What is the area of a triangle formed by lines y=5-x, 2y=3x , y=0?
7.5
8
9
10.5
15
Can some one explain how to approach such problems?
Thanks.
A
Area of triangle
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The triangle will have a base on the x-axis (y=0). The length of the base will be determined by the x-intercepts of the other two lines. Calculate these values by plugging in y=0 to the equations of the other two lines: y=5-x ---> 0=5-x ----> x=5; 2y=3x -----> 2(0)=3x ----> x=0. Thus, the length of the base will be 5.PGMAT wrote:What is the area of a triangle formed by lines y=5-x, 2y=3x , y=0?
7.5
8
9
10.5
15
Can some one explain how to approach such problems?
Thanks.
A
The height of the triangle will be determined by the y-coordinate of the intersection of the two lines other than y=0. We can find this by substitution: y=5-x ---> x=5-y, then plug this into x in the other equation----> 2y=3(5-y) ----> 2y=15-3y ----> 5y=15 ----> y=3, so the height of the triangle is 3. Note that this is true regardless of what the x-coordinate is of the intersection point.
So, the base is 5 and the height is 3. Plug this into the formula for area of a triangle: A=b*h/2 ---> A=3*5/2= 7.5