Leaving Cert Mathematics Co-ordinate Geometry: Find the area of the triangle wi

Question

Find the area of the triangle with vertices (0, 0), (8, -6) and (-1, 5).

Let the vertices be A(0, 0), B(8, -6), and C(-1, 5).

The formula for the area of a triangle given by vertices (x1, y1), (x2, y2), and (x3, y3) is:

$$	ext{Area} = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |$$

Substituting in the coordinates:

$$	ext{Area} = \frac{1}{2} | 0(-6 - 5) + 8(5 - 0) + (-1)(0 + 6) |$$

Calculating each term:

$$ = \frac{1}{2} | 0 + 40 + 6 | = \frac{1}{2} | 46 | = 23.$$

So the area of the triangle is 23 square units.

(b) i is the line 3x - 4y - 15 = 0.

To verify that (1, -3) is a point on l:

Substituting into the equation:

$$3(1) - 4(-3) - 15 = 3 + 12 - 15 = 0.$$

Thus, (1, -3) is indeed on line l.

(ii) To find the x-intercept:

Set y = 0 in the equation:

$$3x - 4(0) - 15 = 0 ightarrow 3x = 15 ightarrow x = 5.$$

So the x-intercept is (5, 0).

(iii) The line l passes through (1, -3) and is perpendicular to l. To find k:

If l has slope m = \frac{4}{3}, then the slope of k is -\frac{3}{4}.

Using point-slope form for point (1, -3):

$$y + 3 = -\frac{3}{4}(x - 1)$$
$$ \rightarrow y = -\frac{3}{4}x + \frac{3}{4} - 3$$
$$ \rightarrow y = -\frac{3}{4}x - \frac{9}{4}.$$

(iv) The equation of line k is:

$$ y = -\frac{3}{4}x - \frac{9}{4} $$

(c) A(2, -1) and B(-4, 7) are two points.

(i) To find |AB|:

The distance formula is:

$$|AB| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Substituting:
$$|AB| = \sqrt{(-4 - 2)^2 + (7 - (-1))^2}$$
$$|AB| = \sqrt{(-6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10.$$

(ii) To find the image of B under the translation (2, -1) -> (-7, 11):

The new coordinates would be:

$$B' = B + T$$
$$B' = (-4, 7) + (-7, 11) = (-4 - 7, 7 + 11) = (-11, 18).$$

(iii) Show that |AB| : |AC| = 2 : 5:
Find |AC|:
$$|AC| = \sqrt{(-1 - 2)^2 + (5 - (-1))^2}$$
$$|AC| = \sqrt{(-3)^2 + (6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}.$$

Now,
$$\frac{|AB|}{|AC|} = \frac{10}{3\sqrt{5}} = \frac{20}{3\sqrt{5}} = \frac{2}{5}.$$

Find the area of the triangle with vertices (0, 0), (8, -6) and (-1, 5) - Leaving Cert Mathematics - Question 2 - 2010

Worked Solution & Example Answer:Find the area of the triangle with vertices (0, 0), (8, -6) and (-1, 5) - Leaving Cert Mathematics - Question 2 - 2010

Find the area of the triangle with vertices (0, 0), (8, -6) and (-1, 5)

Verify that (1, -3) is a point on l

Find the x-axis intercept of P

Show the lines l and k on a co-ordinate diagram

Find the equation of k

Find |AB|

Find C, the image of B under the translation (2, -1) -> (-7, 11)

Show that |AB| : |AC| = 2 : 5

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