kbdevaraj wrote:Hi,

This is about Coordinate Geometry questions on GMAT. Can anyone help me?

Anybody experienced below type of questions on GMAT? Otherwise, can we expect them?

1. Find the area of the triangle ABC given A(6,7), B(2,-9) and C(-4, 1).
2. If the points (-1, 3), (b, -1), (0, 4) are on a line, find value of 'b'.

Regards
KB Devaraj

Whenever GMAT asks the area of a triangle with the coordinates of its three vertices given, in most of the cases plotting the points on the rectangular axes helps us easily find one base and its corresponding altitude so that ½ b h formula can be used. But if it's not so, then dole out the coordinates of three vertices as (x1, y1), (x2, y2), and (x3, y3) and then use the following formula

∆ = ½ |x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)|

1. This way, A (6, 7), B (2, -9) and C (-4, 1) could give

∆ = ½ |6 (-9 - 1) + 2 (1 - 7) - 4 (7 + 9)|

= ½ |-60 - 12 - 64|

= ½ × 136 = [spoiler]68 units[/spoiler]

2. The concept of three points being collinear lies in the fact that no triangle can be formed using three points that are collinear, and if the area of a triangle whose three vertices are given collinear is calculated by using the above mentioned formula, then the result will be zero. See

When (-1, 3), (b, -1), (0, 4) are collinear

∆ = ½ |-1 (-1 - 4) + b (4 - 3) + 0 (3 + 1)| = 0

or, |5 + b + 0| = 0

or b = [spoiler]-5[/spoiler]

Hi Sanju.. Kindly judge my approach

Calculate the base = lenght of BC = distance btw two point BC = sqrt(136)

Calculate the height = distance between point A to line BC

Line BC's equation is 5x+3y-17 = 0
distance of point A to above line = 5*6+3*7-17/sqrt(5^2+3^2) = sqrt(34)

Now area = 1/2*base*height = 1/2*sqrt(136)*sqrt(34) = 34 sq units

Thanks Sanju for the explanation.

But do we get these kind of question on gmat?

Devaraj

kbdevaraj wrote:Thanks Sanju for the explanation.

But do we get these kind of question on gmat?

Devaraj

yes you may

HSPA wrote:Hi Sanju.. Kindly judge my approach

Calculate the base = lenght of BC = distance btw two point BC = sqrt(136)

Calculate the height = distance between point A to line BC

Line BC's equation is 5x+3y-17 = 0
distance of point A to above line = 5*6+3*7-17/sqrt(5^2+3^2) = sqrt(34)

Now area = 1/2*base*height = 1/2*sqrt(136)*sqrt(34) = 34 sq units

This approach is not recommendable when easier and quicker approaches are already there. Your work is absolutely correct ACCEPT a calculation error towards the end (see BOLD above), I am getting the equation of line BC as

5 x + 3 y + 17 = 0

and this would get you the area as [spoiler]68[/spoiler] sq units

1. Find the area of the triangle ABC given by A(6,7), B(2,-9) and C(-4, 1).

2. If the points (-1, 3), (b, -1), (0, 4) are on a line, find the value of 'b'.

Use (-1,3) and (0,4) to determine the slope:
(4-3)/(0-(-1)) = 1/1 = 1.

Thus, (b,-1) and (0,4) must yield a slope of 1:
(b-0)/(-1-4) = 1
b/(-5) = 1.
b = -5.

Thanks for the awesome and quick approaches ... Sanju09 and Mitch .

First question: Thanks for the correction sanju...


For the second question.. My approach
three points are on same line.. so substitute the unknown point in the line equation formed using the other two points.