac{|3\cdot5 - 4\cdot1 + 2|}{\sqrt{9+16}} = \dfrac{|15-4+2|}{5} = \dfrac{13}{5} \).