O lvl E Maths: Coordinate Geometry

RI sec 3 1997 EOY P1 Q9

A line passing through the points (-5, w) and (3, 4) is parallel to 5y - 4x - 14 = 0. Find the value of w.

This line cuts the x-axis at A and y-axis at B. Find the area of ΔOAB where O is the origin.

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Answer:

5y - 4x - 14 = 0
5y = 4x + 14
y = 4x/5 + 14/5
Thus, equation of line is y = 4x/5 + c

Sub x=3, y=4
4 = 4(3)/5 + c
c = 8/5
Thus, equation of line is y = 4x/5 + 8/5

Sub x=-5, y=w
w = 4(-5)/5 + 8/5
w = -12/5



When y = 0, 4x/5 + 8/5 = 0
4x/5 = -8/5
x = -2
==> A: (-2,0)

When x = 0, y = 8/5
==> B: (0, 8/5)

Thus, using shoelace formula,
area =


= 1.6 units²

O lvl E Maths: Coordinate Geometry

The vertices of triangle ABC are A(-1,2), B (1,5) and C(4,3).

(a) Find the lengths of the sides AB, BC and CA
(b) What type of triangle is ABC.
(c) Find the perpendicular distance form B to AC.
(d) Find the coordinates of the point at which the line AC cuts the x-axis

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Answer:

(a) Length of side AB = √[(1 - (-1))² + (5 - 2)²] = √13

Length of side BC = √[(4 - 1)² + (3 - 5)²] = √13

Length of side CA = √[(4 - (-1))² + (3 - 2)²] = √26


(b) 2 sides are equal
Therefore, triangle ABC is an isosceles triangle.


(c) Let the perpendicular distance from B to AC be h

h² + (0.5 √26)² = (√13)²
h² = 13 - 6.5
h² = 6.5
h = 2.55 units


(d) Gradient of AC = (3 - 2) / (4 - (-1)) = 1/5
Eqn of AC: y - 3 = 1/5 (x - 4)
y = x/5 + 11/5

Cut the x-axis:
When y = 0, x = -11
Hence, the coordinates of the point at which the line AC cuts the x-axis is (-11, 0)

O lvl E Maths: Coordinate Geometry

The vertices of the triangle ABC are A(-4,1), B(-4,-2) and C(2,5). Find

(a) the equations of the lines AB, BC, CA
(b) coordinates of the point where AC cuts the y-axis
(c) area of triangle ABC
(d) perpendicular distance from A to BC

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Answer:

(a) Line AB is a vertical line
Eqn of line AB: x = -4

Gradient of BC = (5 - (-2)) / (2 - (-4)) = 7/6
y - 5 = 7/6 (x - 2)
y = 7x/6 + 8/3

Gradient of CA = (5 - 1)/(2 - (-4)) = 2/3
y - 5 = 2/3 (x - 2)
y = 2x/3 + 11/3



(b) Where AC cuts x-axis => x = 0
Hence, y = 11/3
Coordinates of point where AC cuts x-axis = (0, 11/3)


(c) Using shoelace formula,
area of triangle =

= ½ [ (-4)(-2) + (-4)(5) + (2)(1) - (-4)(1) - (2)(-2) - (-4)(5) ]
= ½ [ 8 - 20 + 2 + 4 + 4 + 20 ]
= 9 units²


(d)

length BC = √(7² + 6²) = √(85)

0.5 * length BC * perpendicular distance from A to BC = area of triangle ABC
0.5 * √(85) * perpendicular distance from A to BC = 9 (from part c)

Perpendicular distance from A to BC = 1.95 units