Question 1110560: How do you find the area of a triangle using a base and altitude with vertices (-6,3), (4,13) and (10,-5)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you can pick any of the sides to be the base.
you then need to find a perpendicular to that base that intersect with the intersection of the other 2 sides.
it helps to sketch your triangle first.
pick your base.
i picked the line formed by (-6,3) and (10,-5)
the equation for that line is y = -1/2 * x
now you want to find the line that's perpendicular to that line and intersects at the point of intersection of the other 2 lines.
from your sketch you should see that the point of intersection of the other 2 lines is at (4,13).
the line perpendicular to your base line will have a slope that is a negative reciprocal of the slope of your base line.
that makes the slope equal to 2.
you then use the point (4,13) to find the equation.
the equation for that line will be y = 2x + 5.
you now have 2 lines and you need to find the length of each line.
the length of your base will be the length of the line between the points (-6,3) and (10,-5).
that length will be sqrt((-16)^2 + 8^2) = sqrt(320).
next you need to find the intersection of the line perpendicular to the base with the base.
that requires you to solve those 2 equations simultaneously.
the 2 equations are
y = -1/2 * x
y = 2x + 5
subtract the first equation from the second to get:
0 = 2.5 * x + 5
solve for x to get x = -2
replace x in either of the original equations to get y = 1.
your point of intersection is (-2,1)
you can now solve for the length of the altitude because that line goes between the points (4,13) and (-2,1)
the length is sqrt(6^2 + 12^2) = sqrt(180).
you find the area by using the formula 1/2 * base * height.
this formula becomes area = 1/2 * sqrt(320) * sqrt(180).
solve for area to get area = 120 square units.
here's my sketch drawn by hand and final diagram using the equations of the lines.
in the final diagram, the perpendicular of the triangle is red and the triangle itself is blue.
the lines themselves go on indefinitely, but it's the line segments between the points shown that you want.
to find the equation of a line from 2 points, you first find the slope and then solve for the y-intersection using the slope intercept form of the equation for a straight line.
the general form of the slope intercept equation is y = mx + b.
m is the slope
b is the y-intercept
for the base line, the points were (-6,3) and (10,-5)
the slope was (-5-3) / (10-(-6) which became -8/16 which became -1/2.
you then use one of the points to solve for the y-intercept.
i used the point (-6,3).
replace y with 3 and x with -6 and y = mx + b became 3 = -1/2 * (-6) + b
simplify to get 3 = 3 + b
solve for b to get b = 0
the equation for the base line became y = -1/2 * x
once you know the point where the perpendicular line intersect with the base line, you use the same procedure to solve for the line perpendicular to the base.
the equation for that line was calculated to be y = 2x + 5
the formula for the length of each line is length = sqrt((x2-x1)^2 + (y2-y1)^2)
the line perpendicular to the base line went between the points (4,13) and (-2,1)
the length of the line became sqrt((1-13)^2 + (-2-4)^2).
this became sqrtZ((-12)^2+ (-6)^2) which became sqrt(144+36) which became sqrt(120).
similar procedure was use to find the length of the base line, which was sqrt(320).
finally, the area became 1/2 * base * height which became 1/2 * sqrt(180) * sqrt(320) which became 1/2 * sqrt(180 * 320) which became 1/2 * 240 which became 120.
all you needed was the length of the base line and the length of the line perpendicular to it.
in order to graph the triangle using equations, i needed to find the equation for each of the lines.
it was more work, but the procedure to find the equation for each line was just a repetition of the procedure i showed you for one of the lines.
without the sketch, you might have needed to find the intersections algebraically, so the sketch helped to eliminate the need for additional line equations and was useful to give you a picture of what your triangle would look like generally.
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