Coordinate Geometry

The gradient is the slope of a given object in a x and y plane

Example of finding the gradient of a straight line

Given that a straight line passes through the points A (5, -9) and B (3, 7), Find

(a) The gradient of a straight line passing through A and B.

(b) The equation of the line passing through the points A and B.

(c) The coordinates of a midpoint between A and B.

(d) The distance between points A and B.

Solutions:

(a)

A (5, -9) and B (3, 7)

Formula for gradient

Gradient = $\frac{\mathrm{Y2\; -\; Y1}}{\mathrm{X2\; -\; X1}}$

Y2 = 7, Y1 = -9, X2 = 3, X1 = 5

Gradient = $\frac{\mathrm{7\; -\; (-9)}}{\mathrm{3\; -\; 5}}$

Gradient = $\frac{\mathrm{7\; +\; 9}}{\mathrm{3\; -\; 5}}$

Gradient = $\frac{\mathrm{16}}{-2}$

Answer: Gradient/M = $-8$

(b)

Formula for the equation of a straight line

y - Y1 = m(x - X1)

A (5, -9)

X1 = 5, Y1 = -9

replace the values X1 and Y1 in the equation

y - (-9) = -8(x - 5)

y + 9 = -8(x - 5)

y + 9 = -8x + 40

Format the solution in this format y = mx + c

y = -8x + 40 - 9

Answer: y = -8x + 31

(c)

Formula for Midpoint

M.p = $(\frac{\mathrm{X1\; +\; X2}}{2},\frac{\mathrm{Y1\; +\; Y2}}{2})$

A (5, -9) and B (3, 7)

Y2 = 7, Y1 = -9, X2 = 3, X1 = 5

M.p = $(\frac{\mathrm{5\; +\; 3}}{2},\frac{\mathrm{-9\; +\; 7}}{2})$

M.p = $(\frac{8}{2},\frac{\mathrm{-2}}{2})$

Answer: M.p = $(4,\mathrm{-1})$

(d)

Formula for distance between points

AB = $\sqrt{\left(\mathrm{X2-\; X1}\right){}^2+\left(\mathrm{Y2-\; Y1}\right){}^2}$

A (5, -9) and B (3, 7)

Y2 = 7, Y1 = -9, X2 = 3, X1 = 5

AB = $\sqrt{\left(\mathrm{3\; -\; 5}\right){}^2+\left(\mathrm{7\; -\; (-9)}\right){}^2}$

AB = $\sqrt{\left(-2\right){}^2+\left(\mathrm{7\; +\; 9}\right){}^2}$

AB = $\sqrt{\left(-2\right){}^2+\left(16\right){}^2}$

AB = $\sqrt{4+\mathrm{256}}$

AB = $\sqrt{\mathrm{260}}$

Answer: AB = $\mathrm{16.1}$

How to find gradient of the straight line

How to find the Coordinates from a gradient

How to find the midpoint of Coordinates, its distance and equation