# 2019 AMC 10A Problems/Problem 7

*The following problem is from both the 2019 AMC 10A #7 and 2019 AMC 12A #5, so both problems redirect to this page.*

## Contents

## Problem

Two lines with slopes and intersect at . What is the area of the triangle enclosed by these two lines and the line

## Solution 1

Let's first work out the slope-intercept form of all three lines: and implies so , while implies so . Also, implies . Thus the lines are and . Now we find the intersection points between each of the lines with , which are and . Using the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle with base and height , whose area is .

## Solution 2

Like in Solution 1, we determine the coordinates of the three vertices of the triangle. The coordinates that we get are: . Now, using the Shoelace Theorem, we can directly find that the area is .

## Solution 3

Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at and . Then apply Heron's Formula: the semi-perimeter will be , so the area reduces nicely to a difference of squares, making it .

## Solution 4

Like in the other solutions, we find, either using algebra or simply by drawing the lines on squared paper, that the three points of intersection are , , and . We can now draw the bounding square with vertices , , and , and deduce that the triangle's area is .

## Solution 5

Like in other solutions, we find that the three points of intersection are , , and . Using graph paper, we can see that this triangle has boundary lattice points and interior lattice points. By Pick's Theorem, the area is .

## Solution 6

Like in other solutions, we find the three points of intersection. Label these , , and . By the Pythagorean Theorem, and . By the Law of Cosines, Therefore, , so the area is .

## Solution 7

Like in other solutions, we find that the three points of intersection are , , and . The area of the triangle is half the absolute value of the determinant of the matrix determined by these points.

## Solution 8

Like in other solutions, we find the three points of intersection. Label these , , and . Then vectors and . The area of the triangle is half the magnitude of the cross product of these two vectors.

## Solution 9

Like in other solutions, we find that the three points of intersection are , , and . By the Pythagorean theorem, this is an isosceles triangle with base and equal length . The area of an isosceles triangle with base and equal length is . Plugging in and ,

## Solution 10 (Trig)

Like in other solutions, we find the three points of intersection. Label these , , and . By the Pythagorean Theorem, and . By the Law of Cosines, Therefore, . By the extended Law of Sines, Then the area is .

## Solution 11

The area of a triangle formed by three lines, is the absolute value of Plugging in the three lines, the area is the absolute value of Source: Orrick, Michael L. “THE AREA OF A TRIANGLE FORMED BY THREE LINES.” Pi Mu Epsilon Journal, vol. 7, no. 5, 1981, pp. 294–298. JSTOR, www.jstor.org/stable/24336991.

## Solution 12 (Heron's Formula)

Like in other solutions, we find that our triangle is isosceles with legs of and base . Then, the semi - perimeter of our triangle is, Applying Heron's formula, we find that the area of this triangle is equivalent to

~rbcubed13

## Video Solution 1

Education, the Study of Everything

## Video Solution 2

~savannahsolver